The value of ^25 12 - ^2 12 is: - Vidyadip

MCQ

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

A
$\frac12$
✓
$\frac{\sqrt{3}}{2}$
C
$1$
D
$0$

✓

Answer

Correct option: B.

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

$\frac{5\pi}{12}=75^\circ,\frac{\pi}{12}=15^\circ$
$\sin^275^\circ-\sin^215^\circ$
$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.
Now, $\sin75^\circ=\sin(45^\circ+30^\circ)$
$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\cos75^\circ=\cos(45^\circ+30^\circ)$
$=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Hence,
$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$
$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8}$
$=\frac{4\sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$

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