Trigonometric Functions — MATHS STD 11 Science — Question

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

Solution:

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

$\sin^275^\circ-\sin^215^\circ$

$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.

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,

$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$

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

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

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