Trigonometric Identities

Reciprocal and Quotient Identities

$ \displaystyle \sin \theta =\frac{1}{{\csc \theta }}\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac{1}{{\sin \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{1}{{\sec \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac{1}{{\cos \theta }}\,$        $ \displaystyle \tan \theta =\frac{1}{{\cot \theta }}=\frac{{\sin \theta }}{{\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac{1}{{\tan \theta }}=\frac{{\cos \theta }}{{\sin \theta }}$

Pythagorean Identities

$ \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cot }^{2}}\theta +1={{\csc }^{2}}\theta $

Sum and Difference Identities

$ \begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}$

$ \displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}$

From these identities, you may be asked to be familiar with the Odd/Even Identities:

$ \displaystyle \begin{array}{l}\text{EVEN: }\,\,\,\cos \left( {-x} \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ODD:}\,\,\,\,\,\,\sin \left( {-x} \right)=-\sin \left( x \right)\,\,\,\,\,\,\,\tan \left( {-x} \right)=-\tan \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {-x} \right)=-\csc \left( x \right)\,\,\,\,\,\,\,\cot \left( {-x} \right)=-\cot \left( x \right)\end{array}$

…..and the Cofunction Identities in radians (trig function of an angle is equal to the value of the cofunction of the complement):

$ \displaystyle \begin{align}\sin \left( {\frac{\pi }{2}-x} \right)&=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {\frac{\pi }{2}-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{\pi }{2}-x} \right)=\cot \left( x \right)\\\cos \left( {\frac{\pi }{2}-x} \right)&=\sin \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {\frac{\pi }{2}-x} \right)=\csc \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( {\frac{\pi }{2}-x} \right)=\tan \left( x \right)\end{align}$

Double Angle Identities

$ \sin \left( {2A} \right)=2\sin A\cos A$          $ \begin{align}\cos \left( {2A} \right)&={{\cos }^{2}}A-{{\sin }^{2}}A\\&=1-2{{\sin }^{2}}A\\&=2{{\cos }^{2}}A-1\end{align}$               $ \displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}$

Half Angle Identities

  $ \displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}=\pm \sqrt{{\frac{1}{2}\left( {1-\cos A} \right)}};\,\cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}=\pm \sqrt{{\frac{1}{2}\left( {1+\cos A} \right)}}$

$ \displaystyle \tan \left( {\frac{A}{2}} \right)=\frac{{\sin A}}{{1+\cos A}}=\frac{{1-\cos A}}{{\sin A}}$

Note: When solving, to get the correct sign when there’s a $ \pm $, check for the quadrant of the half angle, and see if that trig function is positive or negative. For example, for $ \displaystyle \cos \left( {\frac{A}{2}} \right)$, where $ A={{300}^{{}^\circ }}$, $ \displaystyle \frac{A}{2}=150{}^\circ $, which is in Quadrant 2, where cos is negative. Thus, use $ \displaystyle -\sqrt{{\frac{{1+\cos A}}{2}}}$.

     $ \displaystyle \sin x\cot x\,=\,\cos x$

$ \require{cancel} \displaystyle \cancel{{\sin x}}\left( {\frac{{\cos x}}{{\cancel{{\sin x}}}}} \right)=\cos x\,\surd $

$ \begin{align}\sin \theta \cot \theta +\sin \theta \sec \theta &=\\\cancel{{\sin \theta }}\frac{{\cos \theta }}{{\cancel{{\sin \theta }}}}+\sin \theta \frac{1}{{\cos \theta }}&=\\\cos \theta +\frac{{\sin \theta }}{{\cos \theta }}&=\cos \theta +\tan \theta \,\surd \end{align}$

$ \displaystyle \begin{align}\frac{{1+\displaystyle \frac{1}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\displaystyle \frac{{\displaystyle \frac{{\tan \theta }}{{\tan \theta }}+\displaystyle \frac{1}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\displaystyle \frac{{\displaystyle \frac{{\tan \theta +1}}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\frac{{\cancel{{\tan \theta +1}}}}{{\tan \theta }}\times \displaystyle \frac{1}{{\cancel{{1+\tan \theta }}}}&=\\\displaystyle \frac{1}{{\tan \theta }}&=\cot \theta \,\surd \end{align}$

$ \begin{align}\frac{{\cos x}}{{1-\displaystyle \frac{{\sin x}}{{\cos x}}}}+\frac{{\sin x}}{{1-\displaystyle \frac{{\cos x}}{{\sin x}}}}&=\\\frac{{\cos x}}{{\displaystyle \frac{{\cos x-\sin x}}{{\cos x}}}}+\frac{{\sin x}}{{\displaystyle \frac{{\sin x-\cos x}}{{\sin x}}}}&=\,\,\,\,\\\frac{{{{{\cos }}^{2}}x}}{{\cos x-\sin x}}+\frac{{{{{\sin }}^{2}}x}}{{\sin x-\cos x}}&=\,\,(\text{flip bottom and multiply)}\\\frac{{{{{\cos }}^{2}}x}}{{\cos x-\sin x}}-\frac{{{{{\sin }}^{2}}x}}{{\cos x-\sin x}}&=\\\frac{{\cancel{{\left( {\cos x-\sin x} \right)}}\left( {\cos x+\sin x} \right)}}{{\cancel{{\cos x-\sin x}}}}&=\cos x\,+\sin x\,\,\,\,\surd \end{align}$

• Convert everything to sin and cos.
• Find common denominator to to combine fractions.
• Notice that $ \sin x-\cos x=-\left( {\cos x-\sin x} \right)$.
• Use difference of squares to factor.
• Simplify.

           “Proof“:    $ \displaystyle \begin{align}\sin \theta =\frac{y}{r}\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{x}{r}\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}+{{x}^{2}}=1\,\,\,\,\left( {\text{Pythagorean}} \right)\\{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ={{\left( {\frac{y}{r}} \right)}^{2}}+{{\left( {\frac{x}{r}} \right)}^{2}}={{\left( {\frac{y}{1}} \right)}^{2}}+{{\left( {\frac{x}{1}} \right)}^{2}}={{y}^{2}}+{{x}^{2}}=1\end{align}$

                                  (Note that $ {{\left( {\sin x} \right)}^{2}}$ is written as $ {{\sin }^{2}}x$.)

From the first Pythagorean identity, we can derive the other two:

$ \begin{align}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta &=1\\\frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}&=\frac{1}{{{{{\cos }}^{2}}\theta }}\\{{\tan }^{2}}\theta +1&={{\sec }^{2}}\theta \end{align}$                                $ \begin{align}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta &=1\\\frac{{{{{\sin }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}&=\frac{1}{{{{{\sin }}^{2}}\theta }}\\1+{{\cot }^{2}}\theta &={{\csc }^{2}}\theta \end{align}$

The three Pythagorean Identities are:

$ \displaystyle \begin{array}{c}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\\{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \end{array}$

$ \displaystyle \require{cancel} \begin{align}\frac{1}{{\sin x}}\cdot \frac{1}{{\cos x}}-\frac{{\cos x}}{{\sin x}}&=\\\frac{1}{{\sin x\cos x}}-\frac{{\cos x\cdot \cos x}}{{\sin x\cos x}}&=\\\frac{{1-{{{\cos }}^{2}}x}}{{\sin x\cos x}}&=\\\frac{{{}^{{\sin x}}\cancel{{{{{\sin }}^{2}}x}}}}{{\cancel{{\sin x}}\cos x}}&=\,\tan x\,\,\,\,\,\,\surd \end{align}$

• Convert everything to sin and cos.
• Find common denominator to combine fractions.
• Use Pythagorean identity to turn two terms in the numerator into one.
• Simplify.

               $ \displaystyle \frac{{{{{\cos }}^{2}}\theta }}{{1+\sin \theta }}\,=\,1-\sin \theta $

$ \displaystyle \begin{align}\frac{{1-{{{\sin }}^{2}}\theta }}{{1+\sin \theta }}\,\,\,&=\\\frac{{\left( {1-\sin \theta } \right)\cancel{{(1+\sin \theta )}}}}{{\cancel{{1+\sin \theta }}}}\,&=\,1-\sin \theta \,\,\,\,\,\,\,\,\surd \end{align}$

• Convert everything to sin by using Pythagorean identity.
• Factor numerator (difference of squares).
• Simplify.

