Chapter 7, Problem 73RE

To determine

To find:the exact value of the given expression.

Answer to Problem 73RE

  $\sin {165}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{2}$ .

Explanation of Solution

Given:

The given expression is $\sin {165}^{\circ }$ .

Calculation:

As the given expressionis $\sin {165}^{\circ }=\sin \left({135}^{\circ }+{30}^{\circ }\right)$ .

Now, using the identity $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$ :

  $\begin{array}{l}\sin {165}^{\circ }=\sin \left({135}^{\circ }+{30}^{\circ }\right)\\ =\sin {135}^{\circ }\cos {30}^{\circ }+\cos {135}^{\circ }\sin {30}^{\circ }\\ =\sin {135}^{\circ }\cos {30}^{\circ }+\cos {135}^{\circ }\sin {30}^{\circ }\\ =\sin {\left(180-45\right)}^{\circ }\cos {30}^{\circ }+\cos {\left(180-45\right)}^{\circ }\sin {30}^{\circ }\\ \left\{\text{as}\sin \left(180-\theta \right)=\sin \theta \text{ and cos}\left(180-\theta \right)=-\cos \theta \right\}\\ =\sin {45}^{\circ }\cos {30}^{\circ }+\left(-\cos {\left(45\right)}^{\circ }\sin {30}^{\circ }\right)\\ \left\{\text{substitute the value of sin}\theta \text{ and }\cos \theta \text{ from trigonometry table}\right\}\\ =\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times \frac{1}{2}\\ =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{2}\end{array}$

Hence, the value of $\sin {165}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{2}$ .