Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 58

a.) Differentiate the Double-angle Formula $\cos 2x - \cos^2x - \sin^2x$ to obtain the Double-angle Formula for the sine function.


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\cos 2x) =& \frac{d}{dx} (\cos ^2 x) - \frac{d}{dx} (\sin^2 x)
\\
\\
- \sin 2 x \frac{d}{dx} (2x) =& \frac{d}{dx} (\cos x)^2 - \frac{d}{dx} (\sin x) ^2
\\
\\
- \sin 2 x =& 2 \cos x \frac{d}{dx} (\cos x) - 2 \sin x \frac{d}{dx} (\sin x)
\\
\\
- 2 \sin 2x =& 2 \cos x \sin x - 2 \sin x \cos x
\\
\\
- 2 \sin 2x =& -4 \sin x \cos x
\\
\\
\frac{-\cancel{2} \sin 2x}{-\cancel{2}} =& \frac{-4 \sin x \cos x}{-2}
\\
\\
\sin 2x =& 2 \sin x \cos x

\end{aligned}
\end{equation}
$



b.) Differentiate the Addition Formula $\sin (x + a) = \sin x \cos a + \cos x \sin a $ to obtain the Addition Formula for the cosine function.


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} [\sin (x + a)] =& \frac{d}{dx} (\sin x \cos a) + \frac{d}{dx} (\cos x \sin a)
\\
\\
\cos (x + a) \frac{d}{dx} (x + a) =& \left[ (\sin x) \frac{d}{dx} (\cos a) + (\cos a) \frac{d}{dx} (\sin x) \right] + \left[ \cos x \frac{d}{dx} (\sin a) + (\sin a) \frac{d}{dx} (\cos x)\right]
\\
\\
\cos (x + a)(1) =& [(\sin x) (0) + (\cos a)(\cos x)] + [(\cos x) (0) + (\sin a)(- \sin x)]
\\
\\
\cos (x + a) =& \cos a \cos x + 0 + 0 - \sin a \sin x
\\
\\
\cos (x + a) =& \cos x \cos a - \sin x \sin a

\end{aligned}
\end{equation}
$