$ \require{cancel} \displaystyle \begin{align}&=\frac{{\tan x}}{{\sec x+1}}\left( {\frac{{\tan x}}{{\tan x}}} \right)+\frac{{\sec x+1}}{{\tan x}}\left( {\frac{{\sec x+1}}{{\sec x+1}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\tan }}^{2}}x}}{{\tan x\left( {\sec x+1} \right)}}+\frac{{{{{\left( {\sec x+1} \right)}}^{2}}}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\tan }}^{2}}x}}{{\tan x\left( {\sec x+1} \right)}}+\frac{{{{{\sec }}^{2}}x+2\sec x+1}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{\left( {{{{\tan }}^{2}}x+1} \right)+{{{\sec }}^{2}}x+2\sec x}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\sec }}^{2}}x+{{{\sec }}^{2}}x+2\sec x}}{{\tan x\left( {\sec x+1} \right)}}\,\,=\,\frac{{2\sec x\cancel{{\left( {\sec x+1} \right)}}}}{{\tan x\cancel{{\left( {\sec x+1} \right)}}}}\\\frac{2}{{\sin x}}\,\,&=\frac{2}{{\cancel{{\cos x}}}}\cdot \frac{{\cancel{{\cos x}}}}{{\sin x}}\,\,\,\surd \end{align}$

• Start with the right-hand side, since it’s more complicated.
• Find common denominator to add fractions, since we have to turn two terms into one.
• Convert $ \left( {{{{\tan }}^{2}}x+1} \right)$ into $ {{\sec }^{2}}x$, since we have more sec‘s than tan‘s in the problem at this point.
• Factor out a $ 2\sec x$ in numerator.
• Turn everything into sin and cos to match left-hand side.
• Simplify.

$ \displaystyle \begin{align}\frac{{1-\displaystyle \frac{{{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y}}}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,&=\\\displaystyle \frac{{\displaystyle \frac{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y}}}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,&=\\\frac{{\cancel{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}}}{{{{{\cos }}^{2}}y}}\cdot \displaystyle \frac{1}{{\cancel{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}}}\,\,&=\\\displaystyle \frac{1}{{{{{\cos }}^{2}}y}}\,&=\,\,{{\sec }^{2}}y\,\,\,\surd \end{align}$

• The trick here is that it looks like you should use a Pythagorean identity because of all the squares, but you don’t (you can try, but it won’t get you anywhere).
• Convert everything to sin and cos.
• Find common denominator in the numerator to combine fractions.
• To divide the fractions, flip the bottom (put 1 over it) and multiply.
• Simplify.

$ \displaystyle \begin{align}\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\cancel{{\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)}}}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{{{{{\cos }}^{2}}\theta -\left( {1-{{{\cos }}^{2}}\theta } \right)}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{{2{{{\cos }}^{2}}\theta -1}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{1}{{2{{{\cos }}^{2}}\theta -1}}&=\\\frac{1}{{2\left( {1-{{{\sin }}^{2}}\theta } \right)-1}}&=\frac{1}{{1-2{{{\sin }}^{2}}\theta }}\,\,\,\surd \end{align}$

• We need to get rid of the fourth powers by using the difference of squares. Leave the denominator alone, since there would be a fourth root if we multiplied out.
• Aha! We can get rid of the $ ({{\sin }^{2}}x+{{\cos }^{2}}x)$ since it is 1, and we are multiplying.
• Turn everything we can into cos’s to see if we can simplify; we can!
• Now turn back to sin since that’s on the right-hand side.
• That was difficult! And remember that there are other ways to do this!

                                                                                                                                                  $ \displaystyle \frac{{1-\sin x+\cos x}}{{1+\sin x+\cos x}}\,=\,\frac{{\cos x}}{{1+\sin x}}$

$ \displaystyle \begin{align}\frac{{1-\left( {\sin x-\cos x} \right)}}{{1+\left( {\sin x+\cos x} \right)}}\,\cdot \frac{{1+\left( {\sin x-\cos x} \right)}}{{1+\left( {\sin x-\cos x} \right)}}&=\\\frac{{1-{{{\left( {\sin x-\cos x} \right)}}^{2}}}}{{\left( {1+\sin x+\cos x} \right)\left( {1+\sin x-\cos x} \right)}}&=\\\frac{{1-\left( {{{{\sin }}^{2}}x-2\sin x\cos x+{{{\cos }}^{2}}x} \right)}}{{1+2\sin x+{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}&=\\\frac{{1-\left( {1-2\sin x\cos x} \right)}}{{1+2\sin x+{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}&=\\\frac{{2\sin x\cos x}}{{\cancel{1}+2\sin x+{{{\sin }}^{2}}x-\left( {\cancel{1}-{{{\sin }}^{2}}x} \right)}}&=\\\frac{{2\sin x\cos x}}{{2\sin x+2{{{\sin }}^{2}}x}}&=\\\frac{{\cancel{{2\sin x}}\cos x}}{{\cancel{{2\sin x}}\left( {1+\sin x} \right)}}&=\frac{{\cos x}}{{1+\sin x}}\,\,\,\surd \end{align}$

• This one’s tricky since it’s hard to know where to start. Since we have a “$ 1-$“ and a “$ 1+$” on the left, use the conjugate of the numerator to multiply by 1 (we could have found the conjugate of the denominator instead).
• We then get a difference of squares in the numerator.
• For the denominator, note that   $ \displaystyle \begin{array}{c}\left( {1+\sin x+\cos x} \right)\left( {1+\sin x-\cos x} \right)\\=1+2\sin x+{{\sin }^{2}}x-{{\cos }^{2}}x\end{array}$.
• Keep plugging until you can cancel out factors on top and bottom. Remember that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1$.
• Plug in $ \left( {1-{{{\sin }}^{2}}x} \right)$ for $ {{\cos }^{2}}x$ to help simplify, since there are more sin‘s on the bottom.
• We did it! Don’t worry if you have to erase a lot before solving these!

$ \displaystyle \begin{align}2{{\sin }^{2}}+3\cos x-3&=0\\2\left( {1-{{{\cos }}^{2}}} \right)+3\cos x-3&=0\\2-2{{\cos }^{2}}+3\cos x-3&=0\\-2{{\cos }^{2}}+3\cos x-1&=0\\2{{\cos }^{2}}-3\cos x+1&=0\\\left( {2\cos x-1} \right)\left( {\cos x-1} \right)&=0\end{align}$

$ \displaystyle \cos x=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\cos x=1\,\,\,\,\,\,\,\,\,\,x=\left\{ {0,\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}$

• Since we have a quadratic, set everything to 0 and factor.
• Use Pythagorean Identity to convert $ {{\sin }^{2}}x$ into $ \left( {1-{{{\cos }}^{2}}x} \right)$, since we have a $ 3\cos x$ in the problem (try to match trig functions).
• Now we can factor and solve!
• We can check answers in the graphing calculator.

$ \displaystyle \sec x-\tan x=\cot x$

$ \displaystyle \begin{align}\frac{1}{{\cos x}}-\frac{{\sin x}}{{\cos x}}&=\frac{{\cos x}}{{\sin x}}\\\frac{{1-\sin x}}{{\cos x}}-\frac{{\cos x}}{{\sin x}}&=0\\\frac{{\sin x\left( {1-\sin x} \right)}}{{\sin x\cos x}}-\frac{{{{{\cos }}^{2}}x}}{{\sin x\cos x}}&=0\\\frac{{\left( {\sin x-{{{\sin }}^{2}}x-{{{\cos }}^{2}}x} \right)}}{{\sin x\cos x}}&=0\\\left( {\sin x\cancel{{-{{{\sin }}^{2}}x}}-\left( {1\cancel{{-{{{\sin }}^{2}}x}}} \right)} \right)&=0\\\left( {\sin x-1} \right)&=0\,\\\sin x&=1\\x&=\left\{ {\frac{\pi }{2}} \right\}\end{align}$

This is an extraneous solution, since $ \displaystyle \tan \left( {\frac{\pi }{2}} \right)$ is undefined. The answer is no solution or $ \emptyset$.

• Convert everything to sin and cos, using Reciprocal and Quotient Identities.
• Find common denominator to add terms.
• Turn $ {{\cos }^{2}}x$ into $ \left( {1-{{{\sin }}^{2}}x} \right)$ since we have mainly sin‘s; combine terms.
• We have to check for extraneous solutions by evaluating the solution in the original equation.
• You can also square each side, but be sure to check for extraneous solutions when you do this:

$ \displaystyle \sec x=\cot x+\tan x$                   

$ \require{cancel} \displaystyle \begin{align}{{\sec }^{2}}x&={{\left( {\cot x+\tan x} \right)}^{2}}\\1+{{\tan }^{2}}x&={{\cot }^{2}}x+2\cancel{{\cot x\tan x}}+{{\tan }^{2}}x\\1+{{\tan }^{2}}x&={{\cot }^{2}}x+2+{{\tan }^{2}}x\\1+\cancel{{{{{\tan }}^{2}}x}}&={{\cot }^{2}}x+2+\cancel{{{{{\tan }}^{2}}x}}\\{{\cot }^{2}}x&=-1\end{align}$

No solution, since we can’t square something and have it be a negative number.

$ \displaystyle \begin{array}{c}{{\cot }^{2}}\left( x \right)=3\\\cot \left( x \right)=\pm \sqrt{3}\end{array}$

$ \displaystyle x=\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}$

• Use Quotient Identity to convert $ \displaystyle \frac{{\cos }}{{\sin }}$ into cot.
• Take square roots of both sides, and don’t forget the $ \pm $.
• There are no extraneous solutions.

