Q. If $1 + \cos{x = k}$, where $x$ is acute, then $\sin\frac{x}{2}$ is
A) $\sqrt{\frac{1 - k}{2}}$
B) $\sqrt{2 - k}$
C) $\sqrt{\frac{2 + k}{2}}$
D) $\sqrt{\frac{2 - k}{2}}$
Explanation
(d)
$1 + \cos{x = k}$
$\Rightarrow \ \ 1 + 1 - 2\sin^{2}{\frac{x}{2} = k}$
$\Rightarrow \ \ 1 - \sin^{2}{\frac{x}{2} = \frac{k}{2}}$
$\Rightarrow \ \sin^{2}{\frac{x}{2} = 1 - \frac{k}{2}}$
$\Rightarrow \ \sin{\frac{x}{2} = \sqrt{\frac{2 - k}{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.