Chapter 7, Problem 135RE

To determine

To find:the exact solution of the given equation by using a half-angle formula. Then use a difference formula, show that the answers are same in both the cases.

Answer to Problem 135RE

  $x=0$ .

Explanation of Solution

Given:

The given equation is $\sin {15}^{\circ }$ .

Calculation:

As the given equationis $\sin {15}^{\circ }$

By half angle formula:

The equation can be simplified as:

  $\begin{array}{l}\sin {15}^{\circ }=\sin {\left(\frac{30}{2}\right)}^{\circ }\left\{\because \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}\right\}\\ \Rightarrow \sin {15}^{\circ }=\pm \sqrt{\frac{1-\cos 30}{2}}\\ \sin {15}^{\circ }=\pm \sqrt{\frac{1-\sqrt{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2}}\\ \sin {15}^{\circ }=\pm \sqrt{\frac{1-\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}}\\ \sin {15}^{\circ }=\pm \sqrt{\frac{2-\sqrt{3}}{2\cdot 2}}\\ \sin {15}^{\circ }=\pm \sqrt{\frac{2-\sqrt{3}}{4}}\\ \sin {15}^{\circ }=\pm \frac{\sqrt{2-\sqrt{3}}}{2}\end{array}$

  $\begin{array}{l}\text{on simplifying,}\\ \sin {15}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$

By difference formula:

The equation can be simplified as:

  $\begin{array}{l}\sin {15}^{\circ }=\sin {\left(45-30\right)}^{\circ }\\ \left\{\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B\right\}\\ \Rightarrow \sin {15}^{\circ }=\sin 45\cos 30-\cos 45\sin 30\\ \Rightarrow \sin {15}^{\circ }=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2}\\ \Rightarrow \sin {15}^{\circ }=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}-1}{2}\right)\\ \sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}\\ \text{on simplifying,}\\ \sin {15}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$

Hence, the solution of $\sin {15}^{\circ }$ is $\sin {15}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}$ in both the cases.