Question

$\left(\cos 252^{\circ}-\sin 126^{\circ}\right)\left(\cos 252^{\circ}+\right.$ $\left.\sin 126^{\circ}\right)\left(\sin ^{2} 126^{\circ}+\sin ^{2} 186^{\circ}+\sin ^{2} 66^{\circ}\right)=$

• A. $\frac{3 \sqrt{5}}{8}$
• B. $\frac{-3 \sqrt{5}}{8}$
• C. $\frac{-3 \sqrt{5}}{4}$
• D. $\frac{3 \sqrt{5}}{4}$

Answer 1

Answer: (B)

$\left(\cos 252^{\circ}-\sin 126^{\circ}\right)\left(\cos 252^{\circ}+\sin 126^{\circ}\right) \left(\sin ^{2} 126^{\circ}+\sin ^{2} 186^{\circ}+\sin ^{2} 66^{\circ}\right)$
$=\left(\cos ^{2} 252^{\circ}-\sin ^{2} 126^{\circ}\right)\left(\sin ^{2} 126^{\circ}+\sin ^{2} 186^{\circ}+\sin ^{2} 66^{\circ}\right)$
$=\left[\frac{1+\cos 504^{\circ}}{2}-\frac{1-\cos 252^{\circ}}{2}\right]\left[\frac{1-\cos 252^{\circ}+1-\cos 372^{\circ}+1-\cos 132^{\circ}}{2}\right]$
$= \frac{1}{4}\left(\cos 504^{\circ}+\cos 252^{\circ}\right)\left[3-\cos 252^{\circ}-2 \cos 252^{\circ} \cos 120^{\circ}\right]$
$=\frac{1}{4}\left(2 \cos 378^{\circ} \cos 126^{\circ}\right)\left[3-\cos 252^{\circ}+\cos 252^{\circ}\right]$
$=\frac{3}{2} \cos 18^{\circ}\left(-\sin 36^{\circ}\right)$
$=-\frac{3}{2} \times \frac{\sqrt{10+2 \sqrt{5}}}{4} \times \frac{\sqrt{10-2 \sqrt{5}}}{4}$
$=-\frac{3}{32} \sqrt{100-20}$
$=\frac{-3}{32} \sqrt{80}=-\frac{3}{32}(4 \sqrt{5})$
$=-\frac{3 \sqrt{5}}{8}$