Question

Match the following:
Given $\sin A=\frac{2}{3}$ and $\sin B=\frac{1}{4}$

• A. $\frac{2\sqrt{15}+\sqrt{5}}{12}$
• B. $\frac{2\sqrt{15}-\sqrt{5}}{2+5\sqrt{3}}$
• C. $\frac{5\sqrt{3}+2}{12}$

Answer 1

This is a direct application of $\sin (A+B),\sin (A-B),\cos (A+B)$ etc., or trigonometric ratios of compound angles.

Before we calculate them, we need $\cos A,\cos B,\tan A$ and $\tan B$ to easily calculate the ratios asked. For that, we will construct proper triangles and find the ratios.

$\sin A=\frac{2}{3}$ $\sin B=\frac{1}{4}$

$\cos A=\frac{\sqrt{5}}{3}$ $\cos A=\frac{\sqrt{15}}{4}$

$\tan A=\frac{2}{\sqrt{5}}$ $\tan B=\frac{1}{\sqrt{15}}$

(1) $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$=\frac{2}{3}\times \frac{\sqrt{15}}{4}+\frac{\sqrt{5}}{3}\times \frac{1}{4}$

$=\frac{2\sqrt{15}+\sqrt{5}}{12}$

(2) $\cos (A-B)=\cos A\cos B+\sin A\sin B$

$=\frac{\sqrt{5}}{3}\times \frac{\sqrt{15}}{4}+\frac{2}{3}\times \frac{1}{4}$

$=\frac{5\sqrt{3}+2}{12}$

(3) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$=\frac{\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{15}}}{(1+\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{15}})}$

$=\frac{2\sqrt{15}-\sqrt{5}}{\sqrt{5}\times \sqrt{15}+2}$

$=\frac{2\sqrt{15}-\sqrt{5}}{5\sqrt{3}+2}$