Question

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

Short answer

$\cos \left(2\theta \right)={(\cos \theta )}^2-\underset{⏝}{{(\sin \theta )}^2}=\underset{⏝}{{2(\cos \theta )}^2}-1=1-\underset{⏝}{{2(\sin \theta )}^2}$.

Step-by-step solution

Step 1. Given Data

The given incomplete equation is

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

Step 2. Explanation

Use Sum formula of the cosine function

$$\begin{gathered} \cos (\theta +\theta )=(\cos \theta )(\cos \theta )-(\sin \theta )(\sin \theta ) \\ \cos \left(2\theta \right)={(\cos \theta )}^2-{(\sin \theta )}^2 \\ \cos \left(2\theta \right)=1-{(\sin \theta )}^2-{(\sin \theta )}^2 \\ \cos \left(2\theta \right)=1-{2(\sin \theta )}^2 \end{gathered}$$

and

$$\begin{gathered} \cos \left(2\theta \right)={(\cos \theta )}^2-\left(1-{(\cos \theta )}^2\right) \\ \cos \left(2\theta \right)=2{(\cos \theta )}^2-1 \end{gathered}$$

so$\cos \left(2\theta \right)={(\cos \theta )}^2-\underset{⏝}{{(\sin \theta )}^2}=\underset{⏝}{{2(\cos \theta )}^2}-1=1-\underset{⏝}{{2(\sin \theta )}^2}$