Chapter 13.9, Problem 3ST

To determine

To find: The value of trigonometric function.

Answer to Problem 3ST

The exact value of the given expression is $\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

The given information is $\cos {310}^{\circ }\cos {260}^{\circ }-\sin {310}^{\circ }\sin {260}^{\circ }$ .

Calculation:

Consider,

  $\cos {310}^{\circ }\cos {260}^{\circ }-\sin {310}^{\circ }\sin {260}^{\circ }$ .

The given function is of the form $\cos \alpha \cos \beta -\sin \alpha \sin \beta$ .

We know that $\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos \left(\alpha +\beta \right)$ .

Replace with $\alpha =310°$ and with $\beta =260°$ in the equation.

  $\cos {310}^{\circ }\cos {260}^{\circ }-\sin {310}^{\circ }\sin {260}^{\circ }=\cos \left({310}^{\circ }+{260}^{\circ }\right)$

Simplify,

  $\begin{array}{l}\cos {310}^{\circ }\cos {260}^{\circ }-\sin {310}^{\circ }\sin {260}^{\circ }=\cos \left({570}^{\circ }\right)\\ \cos {310}^{\circ }\cos {260}^{\circ }-\sin {310}^{\circ }\sin {260}^{\circ }=\frac{\sqrt{3}}{2}\end{array}$

Therefore, the exact value of the given expression is $\frac{\sqrt{3}}{2}$ .