$ \displaystyle \sin \left( x \right)=\cos \left( x \right)+1$

Interval $ \left[ {0,2\pi } \right)$

$ \displaystyle \begin{array}{c}{{\left[ {\sin \left( x \right)} \right]}^{2}}={{\left[ {\cos \left( x \right)+1} \right]}^{2}}\\{{\sin }^{2}}\left( x \right)={{\cos }^{2}}\left( x \right)+2\cos \left( x \right)+1\\1-{{\cos }^{2}}\left( x \right)={{\cos }^{2}}\left( x \right)+2\cos \left( x \right)+1\\2{{\cos }^{2}}\left( x \right)+2\cos \left( x \right)=0\\2\cos \left( x \right)\left[ {\cos \left( x \right)+1} \right]=0\\\cos \left( x \right)=0\,\,\,\,\,\,\,\cos \left( x \right)=-1\\\,x=\displaystyle \frac{\pi }{2},\cancel{{\displaystyle \frac{{3\pi }}{2}}}\,\,\,\,\,\,\,\,\,\,\,x=\pi \,\,\,\,\,\,\,\,\,x=\left\{ {\displaystyle \frac{\pi }{2},\pi } \right\}\end{array}$

$ \displaystyle \frac{{1-\cos x}}{{\sin x}}=\frac{{\sin x}}{{1+\cos x}}$

General Solution (over Reals)

$ \displaystyle \begin{align}\frac{{1-\cos x}}{{\sin x}}-\frac{{\sin x}}{{1+\cos x}}&=0\\\frac{{\left( {1-\cos x} \right)\left( {1+\cos x} \right)}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{\left( {\sin x} \right)\left( {\sin x} \right)}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\\frac{{1-{{{\cos }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\0&=0\end{align}$

This appears to be all real numbers, but since we have denominators that can’t be 0, we have to check for those cases.

The answer, over the reals, is all real numbers except 0 and $ \boldsymbol{\pi}$, or $ \displaystyle \left\{ {x|\,x\ne 0+2\pi k,\,\pi +2\pi k} \right\}$, which is the same as $ \displaystyle \left\{ {x|\,x\ne \pi k} \right\}$.

• Get all terms on one side.
• Find common denominator to add terms.
• Turn $ \left( {1-{{{\cos }}^{2}}x} \right)$ into $ {{\sin }^{2}}x$.
• We get $ 0=0$, which means it looks like it works for all real numbers, or $ \mathbb{R}$ (The two terms are an identity, since they equal each other).
• However, since we have denominators of $ \sin x$ and $ 1+\cos x$, we have to check for extraneous solutions, where the denominators are 0:  $ \displaystyle \begin{array}{l}\sin x=0\,\,\,\,\,\,\,\,\cos x=-1\\\,\,\,\,\,0,\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\pi \end{array}$.

$ \begin{array}{l}\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}$

$ \begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\end{array}$

$ \displaystyle \tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}$

$ \displaystyle \begin{array}{l}\text{EVEN: }\,\,\,\text{ }\cos \left( {-x} \right)=\cos \left( x \right)\,\,\,\,\,\,\,\text{ODD:}\,\,\,\,\,\,\sin \left( {-x} \right)=-\sin \left( x \right)\,\,\,\,\,\,\,\tan \left( {-x} \right)=-\tan \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {-x} \right)=-\csc \left( x \right)\,\,\,\,\,\,\,\cot \left( {-x} \right)=-\cot \left( x \right)\end{array}$

…..and the Cofunction Identities in radians (trig functions of an angle is equal to the value of the cofunction of the complement). Try to prove these using the identities above!

$ \displaystyle \begin{align}\sin \left( {\frac{\pi }{2}-x} \right)&=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {\frac{\pi }{2}-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{\pi }{2}-x} \right)=\cot \left( x \right)\\\cos \left( {\frac{\pi }{2}-x} \right)&=\sin \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {\frac{\pi }{2}-x} \right)=\csc \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( {\frac{\pi }{2}-x} \right)=\tan \left( x \right)\end{align}$

Use a sum and difference identity to show that $ \displaystyle \sin \left( {-x} \right)=-\sin \left( x \right)$

Since $ \sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B$, we have:

$ \displaystyle \begin{align}\sin (-x)&=\sin \left( {0-x} \right)=\sin \left( 0 \right)\cos x-\cos \left( 0 \right)\sin x\\&=0\cdot \cos x-1\cdot \sin x\\&=-\sin x\end{align}$

Use a sum and difference to show that $ \displaystyle \tan \left( {\frac{\pi }{2}-x} \right)=\cot \left( x \right)$

We can’t use the tan difference formula, since $ \displaystyle \tan \left( {\frac{\pi }{2}} \right)$ doesn’t exist. But we can use the sin and cos difference formulas:

$ \displaystyle \begin{align}\tan \left( {\frac{\pi }{2}-x} \right)&=\frac{{\sin \left( {\displaystyle \frac{\pi }{2}-x} \right)}}{{\cos \left( {\displaystyle \frac{\pi }{2}-x} \right)}}\\&=\frac{{\sin \left( {\displaystyle \frac{\pi }{2}} \right)\cos x-\cos \left( {\displaystyle \frac{\pi }{2}} \right)\sin x}}{{\cos \left( {\displaystyle \frac{\pi }{2}} \right)\cos x+\sin \left( {\displaystyle \frac{\pi }{2}} \right)\sin x}}\\&=\frac{{1\cdot \cos x-0\cdot \sin x}}{{0\cdot \cos x+1\cdot \sin x}}=\frac{{\cos x}}{{\sin x}}=\cot x\end{align}$

Simplify $ \displaystyle \sin \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{2\pi }}{7}} \right)+\cos \left( {\frac{{4\pi }}{5}} \right)\sin \left( {\frac{{2\pi }}{7}} \right)$

Since $ \sin A\cos B+\cos A\sin B=\sin \left( {A+B} \right)$, we have:

$ \displaystyle \begin{align}\sin \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{2\pi }}{7}} \right)+\cos \left( {\frac{{4\pi }}{5}} \right)\sin \left( {\frac{{2\pi }}{7}} \right)\\=\sin \left( {\frac{{4\pi }}{5}+\frac{{2\pi }}{7}} \right)=\sin \left( {\frac{{38\pi }}{{35}}} \right)\end{align}$

Find exact value of $ \tan \left( {165{}^\circ } \right)$

Since $ \displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}$, we can find two values on the Unit Circle whose sum is 165°:  120° and 45°:

$ \begin{align}\tan \left( {165} \right)=\tan \left( {120+45} \right)&=\frac{{\tan 120+\tan 45}}{{1-\left( {\tan 120} \right)\left( {\tan 45} \right)}}\\&=\frac{{-\sqrt{3}+1}}{{1-\left( {-\sqrt{3}} \right)\left( 1 \right)}}\\&=\frac{{1-\sqrt{3}}}{{1+\sqrt{3}}}\end{align}$

Find the exact value of $ \displaystyle \cos \left( {\frac{\pi }{{12}}} \right)$

Divide up $ \displaystyle \frac{\pi }{{12}}$ into two fractions that can be found on the Unit Circle:

$ \displaystyle \begin{align}\cos \left( {\frac{\pi }{{12}}} \right)&=\cos \left( {\frac{{4\pi }}{{12}}-\frac{{3\pi }}{{12}}} \right)=\cos \left( {\frac{\pi }{3}-\frac{\pi }{4}} \right)\\&=\cos \left( {\frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{4}} \right)+\sin \left( {\frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{4}} \right)\\&=\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt{2}}}{2}} \right)+\left( {\frac{{\sqrt{3}}}{2}} \right)\left( {\frac{{\sqrt{2}}}{2}} \right)\\&=\frac{{\sqrt{2}+\sqrt{6}}}{4}\end{align}$

Simplify $ \displaystyle \frac{{\tan \displaystyle \frac{\pi }{{12}}-\tan \displaystyle \frac{{5\pi }}{4}}}{{1+\left( {\tan \displaystyle \frac{{5\pi }}{4}} \right)\left( {\tan \displaystyle \frac{\pi }{{12}}} \right)}}$

Since $ \displaystyle \frac{{\tan A-\tan B}}{{1+\tan A\tan B}}=\tan \left( {A-B} \right)$, we have:

$ \displaystyle \begin{align}\frac{{\tan \displaystyle \frac{\pi }{{12}}-\tan \displaystyle \frac{{5\pi }}{4}}}{{1+\left( {\tan \displaystyle \frac{{5\pi }}{4}} \right)\left( {\tan \displaystyle \frac{\pi }{{12}}} \right)}}&=\tan \left( {\displaystyle \frac{\pi }{{12}}-\frac{{5\pi }}{4}} \right)\\&=\,\,\tan \left( {\displaystyle \frac{{-7\pi }}{6}} \right)\end{align}$

Find the exact value of $ \displaystyle \sec \left( {-\frac{\pi }{{12}}} \right)$

