Question

Find the value of :
$\left(i\right)\sin {75}^{\circ }\left(ii\right)\tan {15}^{\circ }$

Answer 1

$\left(i\right)\sin {75}^{\circ }$
$\sin {75}^{\circ }=\sin ({45}^{\circ }+{30}^{\circ })$
$\{\because \sin (A+B)=\sin A\cos B+\cos A\sin B\}$
$=\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }$
$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\therefore \sin {75}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\left(ii\right)\tan {15}^{\circ }$
$\tan {15}^{\circ }=\tan ({45}^{\circ }-{30}^{\circ })$
$=\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\tan {45}^{\circ }\cdot \tan {30}^{\circ }}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
Do rationalization
Multiply and divide by $\sqrt{3}-1$
$\Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-(1{)}^2}$
$=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}$
$=\frac{2(2-\sqrt{3})}{2}=2-\sqrt{3}$
$\therefore \tan {15}^{\circ }=2-\sqrt{3}$