Chapter 14, Problem 2STP

To determine

To calculate:The Value of $\cos {75}^{\circ }$ using sum difference identity

Answer to Problem 2STP

  $H)\frac{\sqrt{6}-\sqrt{2}}{4}$

Explanation of Solution

Given Information: sum identity can be used to find $\cos {75}^{\circ }$ which is

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

Formula used: $\cos (A+B)=\cos A\cos B-\sin A\sin B$

Calculation:

  $\begin{array}{l}PutA={45}^{\circ }andB={30}^{\circ },\\ \cos ({45}^{\circ }+{30}^{\circ })=\cos {45}^{\circ }\cos {30}^{\circ }-\sin {45}^{\circ }\sin {30}^{\circ }\\ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}\\ =\frac{\sqrt{3}-1}{2\sqrt{2}}\\ =\frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ =\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$