Divide up $ \displaystyle -\frac{\pi }{{12}}$ into two fractions that can be found on the Unit Circle; use cos first, and then take reciprocal:

$ \displaystyle \begin{align}\cos \left( {-\frac{\pi }{{12}}} \right)&=\cos \left( {\frac{{3\pi }}{{12}}-\frac{{4\pi }}{{12}}} \right)=\cos \left( {\frac{\pi }{4}-\frac{\pi }{3}} \right)\\&=\cos \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{3}} \right)+\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{3}} \right)\\&=\left( {\frac{{\sqrt{2}}}{2}} \right)\left( {\frac{1}{2}} \right)+\left( {\frac{{\sqrt{2}}}{2}} \right)\left( {\frac{{\sqrt{3}}}{2}} \right)\\&=\frac{{\sqrt{2}+\sqrt{6}}}{4}\end{align}$

$ \displaystyle \begin{align}\sec \left( {-\frac{\pi }{{12}}} \right)&=\frac{1}{{\cos \left( {-\displaystyle \frac{\pi }{{12}}} \right)}}\\&=\frac{4}{{\sqrt{2}+\sqrt{6}}}\left( {\frac{{\sqrt{2}-\sqrt{6}}}{{\sqrt{2}-\sqrt{6}}}} \right)\\&=\sqrt{6}-\sqrt{2}\end{align}$

Find the exact value of $ \displaystyle \cot \left( {-\frac{{5\pi }}{{12}}} \right)$

Divide up $ \displaystyle -\frac{{5\pi }}{{12}}$ into two fractions that can be found on the Unit Circle; use tan first, and then take the reciprocal:

$ \displaystyle \begin{align}\tan \left( {-\frac{{5\pi }}{{12}}} \right)&=\tan \left( {-\displaystyle \frac{{2\pi }}{{12}}+-\frac{{3\pi }}{{12}}} \right)\\&=\tan \left( {-\displaystyle \frac{\pi }{6}+-\displaystyle \frac{\pi }{4}} \right)\\&=\frac{{\tan \left( {-\displaystyle \frac{\pi }{6}} \right)+\tan \left( {-\displaystyle \frac{\pi }{4}} \right)}}{{1-\tan \left( {-\displaystyle \frac{\pi }{6}} \right)\tan \left( {-\displaystyle \frac{\pi }{4}} \right)}}\\&=\displaystyle \frac{{\left( {\displaystyle \frac{{-1}}{{\sqrt{3}}}} \right)+\left( {-1} \right)}}{{1-\left( {\displaystyle \frac{{-1}}{{\sqrt{3}}}} \right)\left( {-1} \right)}}\\&=\frac{{-\displaystyle \frac{1}{{\sqrt{3}}}+-\frac{{\displaystyle \sqrt{3}}}{{\sqrt{3}}}}}{{\displaystyle \frac{{\sqrt{3}}}{{\sqrt{3}}}-\displaystyle \frac{1}{{\sqrt{3}}}}}=\frac{{-\displaystyle \sqrt{3}-1}}{{\sqrt{3}-1}}\end{align}$

$ \displaystyle \cot \left( {-\frac{{5\pi }}{{12}}} \right)=\frac{1}{{\tan \left( {-\displaystyle\frac{{5\pi }}{{12}}} \right)}}=\frac{{1-\sqrt{3}}}{{1+\sqrt{3}}}$

$ \displaystyle \begin{align}\cos x\cos \pi -\sin x\sin \pi &=\\\left( {\cos x} \right)\left( {-1} \right)-\left( {\sin x} \right)\left( 0 \right)&=\\\left( {\cos x} \right)\left( {-1} \right)-0&=-\cos x\,\,\surd \end{align}$

Note:  Evaluate any expressions that turn into constants (like $ \sin \pi $).

$ \displaystyle \begin{align}\left( {\sin \alpha \cos \beta +\cos \alpha \sin \beta } \right)\left( {\sin \alpha \cos \beta -\cos \alpha \sin \beta } \right)&=\\{{\left( {\sin \alpha \cos \beta } \right)}^{2}}-{{\left( {\cos \alpha \sin \beta } \right)}^{2}}&=\\\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\cos }}^{2}}\beta } \right)-\left( {{{{\cos }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)&=\\\left( {{{{\sin }}^{2}}\alpha } \right)\left( {1-{{{\sin }}^{2}}\beta } \right)-\left( {1-{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)&=\\{{\sin }^{2}}\alpha -\cancel{{\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\-\,\,{{\sin }^{2}}\beta +\cancel{{\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)}}&={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta \,\,\surd \end{align}$

Note: We knew to eventually turn everything into sin, since the right-hand side of the identity only had sin in it.

$ \displaystyle \frac{{\sin \left( {x+y} \right)}}{{\sin x\cos y}}=\cot x\tan y+1$

$ \displaystyle \begin{align}\frac{{\sin x\cos y+\cos x\sin y}}{{\sin x\cos y}}&=\\\frac{{\sin x\cos y}}{{\sin x\cos y}}+\frac{{\cos x\sin y}}{{\sin x\cos y}}&=\\1+\frac{{\cos x}}{{\sin x}}\cdot \frac{{\sin y}}{{\cos y}}&=\cot x\tan y+1\,\,\surd \end{align}$

Note: We had to split the fraction on the left to get to the two terms on the right.

$ \displaystyle \sec \left( {x-y} \right)=\frac{{\sec x\sec y}}{{\tan x\tan y+1}}$

$ \displaystyle \begin{align}&=\frac{{\displaystyle \frac{1}{{\cos x}}\cdot \displaystyle \frac{1}{{\cos y}}}}{{\displaystyle \frac{{\sin x}}{{\cos x}}\cdot \displaystyle \frac{{\sin y}}{{\cos y}}+1}}\\&=\frac{{\displaystyle \frac{1}{{\cos x\cos y}}}}{{\displaystyle \frac{{\sin x\sin y}}{{\cos x\cos y}}+1}}=\displaystyle \frac{{\displaystyle \frac{1}{{\cos x\cos y}}}}{{\displaystyle \frac{{\sin x\sin y}}{{\cos x\cos y}}+\frac{{\cos x\cos y}}{{\cos x\cos y}}}}\\\text{(flip & multiply)}\,\,\,\,&=\displaystyle \frac{1}{{\cancel{{\cos x\cos y}}}}\cdot \displaystyle \frac{{\cancel{{\cos x\cos y}}}}{{\sin x\sin y+\cos x\cos y}}\\&=\displaystyle \frac{1}{{\sin x\sin y+\cos x\cos y}}\\\sec \left( {x-y} \right)&=\displaystyle \frac{1}{{\cos \left( {x-y} \right)}}\,\,\surd \end{align}$

Note: Start with the more complicated side and turn everything into sin and cos.

$ \displaystyle \cos \left( {3x} \right)\cos 15{}^\circ +\sin \left( {3x} \right)\sin 15{}^\circ =\frac{1}{2}$, interval $ \left[ {0,360{}^\circ } \right)$

$ \displaystyle \cos \left( {3x-15} \right)=\frac{1}{2}$

First solve over reals and then see how many of these solutions are in $ \left[ {0,360{}^\circ } \right)$:

$ \displaystyle \begin{array}{c}3x-15=60{}^\circ +360{}^\circ k\,\,\,\,\,\,\,\,3x-15=300{}^\circ +360{}^\circ k\\3x=75{}^\circ +360{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,3x=315{}^\circ +360{}^\circ k\\\,x=25{}^\circ +120{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,x=105{}^\circ +120{}^\circ k\\\\\left\{ {x|x=25{}^\circ ,105{}^\circ ,145{}^\circ ,225{}^\circ ,265{}^\circ ,345{}^\circ } \right\}\end{array}$

$ \displaystyle \sin \left( {\frac{x}{4}-x} \right)=-1$

$ \begin{align}-\frac{{3x}}{4}&=\frac{{3\pi }}{2}+2\pi k\\-3x&=6\pi +8\pi k\\x&=\frac{{6\pi }}{{-3}}+\frac{{8\pi }}{{-3}}k\end{align}$

$ \displaystyle \left\{ {x|x=-2\pi -\frac{{8\pi }}{3}k} \right\}$

$ \displaystyle \tan \left( {3x} \right)-\tan 63{}^\circ =\sqrt{3}+\sqrt{3}\tan \left( {3x} \right)\tan {{63}^{{}^\circ }}$, over the reals

$ \displaystyle \begin{align}\tan \left( {3x} \right)-\tan {{63}^{{}^\circ }}&=\sqrt{3}\left( {1+\tan \left( {3x} \right)\tan {{{63}}^{{}^\circ }}} \right)\\\frac{{\tan \left( {3x} \right)-\tan {{{63}}^{{}^\circ }}}}{{1+\tan \left( {3x} \right)\tan {{{63}}^{{}^\circ }}}}&=\sqrt{3}\\\tan \left( {3x-{{{63}}^{{}^\circ }}} \right)&=\sqrt{3}\\\\3x-{{63}^{{}^\circ }}&={{60}^{{}^\circ }}+{{180}^{{}^\circ }}k\\3x&={{123}^{{}^\circ }}+{{180}^{{}^\circ }}k\\x&=\frac{{{{{123}}^{{}^\circ }}}}{3}+\frac{{{{{180}}^{{}^\circ }}k}}{3}\end{align}$

