Chapter 14.4, Problem 40HP

To determine

To Find:Derived the formula of $\sin 2\theta$ and $\cos 2\theta$

Answer to Problem 40HP

  $\begin{array}{l}\sin 2\theta =2\sin \theta \cos \theta \\ \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \end{array}$

Explanation of Solution

Given information:

We know that, used formula of $\sin (A+B)and\cos (A+B)$

Formula used:

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

Calculation:

  $\begin{array}{l}\sin (2\theta )=\sin (\theta +\theta )=\sin \theta \cos \theta +\cos \theta \sin \theta \\ \sin 2\theta =2\sin \theta \cos \theta \\ \because \cos (2\theta )=\cos (\theta +\theta )\\ =\cos \theta \cos \theta +\sin \theta \sin \theta \\ ={\cos }^2\theta -{\sin }^2\theta \end{array}$