Chapter 13.7, Problem 33WE

To determine

To Prove: The given trigonometric function.

Explanation of Solution

Given:

The given function is $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ .

Calculation:

Consider the given function is,

  $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

Rewrite the above equation as,

  $\begin{array}{l}\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \sin \left[\alpha +\left(-\beta \right)\right]=\sin \alpha \cos \left(-\beta \right)+\cos \alpha \sin \left(-\beta \right)\end{array}$

In the above function $\cos \left(-\beta \right)=\cos \beta$ and $\sin \left(-\beta \right)=-\sin \beta$ .

Then,

  $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

Hence, proved.