$ \displaystyle \left\{ {x|x=41{}^\circ +60{}^\circ k} \right\}$

    $ \displaystyle 2\cos \left( {x+\frac{\pi }{6}} \right)\cos \left( {x-\frac{\pi }{6}} \right)=1$, interval $ \left[ {0,2\pi } \right)$

$ \displaystyle \begin{align}\cos \left( {x+\frac{\pi }{6}} \right)\cos \left( {x-\frac{\pi }{6}} \right)&=\frac{1}{2}\\\left( {\cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}} \right)\left( {\cos x\cos \frac{\pi }{6}+\sin x\sin \frac{\pi }{6}} \right)&=\frac{1}{2}\\\left( {\frac{{\sqrt{3}}}{2}\cos x-\frac{1}{2}\sin x} \right)\left( {\frac{{\sqrt{3}}}{2}\cos x+\frac{1}{2}\sin x} \right)&=\frac{1}{2}\\\frac{3}{4}{{\cos }^{2}}x-\frac{1}{4}{{\sin }^{2}}x&=\frac{1}{2}\\\frac{3}{4}\left( {1-{{{\sin }}^{2}}x} \right)-\frac{1}{4}{{\sin }^{2}}x&=\frac{1}{2}\\3-3{{\sin }^{2}}x-{{\sin }^{2}}x&=2\end{align}$

$ \displaystyle \begin{array}{c}-4{{\sin }^{2}}x=-1;\,\,{{\sin }^{2}}x=\displaystyle \frac{1}{4}\\\sin x=\pm \displaystyle\frac{1}{2}\end{array}$

$ \displaystyle x=\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}$

Here’s one with a coterminal identity:

$ \displaystyle \frac{{\cot \left( {90{}^\circ -x} \right)}}{{1-{{{\tan }}^{2}}x}}=\frac{{\sqrt{3}}}{2}$, interval $ \left[ {0,360{}^\circ } \right)$

$ \begin{align}\frac{{\cot \left( {90{}^\circ -x} \right)}}{{1-{{{\tan }}^{2}}x}}&=\frac{{\tan x}}{{1-{{{\tan }}^{2}}x}}=\frac{{\sqrt{3}}}{2}\\\,\,\frac{{2\tan x}}{{1-{{{\tan }}^{2}}x}}&=\sqrt{3}\\\tan \left( {2x} \right)&=\sqrt{3}\,\,\,\,\left( {\text{(double angle identity}} \right)\end{align}$

$ \begin{array}{c}2x=60{}^\circ +180{}^\circ k\\x=30{}^\circ +90{}^\circ k\\\\\left\{ {x|x=30{}^\circ ,120{}^\circ ,210{}^\circ ,300{}^\circ } \right\}\end{array}$

Here’s one with odd/even identities:

  $ \cos \left( {-4x} \right)+\sin \left( {-2x} \right)=0$, over the reals

$ \displaystyle \begin{array}{c}\cos \left( {-4x} \right)+\sin \left( {-2x} \right)=\cos \left( {4x} \right)-\sin \left( {2x} \right)=0\\1-2{{\sin }^{2}}\left( {2x} \right)-\sin \left( {2x} \right)=0\,\,\,\,\left( {\text{use identity with sin in it}} \right)\\2{{\sin }^{2}}\left( {2x} \right)+\sin \left( {2x} \right)-1=0\\\left( {2\sin \left( {2x} \right)-1} \right)\left( {\sin \left( {2x} \right)+1} \right)=0\end{array}$

$ \displaystyle \sin \left( {2x} \right)=\frac{1}{2};\,\,\,\,\sin \left( {2x} \right)=-1$

$ \displaystyle 2x=\frac{\pi }{6}+2\pi k\,\,\,\,\,\,\,\,\,2x=\frac{{5\pi }}{6}+2\pi k\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{2}+2\pi k$

$ \displaystyle \left\{ {x|x=\frac{\pi }{{12}}+\pi k,\,\,\frac{{5\pi }}{{12}}+\pi k,\,\,\frac{{3\pi }}{4}+\pi k\,\,} \right\}$

$ \cos \left( {2A} \right)$ is a little more complicated, especially since it can be written in three ways. I memorize the $ {{\cos }^{2}}A-{{\sin }^{2}}A$ part  and then remember that I can use the Pythagorean Identity $ {{\sin }^{2}}A+{{\cos }^{2}}A=1$ to substitute and do the algebra to arrive at the other expressions.

For $ \tan \left( {2A} \right)$, the identity only has tan’s in it, with the “2” in the numerator, and the $ 1-{{\tan }^{2}}A$ in the denominator (or you can derive it from the sum identity above: $ \tan \left( {2A} \right)=\tan \left( {A+A} \right)=\displaystyle \frac{{\tan A+\tan A}}{{1-\left( {\tan A} \right)\left( {\tan A} \right)}}=\displaystyle \frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}$.)

The tricky thing is that the sign of the whole identity depends on the quadrant where the angle $ \displaystyle \boldsymbol{\frac{A}{2}}$ is (not angle $ A$, but $ \displaystyle \frac{A}{2}$). For example, for $ \displaystyle \sin \left( {\frac{A}{2}} \right)$, if $ \displaystyle \frac{A}{2}$ is in the first or second quadrants, use $ \displaystyle +\sqrt{{\frac{{1-\cos A}}{2}}}$, and if $ \displaystyle \frac{A}{2}$ is in the third or fourth quadrants, use $ \displaystyle -\sqrt{{\frac{{1-\cos A}}{2}}}$. Similarly for $ \displaystyle \cos \left( {\frac{A}{2}} \right)$, if $ \displaystyle \frac{A}{2}$ is in the first or fourth quadrants, use $ \displaystyle +\sqrt{{\frac{{1+\cos A}}{2}}}$, and if $ \displaystyle \frac{A}{2}$ is in the second or third quadrants, use $ \displaystyle -\sqrt{{\frac{{1+\cos A}}{2}}}$.

For $ \displaystyle \tan \left( {\frac{A}{2}} \right)$, you’ll want to memorize both parts of the identity, but you can derive the second part by multiplying the first by $ \displaystyle \frac{{1-\cos A}}{{1-\cos A}}$ (use conjugate of denominator).

$ \displaystyle \frac{{\sin \left( {2x} \right)}}{2}=\frac{{\tan x}}{{{{{\tan }}^{2}}x+1}}$                     

$ \require{cancel} \displaystyle \begin{align}&=\frac{{\tan x}}{{{{{\sec }}^{2}}x}}\\&=\frac{{\sin x}}{{\cancel{{\cos x}}}}\cdot \frac{{{{{\cancel{{{{{\cos }}^{2}}x}}}}^{{\cos x}}}}}{1}\\&=\sin x\cos x\\&=\sin x\cos x\cdot \frac{2}{2}\\\frac{{\sin \left( {2x} \right)}}{2}&=\frac{{2\sin x\cos x}}{2}\,\,\,\,\,\surd \end{align}$

Note: To get $ \displaystyle \sin x\cos x$ into the right form for the identity $ \displaystyle (2\sin x\cos x)$, we had to multiply by $ \displaystyle \frac{2}{2}$.

 $ \displaystyle \frac{{1-\cos \left( {2A} \right)}}{{\sin \left( {2A} \right)}}=\tan A$

$ \displaystyle \begin{align}\frac{{1-\left( {1-2{{{\sin }}^{2}}A} \right)}}{{\sin \left( {2A} \right)}}&=\\\frac{{2{{{\sin }}^{2}}A}}{{\sin \left( {2A} \right)}}&=\\\frac{{{}^{{\sin A}}\cancel{{2{{{\sin }}^{2}}A}}}}{{\cancel{{2\sin A}}\cos A}}&=\tan A\,\,\,\,\,\surd \end{align}$

Note: We knew to use the $ \displaystyle 1-2{{\sin }^{2}}A$ (instead of $ \displaystyle 2{{\cos }^{2}}A-1$) for $ \cos 2A$ to get rid of the 1’s.

  $ \displaystyle \tan \left( {\frac{1}{2}\theta } \right)=\csc \theta -\cot \theta $      

$ \displaystyle \begin{align}&=\frac{1}{{\sin \theta }}-\frac{{\cos \theta }}{{\sin \theta }}\\\tan \left( {\frac{1}{2}\theta } \right)&=\frac{{1-\cos \theta }}{{\sin \theta }}\,\,\,\,\,\surd \end{align}$

Note: Start from the right-hand side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos.

