Question

If $\sin 27^{\circ }=p$, then the value of $\sqrt{1+\sin 36^{\circ }}$ is

• A. $p$
• B. $\frac{p}{2}$
• C. $\sqrt{1-p^2}$
• D. $\sqrt{2-2p^2}$

Answer 1

Answer: (D)

$\sqrt{1+\sin 36^{\circ }}=\sqrt{{\sin }^{2}18^{\circ }+{\cos }^{2}18^{\circ }+2\sin 18^{\circ }\cos 18^{\circ }}$
$=\sin 18^{\circ }+\cos 18^{\circ }=\sqrt{2}\left[\sin \left(45^{\circ }+18^{\circ }\right)\right]$
$=\sqrt{2}\sin 63^{\circ }=\sqrt{2}\cos 27^{\circ }$
$=\sqrt{2}\sqrt{1-p^2}=\sqrt{2-2p^2}$