Chapter 7.6, Problem 1AYU

$\cos \left(2\theta \right)={\cos }^2\theta -\_\_\_\_\_\_=\_\_\_\_\_\_-1=1-\_\_\_\_\_\_\_$

To determine

To fill: $\cos \left(2\theta \right)={\text{ cos}}^2\theta -\_\_\_\_\_\_=\_\_\_\_\_\_-1=1-\_\_\_\_\_\_\_\_$

Answer to Problem 1AYU

$\cos \left(2\theta \right)={\text{ cos}}^2\theta -{\text{ sin}}^2\theta =2{\text{cos}}^2\theta -1=1-2{\text{sin}}^2\theta \_\_\_\_\_\_\_\_$

Explanation of Solution

Given:

$\cos \left(2\theta \right)={\text{ cos}}^2\theta -\_\_\_\_\_\_=\_\_\_\_\_\_-1=1-\_\_\_\_\_\_\_\_$

Formula used:

Sum formula of cos

$\cos \left(a+b\right)=\cos a\cos b-\sin a\sin b$

Calculation:

Let $a=b=\theta$

$\cos \left(\theta +\theta \right)=\cos \left(2\theta \right)=\cos \theta \cos \theta -\sin \theta \sin \theta$

$\cos \left(\theta +\theta \right)=\cos \left(2\theta \right)={\text{ cos}}^2\theta -{\text{ sin}}^2\theta$

By Pythagorean identity

${\text{cos}}^2\theta +{\text{ sin}}^2\theta =1$

${\text{cos}}^2\theta -(1-{\text{ cos}}^2\theta )=2{\text{cos}}^2\theta -1$

$(1-{\text{ sin}}^2\theta )-{\text{ sin}}^2\theta =1-2{\text{sin}}^2\theta$