                       $ \displaystyle \frac{{\cos \left( {2\theta } \right)}}{{\cos \theta -\sin \theta }}=\cos \theta +\sin \theta $

$ \displaystyle \begin{align}\frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{\cos \theta -\sin \theta }}&=\\\frac{{\cancel{{\left( {\cos \theta -\sin \theta } \right)}}\left( {\cos \theta +\sin \theta } \right)}}{{\cancel{{\cos \theta -\sin \theta }}}}&=\cos \theta +\sin \theta \,\,\,\,\,\surd \end{align}$

Note: We knew to use $ \displaystyle {{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ and difference of squares for $ \cos 2\theta $ since the denominator contains both cos and sin.

$ \displaystyle \frac{{1-\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}=\frac{{1-\tan \theta }}{{1+\tan \theta }}$                                                   

$ \displaystyle \begin{align}&=\frac{{1-\displaystyle \displaystyle \frac{{\sin \theta }}{{\cos \theta }}}}{{1+\displaystyle \frac{{\sin \theta }}{{\cos \theta }}}}=\displaystyle \frac{{\displaystyle \frac{{\cos \theta }}{{\cos \theta }}-\displaystyle \frac{{\sin \theta }}{{\cos \theta }}}}{{\displaystyle \frac{{\cos \theta }}{{\cos \theta }}+\displaystyle \frac{{\displaystyle \sin \theta }}{{\cos \theta }}}}=\displaystyle \frac{{\displaystyle \frac{{\cos \theta -\sin \theta }}{{\cos \theta }}}}{{\displaystyle \frac{{\cos \theta +\sin \theta }}{{\cos \theta }}}}\\&=\displaystyle \frac{{\cos \theta -\sin \theta }}{{\cos \theta +\sin \theta }}\cdot \displaystyle \frac{{\cos \theta -\sin \theta }}{{\cos \theta -\sin \theta }}\\&=\displaystyle \frac{{{{{\cos }}^{2}}\theta -2\sin \theta \cos \theta +{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}\\&=\displaystyle \frac{{1-2\sin \theta \cos \theta }}{{\cos \left( {2\theta } \right)}}\\\displaystyle \frac{{1-\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}&=\displaystyle \frac{{1-\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}\,\,\,\,\,\,\,\,\surd \end{align}$

Note: The right-hand side looked a little more complicated, so we started there. We turned tan into sin and cos and simplified first. Then we multiplied by 1, using the conjugate of the bottom; this got us the double angle identity for $ \displaystyle \cos \left( {2\theta } \right)$ in the denominator. We also used the Pythagorean Identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1$.

         $ \displaystyle \frac{{\cos \left( {2x} \right)+\cos x}}{{\sin \left( {2x} \right)-\sin x}}=\frac{{\sin x}}{{1-\cos x}}$

$ \displaystyle \begin{align}\frac{{2{{{\cos }}^{2}}x-1+\cos x}}{{2\sin x\cos x-\sin x}}&=\\\frac{{\cancel{{\left( {2\cos x-1} \right)}}\left( {\cos x+1} \right)}}{{\sin x\cancel{{\left( {2\cos x-1} \right)}}}}&=\\\frac{{\cos x+1}}{{\sin x}}\cdot \frac{{\cos x-1}}{{\cos x-1}}&=\\\frac{{{{{\cos }}^{2}}x-1}}{{\sin x\left( {\cos x-1} \right)}}&=\\\frac{{{{{\cancel{{-{{{\sin }}^{2}}x}}}}^{{-\sin x}}}}}{{\cancel{{\sin x}}\left( {\cos x-1} \right)}}&=\frac{{\sin x}}{{1-\cos x}}\,\,\,\,\,\,\surd \end{align}$

Note: We knew to use the $ \displaystyle 2{{\cos }^{2}}A-1$ for $ \cos 2A$ because of the cos in the numerator; then we could factor. We multiplied by 1, using the conjugate of the numerator, and had to use the Pythagorean identity to cancel out the sin’s. Tricky!

 $ \displaystyle -\sec \left( {2x} \right)=\frac{{{{{\sec }}^{2}}x}}{{\sec {{x}^{2}}-2}}$                                                  

$ \displaystyle \begin{align}&=\frac{{\displaystyle \frac{1}{{{{{\cos }}^{2}}x}}}}{{\displaystyle \frac{1}{{{{{\cos }}^{2}}x}}-2}}\\&=\displaystyle \frac{{\displaystyle \frac{1}{{\cancel{{{{{\cos }}^{2}}x}}}}}}{{\displaystyle \frac{{1-2{{{\cos }}^{2}}x}}{{\cancel{{{{{\cos }}^{2}}x}}}}}}\\&=\displaystyle \frac{1}{{1-2{{{\cos }}^{2}}x}}=\,-\displaystyle \frac{1}{{2{{{\cos }}^{2}}x-1}}\\\,-\sec \left( {2x} \right)&=-\displaystyle \frac{1}{{\displaystyle \frac{1}{{2{{{\cos }}^{2}}x-1}}}}\,\,\,\surd \end{align}$

Note: We knew to put everything in terms of cos on the right and then use the reciprocal of the $ \cos \left( {2x} \right)$ identity to get $ \sec \left( {2x} \right)$.

         $ \displaystyle \frac{{\cos 2x-4\cos x-5}}{{\sin 2x-6\sin x}}=\cot x+\csc x$    

$ \displaystyle \begin{align}\frac{{2{{{\cos }}^{2}}x-1-4\cos x-5}}{{2\sin x\cos x-6\sin x}}&=\\\frac{{2{{{\cos }}^{2}}x-4\cos x-6}}{{2\sin x\left( {\cos x-3} \right)}}&=\\\frac{{\left( {2\cos x+2} \right)\cancel{{\left( {\cos x-3} \right)}}}}{{2\sin x\cancel{{\left( {\cos x-3} \right)}}}}&=\\\frac{{2\cos x}}{{2\sin x}}+\frac{2}{{2\sin x}}&=\cot x+\csc x\,\,\,\surd \end{align}$

Note: We knew to put everything in terms of cos on the top so we could factor; it worked out that we could simplify after factoring. Pretty cool!

$ \displaystyle \begin{align}\color{#804040}{{\frac{1}{{\tan \left( {\displaystyle \frac{x}{2}} \right)}}}}&=\frac{{\displaystyle \frac{{\sin x}}{{\cancel{{\cos x}}}}\cancel{{\cos x}}}}{{2\cdot {{{\left( {\pm \sqrt{{\displaystyle \frac{{1-\cos x}}{2}}}} \right)}}^{2}}}}\\\color{#804040}{{\frac{{\sin x}}{{1-\cos x}}}}&=\frac{{\sin x}}{{\cancel{2}\cdot \displaystyle \frac{{1-\cos x}}{{\cancel{2}}}}}\\&=\displaystyle \frac{{\sin x}}{{1-\cos x}}\\&=\displaystyle \frac{1}{{\tan \left( {\displaystyle \frac{x}{2}} \right)}}=\cot \left( {\displaystyle \frac{x}{2}} \right)\,\,\,\,\,\,\surd \end{align}$

• Both sides are complicated, so let’s start down both sides to see if it helps.
• $ \displaystyle \cot \left( {\frac{x}{2}} \right)$ can turn into $ \displaystyle \frac{{\sin \left( x \right)}}{{1-\cos \left( x \right)}}$, since it’s the reciprocal of $ \displaystyle \tan \left( {\frac{x}{2}} \right)$.
• When we square the half angle identity for sin, the $ \pm $ goes away.
• Aha! We see how the identity is proven, so let’s continue down the right-hand side, using what we know on the left-hand side!

Since $ \displaystyle \cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}$, we need to get $ \cos A$, and then find the quadrant of $ \displaystyle \cos \left( {\frac{A}{2}} \right)$ to get the correct sign.             

Draw the triangle in the 4^th Quadrant, making sure the signs of the values are correct. From the triangle, $ \displaystyle \pm \sqrt{{\frac{{1+\cos A}}{2}}}=\pm \sqrt{{\frac{{1+\frac{4}{5}}}{2}}}=\pm \sqrt{{\frac{9}{{10}}}}$.

But since angle A is between $ \displaystyle \frac{{3\pi }}{2}$ and $ \displaystyle 2\pi $, angle $ \displaystyle \frac{A}{2}$ will be between $ \displaystyle \frac{{3\pi }}{4}$ and $ \displaystyle \pi $, which is in the 2^nd quadrant (you have to think of these positive angles going counter-clockwise from 0, or the positive $ x$-axis). In the 2^nd quadrant, cos is negative. So, $ \displaystyle \cos \left( {\frac{A}{2}} \right)=-\sqrt{{\frac{9}{{10}}}}$.

Since $ \displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}$, $ \displaystyle \cot \left( {2A} \right)=\frac{1}{{\displaystyle \frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}}}=\frac{{1-{{{\tan }}^{2}}A}}{{2\tan A}}$ (the reciprocal). We can just use $ \tan A$ and plug in.               

Draw the triangle in the 3^rd Quadrant, making sure the signs of the values are correct.

