Chapter 13.3, Problem 1AGP

To determine

To find : The given equation is $\sin 15°.$

Answer to Problem 1AGP

  $\frac{\sqrt{3}-1}{2\sqrt{2}}$

Explanation of Solution

Given information :

We know that, $\sin 15°$

Calculation :

  $\begin{array}{l}\sin 15°=\sin (45°-30°)[\because \sin (A-B)=\sin A\cos B-\sin B\cos A]\\ =\sin 45°\cdot \cos 30°-\sin 30°\cos 45°\\ =\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\\ =\frac{\sqrt{3}-1}{2\sqrt{2}}\end{array}$