Question

The value of ${\sin }^2\frac{5\mathrm{\pi }}{12}-{\sin }^2\frac{\mathrm{\pi }}{12}$ is
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) 1
(d) 0

Answer 1

(b) $\frac{\sqrt{3}}{2}$
$\frac{5\pi }{12}=75°,\frac{\pi }{12}=15°$

$$\begin{gathered} {\sin }^{2}75°-{\sin }^{2}15° \\ ={\sin }^{2}75°-{\cos }^{2}75°\left[\sin \left(90°-\theta \right)=\cos \theta \right] \\ \mathrm{Now},\sin 75°=\sin (45°+30°) \\ =\sin 45°\cos 30°+\cos 45°\sin 30° \\ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2} \\ =\frac{\sqrt{3}+1}{2\sqrt{2}} \\ \cos 75°=\cos (45°+30°) \\ =\cos 45°\cos 30°-\sin 45°\sin 30° \\ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2} \\ =\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \mathrm{Hence}, \\ {\sin }^{2}75°-{\cos }^{2}75°={\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^2-{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2 \\ =\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8} \\ =\frac{4\sqrt{3}}{8} \\ =\frac{\sqrt{3}}{2} \end{gathered}$$