Plug in the values for tan, noting that cot should be is positive in the 3^rd  quadrant:  $ \displaystyle \cot \left( {2A} \right)=\frac{{1-{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{2}}}}{{2\left( {\displaystyle \frac{2}{7}} \right)}}=\frac{{45}}{{28}}$.

Since $ \displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}$, we need to get $ \cos A$, and then find the quadrant of $ \displaystyle \sin \left( {\frac{A}{2}} \right)$ to get the correct sign.           

Draw the triangle in the 2^nd Quadrant, making sure the signs of the values are correct.

From the triangle, $ \displaystyle \begin{align}\sin \left( {\frac{A}{2}} \right)&=\pm \sqrt{{\frac{{1-\cos A}}{2}}}=\pm \sqrt{{\frac{{1-\displaystyle \frac{{-5}}{{\sqrt{{26}}}}}}{2}}}=\pm \sqrt{{\frac{{\displaystyle \frac{{\sqrt{{26}}}}{{\sqrt{{26}}}}-\frac{{-5}}{{\sqrt{{26}}}}}}{2}}}\\&=\pm \sqrt{{\displaystyle \frac{{\sqrt{{26}}+5}}{{2\sqrt{{26}}}}}}=\pm \sqrt{{\displaystyle \frac{{\sqrt{{26}}+5}}{{2\sqrt{{26}}}}}}\cdot \displaystyle \frac{{\sqrt{{26}}}}{{\sqrt{{26}}}}=\pm \sqrt{{\frac{{26+5\sqrt{{26}}}}{{52}}}}\end{align}$. Since angle A is between $ \displaystyle \frac{\pi }{2}$ and $ \displaystyle \pi $, angle $ \displaystyle \frac{A}{2}$ will be between $ \displaystyle \frac{\pi }{4}$ and $ \displaystyle \frac{\pi }{2}$, which is in the 1^st quadrant (you have to think of these positive angles going counter-clockwise from 0, or the positive $ x$-axis). So, $ \displaystyle \sin \left( {\frac{A}{2}} \right)=\,\,\sqrt{{\frac{{26+5\sqrt{{26}}}}{{52}}}}$.

   $ \displaystyle {{\cos }^{2}}\left( {\frac{x}{2}} \right)=\cos x+1$, over the reals

$ \displaystyle \begin{align}{{\cos }^{2}}\left( {\frac{x}{2}} \right)-\cos x-1&=0\\{{\left( {\pm \sqrt{{\frac{{1+\cos x}}{2}}}} \right)}^{2}}-\cos x-1&=0\\\frac{{1+\cos x}}{2}-\cos x-1&=0\\1+\cos x-2\cos x-2&=0\\\cos x&=-1\end{align}$

$ \displaystyle \left\{ {x|x=\pi +2\pi k} \right\}$

$ \displaystyle \begin{align}\frac{1}{{\left( {\displaystyle \frac{{\sin \theta }}{{1+\cos \theta }}} \right)}}&=-1\\\frac{1}{{\tan \left( {\displaystyle \frac{\theta }{2}} \right)}}&=-1;\,\,\,\cot \left( {\frac{\theta }{2}} \right)=-1\end{align}$

$ \displaystyle \frac{\theta }{2}=\frac{{3\pi }}{4}+\pi k\,\,\,\,\,\,\,\,\theta =\frac{{3\pi }}{2}+2\pi k$

$ \displaystyle \theta =\left\{ {\frac{{3\pi }}{2}} \right\}$

Remember from Trig Solving of Multiple Angles, if the coefficient in the trig argument isn’t 1 (we have $ \displaystyle \frac{1}{2}$), we have to check for reals first, and then narrow down to $ \left[ {0,2\pi } \right)$ interval.

   $ \displaystyle \cos \left( {2A} \right)+\cos A=0$, over the reals

$ \displaystyle \begin{align}2{{\cos }^{2}}A-1+\cos A&=0\\2{{\cos }^{2}}A+\cos A-1&=0\\\left( {2\cos A-1} \right)\left( {\cos A+1} \right)&=0\end{align}$

$ \displaystyle \cos A=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos A=-1$

$ \displaystyle \left\{ {A|A=\frac{\pi }{3}+2\pi k,\,\,\frac{{5\pi }}{3}+2\pi k,\,\,\pi +2\pi k} \right\}$

$ \displaystyle \begin{align}2\sin x\cos x-\cos x&=0\\\cos x\left( {2\sin x-1} \right)&=0\end{align}$

$ \displaystyle \cos x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\sin x=\frac{1}{2}$

$ \displaystyle x=\left\{ {\frac{\pi }{2},\frac{{3\pi }}{2},\frac{\pi }{6},\frac{{5\pi }}{6}} \right\}$

    $ \displaystyle \tan \left( {\frac{A}{2}} \right)\sin A=1-\cos A$, over the reals

$ \require{cancel} \displaystyle \begin{align}\left( {\frac{{1-\cos A}}{{\cancel{{\sin A}}}}} \right)\cancel{{\sin A}}&=1-\cos A\\1-\cos A&=1-\cos A\end{align}$

This would appear to be all real numbers, but we have to restrict the answers by the asymptotes of tan, which are $ \displaystyle \frac{\pi }{2}+\pi k$ for the parent function. Since we have $ \displaystyle \tan \left( {\frac{A}{2}} \right)$, the asymptotes are $ \pi +2\pi k$. So, the answer is $ \displaystyle \left\{ {A|A\ne \pi +2\pi k} \right\}$. Tricky!

    $ \displaystyle 4\sin x\cos x=1$, interval $ \left[ {0,2\pi } \right)$

$ \displaystyle \begin{array}{c}2\left( {2\sin x\cos x} \right)=1\\2\left( {\sin \left( {2x} \right)} \right)=1\end{array}$

$ \displaystyle \sin \left( {2x} \right)=\frac{1}{2}$

$ \displaystyle \,2x=\frac{\pi }{6}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{5\pi }}{6}+2\pi k$

$ \displaystyle x=\frac{\pi }{{12}}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5\pi }}{{12}}+\pi k$

In Unit Circle, this will be $ \displaystyle x=\left\{ {\frac{\pi }{{12}},\frac{{5\pi }}{{12}},\frac{{13\pi }}{{12}},\frac{{17\pi }}{{12}}} \right\}$.

$ \displaystyle 1-{{\sin }^{2}}\theta -\cos \left( {2\theta } \right)-\frac{1}{2}=0$, over the reals

$ \displaystyle 1-{{\sin }^{2}}\theta -\left( {1-2{{{\sin }}^{2}}\theta } \right)-\frac{1}{2}=0$

$ \displaystyle {{\sin }^{2}}\theta =\frac{1}{2};\,\,\,\,\,\sin \theta =\pm \sqrt{{\frac{1}{2}}}=\pm \frac{1}{{\sqrt{2}}}=\pm \frac{{\sqrt{2}}}{2}$

$ \displaystyle \left\{ {\theta |\theta =\frac{\pi }{4}+\pi k,\,\,\,\,\frac{{3\pi }}{4}+\pi k} \right\}\text{,}\,\text{or even}$

$ \displaystyle \left\{ {\theta |\theta =\frac{\pi }{4}+\frac{\pi }{2}k} \right\}$

$ \displaystyle \tan \left( {2x} \right)\,-\cot x=0$, interval $ \left[ {0,2\pi } \right)$

$ \displaystyle \begin{align}\tan \left( {2x} \right)\,-\cot x&=0\\\frac{{2\tan x}}{{1-{{{\tan }}^{2}}x}}-\cot x&=0\\2\tan x-\cot x\left( {1-{{{\tan }}^{2}}x} \right)&=0\\2\cot x{{\tan }^{2}}x-\cot x\left( {1-{{{\tan }}^{2}}x} \right)&=0&\,\\\cot x\left( {3{{{\tan }}^{2}}x-1} \right)=0\\\cot x&=0\,\,\,\,\,\,\,\,\,\tan x=\pm \frac{1}{{\sqrt{3}}}\\x&=\left\{ {\frac{\pi }{2},\frac{{3\pi }}{2},\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}\end{align}$

Reciprocal and Quotient Identities

$ \displaystyle \sin \theta =\frac{1}{{\csc \theta }}\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac{1}{{\sin \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{1}{{\sec \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac{1}{{\cos \theta }}\,$        $ \displaystyle \tan \theta =\frac{1}{{\cot \theta }}=\frac{{\sin \theta }}{{\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac{1}{{\tan \theta }}=\frac{{\cos \theta }}{{\sin \theta }}$

Pythagorean Identities

$ \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cot }^{2}}\theta +1={{\csc }^{2}}\theta $

Sum and Difference Identities

$ \begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}$

$ \displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}$

From these identities, you may be asked to be familiar with the Odd/Even Identities:

$ \displaystyle \begin{array}{l}\text{EVEN: }\,\,\,\cos \left( {-x} \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ODD:}\,\,\,\,\,\,\sin \left( {-x} \right)=-\sin \left( x \right)\,\,\,\,\,\,\,\tan \left( {-x} \right)=-\tan \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {-x} \right)=-\csc \left( x \right)\,\,\,\,\,\,\,\cot \left( {-x} \right)=-\cot \left( x \right)\end{array}$

