# What is the solution for 2-cos^2(35)-cos^2(55)=? with trigonometry

## I know that for example sin(y) is sin(90-x) = cos(x) if 90-x=y

Feb 26, 2018

$y = 2 - {\cos}^{2} \left({35}^{\circ}\right) - {\cos}^{2} \left({55}^{\circ}\right) = 1$

#### Explanation:

We want to evalutae

$y = 2 - {\cos}^{2} \left({35}^{\circ}\right) - {\cos}^{2} \left({55}^{\circ}\right)$

We will use the trigonometric identities

• ${\cos}^{2} \left(x\right) = \frac{1}{2} \left(1 + \cos \left(2 x\right)\right)$

• $\cos \left(x\right) = - \cos \left(180 - x\right)$

Thus

$y = 2 - \left(\frac{1}{2} \left(1 + \cos \left({70}^{\circ}\right)\right)\right) - \left(\frac{1}{2} \left(1 + \cos \left({110}^{\circ}\right)\right)\right)$

$= 2 - \left(\frac{1}{2} + \frac{1}{2} \cos \left({70}^{\circ}\right)\right) - \left(\frac{1}{2} + \frac{1}{2} \cos \left({110}^{\circ}\right)\right)$

$= 2 - \frac{1}{2} - \frac{1}{2} \cos \left({70}^{\circ}\right) - \frac{1}{2} - \frac{1}{2} \cos \left({110}^{\circ}\right)$

$= 1 - \frac{1}{2} \cos \left({70}^{\circ}\right) - \frac{1}{2} \cos \left({110}^{\circ}\right)$

Use $\cos \left({110}^{\circ}\right) = - \cos \left({180}^{\circ} - {110}^{\circ}\right) = - \cos \left({70}^{\circ}\right)$

$y = 1 - \frac{1}{2} \cos \left({70}^{\circ}\right) - \frac{1}{2} \left(- \cos \left({70}^{\circ}\right)\right)$

$= 1 - \frac{1}{2} \cos \left({70}^{\circ}\right) + \frac{1}{2} \cos \left({70}^{\circ}\right)$

$= 1$