# Find the value of: (i) sin 75°

Question:

Find the value of:

(i) $\sin 75^{\circ}$

(ii) $\tan 15^{\circ}$

Solution:

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

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

$[\sin (x+y)=\sin x \cos y+\cos x \sin y]$

$=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$

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

(ii) $\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)$

$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \quad\left[\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]$

$=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left(\frac{1}{\sqrt{3}}\right)}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2 \sqrt{3}}{(\sqrt{3})^{2}-(1)^{2}}$

$=\frac{4-2 \sqrt{3}}{3-1}=2-\sqrt{3}$