The value of sin

Question:
The value of $\sin ^{2} \frac{5 \pi}{12}-\sin ^{2} \frac{\pi}{12}$ is

(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

(c) 1

(d) 0

Solution:

(b) $\frac{\sqrt{3}}{2}$

$\frac{5 \pi}{12}=75^{\circ}, \frac{\pi}{12}=15^{\circ}$

$\sin ^{2} 75^{\circ}-\sin ^{2} 15^{\circ}$

$=\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ} \quad\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$

Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$

$=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

$\cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right)$

$=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

Hence,

$\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ}=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^{2}-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2}$

$=\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{8}$

$=\frac{4 \sqrt{3}}{8}$

$=\frac{\sqrt{3}}{2}$