Question

The maximum value of $(\sin \theta \cos \theta {)}^{42}$ is

• A. $\frac{1}{2^{41}}$
• B. $\frac{1}{2^{42}}$
• C. $\frac{1}{2^{43}}$
• D. $\frac{1}{2^{40}}$

Answer 1

The correct option is B $\frac{1}{2^{42}}$

$(\sin \theta \cos \theta {)}^{42}={\left(\frac{2\sin \theta \cos \theta }{2}\right)}^{42}={\left(\frac{\sin 2\theta }{2}\right)}^{42}=\frac{1}{2^{42}}{\sin }^{42}2\theta$
As $\sin 2\theta \in [-1,1]$, so the maximum value of the given expression is $\frac{1}{2^{42}}$