Trig Review - Identities & Sum/Difference Formulas

 $1+\tan^2\theta=\sec^2\theta$  

 $1+\cot^2\theta=\sec^2\theta$  

 $1+\tan^2\theta=\csc^2\theta$  

 $1+\cot^2\theta=\csc^2\theta$  

 $\sin\left(-x\right)=\sin\left(x\right)$  

 $\cos\left(-x\right)=\cos\left(x\right)$  

 $\tan\left(-x\right)=-\tan\left(x\right)$  

 $\sin\left(-x\right)=-\sin\left(x\right)$  

 $\cos\left(-x\right)=-\cos\left(x\right)$  

$\sin^2\theta=\cos^2\theta-1$

$\cos^2\theta=1-\sin^2\theta$

$\cos^2\theta=1+\sin^2\theta$

$\sin^2\theta=1-\cos^2\theta$

Multiply the numerator and denominator by the conjugate  $\left(1-\cos x\right)$  

Square the numerator and denominator

Use a Pythagorean Identity for the denominator to change the  $\left(1+\cos x\right)$  to  $\left(\sin^2x\ +\cos^2x+\cos x\right)$  

Split up the denominator

Split up the  $\left(\sin^4x\right)$  as  $\left(\sin^2x\right)\left(\sin^2x\right)$  to set up a Pythagorean Identity substitution 

Split up the  $\left(\cos^4x\right)$  as  $\left(\cos^2x\right)\left(\cos^2x\right)$  to set up a Pythagorean Identity substitution 

Factor the left side using the difference of squares

Use a Pythagorean Identity on the right side to change  $\left(2\cos^2x-1\right)$  to  $\left(2\left(1-\sin^2x\right)-1\right)$  

 $\frac{\sqrt{2}+\sqrt{6}}{4}$  

 $\frac{\sqrt{2}-\sqrt{3}}{2}$  

 $\frac{\sqrt{6}-\sqrt{2}}{4}$  

 $\frac{\sqrt{2}-\sqrt{6}}{4}$  

 $\sin150\degree=\frac{\sqrt{3}}{2}$  

 $\sin150\degree=\frac{1}{2}$  

 $\sin\left(-76\degree\right)\approx-0.9703$  

 $\sin\left(-76\degree\right)\approx0.2419$