Question

If sin $\alpha =\frac{1}{2}$ and $\alpha$ is acute, then (3 cos $\alpha -4co{s}^3\alpha )$ is equal to:

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{6}$
• D. $-1$

Answer 1

The correct option is A $0$

$\sin \alpha =\frac{1}{2}$

$\Rightarrow \alpha =30°$

$\Rightarrow 3\cos \alpha -4\cos \alpha =-\cos 3\alpha$

$=-\cos 90°$

$=0$

Hence, the answer is $0.$