…..and the Cofunction Identities in radians (trig function of an angle is equal to the value of the cofunction of the complement):

$ \displaystyle \begin{align}\sin \left( {\frac{\pi }{2}-x} \right)&=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {\frac{\pi }{2}-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{\pi }{2}-x} \right)=\cot \left( x \right)\\\cos \left( {\frac{\pi }{2}-x} \right)&=\sin \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {\frac{\pi }{2}-x} \right)=\csc \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( {\frac{\pi }{2}-x} \right)=\tan \left( x \right)\end{align}$

Double Angle Identities

$ \sin \left( {2A} \right)=2\sin A\cos A$          $ \begin{align}\cos \left( {2A} \right)&={{\cos }^{2}}A-{{\sin }^{2}}A\\&=1-2{{\sin }^{2}}A\\&=2{{\cos }^{2}}A-1\end{align}$               $ \displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}$

Half Angle Identities

$ \displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}=\pm \sqrt{{\frac{1}{2}\left( {1-\cos A} \right)}};\,\cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}=\pm \sqrt{{\frac{1}{2}\left( {1+\cos A} \right)}}$

$ \displaystyle \tan \left( {\frac{A}{2}} \right)=\frac{{\sin A}}{{1+\cos A}}=\frac{{1-\cos A}}{{\sin A}}$

Note: When solving, to get the correct sign when there’s a $ \pm $, check for the quadrant of the half angle, and see if that trig function is positive or negative. For example, for $ \displaystyle \cos \left( {\frac{A}{2}} \right)$, where $ A={{300}^{{}^\circ }}$, $ \displaystyle \frac{A}{2}=150{}^\circ $, which is in Quadrant 2, where cos is negative. Thus, use $ \displaystyle -\sqrt{{\frac{{1+\cos A}}{2}}}$.

$ \displaystyle \begin{align}\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)}}{{{{{\cos }}^{2}}\theta +2\cos \theta \sin \theta +{{{\sin }}^{2}}\theta }}&=\\\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\left( 1 \right)}}{{2\cos \theta \sin \theta +1}}&=\frac{{\cos 2\theta }}{{1+\sin 2\theta }}\,\,\,\,\,\,\,\,\,\,\surd \end{align}$

• Factor difference of squares in numerator.
• Multiply binomial in denominator.
• Use Pythagorean Identities to simplify.
• Use Double Angle Identities.

$ \displaystyle \begin{align}\left( {\frac{1}{{1-\sin \theta }}} \right)\left( {\frac{{1+\sin \theta }}{{1+\sin \theta }}} \right)&=\\\frac{{1+\sin \theta }}{{1-{{{\sin }}^{2}}\theta }}&=\\\frac{{1+\sin \theta }}{{{{{\cos }}^{2}}\theta }}&=\\\frac{1}{{{{{\cos }}^{2}}\theta }}+\frac{{\sin \theta }}{{{{{\cos }}^{2}}\theta }}&=\\ {{\sec }^{2}}\theta +\left( {\frac{1}{{\cos \theta }}} \right)\tan \theta &=\sec \theta \tan \theta +{{\sec }^{2}}\theta \,\,\,\,\,\,\,\surd \end{align}$

• Multiply by 1 using conjugate of the denominator.
• Use Pythagorean Identity to simplify.
• Separate fraction into two terms, since there are two terms on the other side.
• Use Reciprocal and Quotient Identities to simplify.

$ \displaystyle \begin{align}&=\frac{{\left( {\displaystyle \frac{{\cos y}}{{\sin y}}+1} \right)}}{{\left( {\displaystyle \frac{{\cos y}}{{\sin y}}-1} \right)}}\\&=\frac{{\left( {\displaystyle \frac{{\cos y+\sin y}}{{\sin y}}} \right)}}{{\left( {\displaystyle \frac{{\cos y-\sin y}}{{\sin y}}} \right)}}\\&=\displaystyle \frac{{\cos y+\sin y}}{{\cos y-\sin y}}\\&=\frac{{\cos y+\sin y}}{{\cos y-\sin y}}\left( {\displaystyle \frac{{\cos y+\sin y}}{{\cos y+\sin y}}} \right)\\\displaystyle \frac{{1+\sin \left( {2y} \right)}}{{\cos \left( {2y} \right)}}&=\displaystyle \frac{{{{{\cos }}^{2}}y+2\sin y\cos y+{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,\,\,\,\,\,\,\,\surd \end{align}$

• Even though the left-hand side looks more complicated, when I tried it, it didn’t go anywhere.
• It turns out the right-hand side can be simplified more easily; turn everything into sin and cos.
• Find common denominators for the fractions, and simplify them.
• Since we can use a $ {{\cos }^{2}}y-{{\sin }^{2}}y$ on the bottom for the $ \cos \left( {2y} \right)$ on the left, multiply by 1 using conjugate of the denominator.
• Use a Pythagorean Identity, and then Double Angle Identities to simplify.

$ \require{cancel} \displaystyle \begin{align}\frac{{2{{{\cos }}^{2}}x-1+\cos x-20}}{{2\cos x\sin x-6\sin x}}&=\\\frac{{2\cos {{x}^{2}}+\cos x-21}}{{2\sin x\left( {\cos x-3} \right)}}&=\\\frac{{\left( {2\cos x+7} \right)\cancel{{\left( {\cos x-3} \right)}}}}{{2\sin x\cancel{{\left( {\cos x-3} \right)}}}}&=\\\frac{{\cancel{2}\cos x}}{{\cancel{2}\sin x}}+\frac{7}{{2\sin x}}&=\cot x+\frac{7}{2}\csc x\,\,\,\,\,\,\,\,\,\surd \end{align}$

• Use Double Angle identities in numerator and denominator.
• Factor numerator and denominator and we can simplify!
• Separate fraction since we have two terms on the right-hand side.
• Use Reciprocal and Quotient Identities to simplify.

$ \begin{align}\frac{{\left( {\tan A} \right)\left( {\tan A} \right)}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}+\frac{{\left( {\sec A+1} \right)\left( {\sec A+1} \right)}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{{{{\tan }}^{2}}A+{{{\sec }}^{2}}A+2\sec A+1}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{{{{\sec }}^{2}}A+{{{\sec }}^{2}}A+2\sec A}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{2{{{\sec }}^{2}}A+2\sec A}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{2\sec A\cancel{{\left( {\sec A+1} \right)}}}}{{\cancel{{\left( {\sec A+1} \right)}}\left( {\tan A} \right)}}&=\\\frac{{2\left( {\displaystyle \frac{1}{{\cos A}}} \right)}}{{\displaystyle \frac{{\sin A}}{{\cos A}}}}=\frac{2}{{\sin A}}\,\,\,\,\,\,\,\,\,\,\surd \end{align}$

• Find common denominators for the fractions, and combine them, since we only have one term on the right.
• Turn $ {{\tan }^{2}}A+1$ into $ {{\sec }^{2}}A$ and combine with other $ {{\sec }^{2}}A$.
• Factor the numerator and simplify.
• Use Reciprocal and Quotient Identities to simplify.

$ \require{cancel} \displaystyle \begin{align}\frac{{1-\displaystyle \frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\displaystyle \frac{{{{{\cos }}^{2}}x}}{{{{{\cos }}^{2}}x}}-\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\displaystyle \frac{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\cancel{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}}}{{{{{\cos }}^{2}}x}}\cdot \displaystyle \frac{1}{{\cancel{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}}}&=\\\frac{1}{{{{{\cos }}^{2}}x}}&={{\sec }^{2}}x\,\,\,\,\,\,\,\surd \end{align}$

• Even though it looks like we should use the $ {{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$ identity, we don’t have a “$ 2x$” angle anywhere else, so it probably wouldn’t help.
• Turn everything into sin and cos and get common denominator to subtract terms in numerator.
• Divide fraction and use Reciprocal Identity.

$ \displaystyle \begin{align}\frac{{{{{\sec }}^{2}}\theta }}{{1-{{{\tan }}^{2}}\theta }}&=\\\frac{{\displaystyle \displaystyle \displaystyle \frac{1}{{{{{\cos }}^{2}}\theta }}}}{{\displaystyle \frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}-\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}&=\\\displaystyle \frac{{\displaystyle \displaystyle \frac{1}{{{{{\cos }}^{2}}\theta }}}}{{\displaystyle \frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}&=\\\displaystyle \frac{1}{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}&=\\\displaystyle \frac{1}{{\cos \left( {2\theta } \right)}}&=\sec \left( {2\theta } \right)\,\,\,\,\,\,\surd \end{align}$

• Use a Pythagorean Identity, and then turn everything into sin and cos.
• Get common denominator to subtract terms in denominator.
• Divide fraction and use Double Angle and Reciprocal Identities.