Question

$\sin (\alpha -\beta )=\sin \alpha \cos \beta \_\_\_\cos \alpha \sin \beta$

Short answer

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

Step-by-step solution

Step 1. Given information

We have been given $\sin (\alpha -\beta )=\sin \alpha \cos \beta \_\_\_\cos \alpha \sin \beta$.

We have to fill in the blank.

Step 2. Use the sum formula for the sine function

Sum of the sine function is $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$.

Substitute $\beta$with $-\beta$.

$$\begin{gathered} \sin (\alpha +(-\beta \left)\right)=\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta ) \\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta \left(\cos \right(-\mathrm{\beta })=\mathrm{cos\beta }\mathrm{and}\sin (-\mathrm{\beta })=-\mathrm{sin\beta }) \end{gathered}$$

Therefore,$\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$