Question

1-22 Differentiate.

8. \(y = \sin \theta \cos \theta \)

Short answer

The differentiation of the function \(y = \sin \theta \cos \theta \) is \(y' = \cos 2\theta \).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions and the product rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\\\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\end{aligned}\)

The product rule: If \(f\) and \(g\) are differentiable functions, then\(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

Find the differentiation of the function

Consider the function \(y = \sin \theta \cos \theta \). Differentiate the function w.r.t \(\theta \) by using the derivatives of trigonometric functions and the product rule.

\(\begin{aligned}\frac{{d\left( y \right)}}{{d\theta }} &= \frac{{d\left( {\sin \theta \cos \theta } \right)}}{{d\theta }}\\ &= \sin \theta \frac{d}{{d\theta }}\left( {\cos \theta } \right) + \cos \theta \frac{d}{{d\theta }}\left( {\sin \theta } \right)\\ &= \sin \theta \left( { - \sin \theta } \right) + \cos \theta \left( {\cos \theta } \right)\\ &= - {\sin ^2}\theta + {\cos ^2}\theta \end{aligned}\)

Furthermore, use \({\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta \).

\(\begin{aligned}y' &= - {\sin ^2}\theta + {\cos ^2}\theta \\ &= \cos 2\theta \end{aligned}\)

Thus, the derivative of the function\(y = \sin \theta \cos \theta \) is \(y' = \cos 2\theta \).