Chapter 14.6, Problem 3LC

Solution

To find: the exact value of the given expression.

  $\cos {\left(-315\right)}^{\circ }=\frac{\sqrt{2}}{2}$ .

Given:

  $\cos {\left(-315\right)}^{\circ }$ .

Concept used:

Angle Sum identity:

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

Negative Angle identity:

  $\cos \left(-x\right)=\cos x$

Calculation:

Consider,

  $\cos {\left(-315\right)}^{\circ }$

By using the Negative Angle identity, it can be written as

  $\cos {\left(-315\right)}^{\circ }=\cos {315}^{\circ }$

Now it can be rewritten as,

  $\begin{array}{c}\cos {\left(-315\right)}^{\circ }=\cos {315}^{\circ }\\ =\cos \left({270}^{\circ }+{45}^{\circ }\right)\end{array}$

Now by using the Angle Sum identity, it follows that

  $\begin{array}{c}\cos {\left(-315\right)}^{\circ }=\cos \left({270}^{\circ }+{45}^{\circ }\right)\\ =\cos {270}^{\circ }\cos {45}^{\circ }-\sin {270}^{\circ }\sin {45}^{\circ }\\ =0\left(\frac{\sqrt{2}}{2}\right)-\left(-1\right)\left(\frac{\sqrt{2}}{2}\right)\\ =\frac{\sqrt{2}}{2}\end{array}$