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. $\sqrt{2-2 p^{2}}$

Answer 1

Answer: (D)

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