### Home > PC > Chapter 11 > Lesson 11.2.2 > Problem11-89

11-89.

Show $\frac { r _ { 1 } ( \operatorname { cos } a + i \operatorname { sin } a ) } { r _ { 2 } ( \operatorname { cos } b + i \operatorname { sin } b ) } = \frac { r _ { 1 } } { r _ { 2 } } ( \operatorname { cos } ( a - b ) + i \operatorname { sin } ( a - b ) )$. (Hint: Multiply the numerator and denominator by the conjugate of the denominator.)

$\frac{\textit{r}_1(\cos \textit{a}+\textit{i}\sin \textit{a})}{\textit{r}_2(\cos \textit{b}+\textit{i}\sin \textit{b})}·\frac{(\cos \textit{b}-\textit{i}\sin \textit{b})}{(\cos \textit{b}-\textit{i}\sin \textit{b})}$

$\frac{r_1(\cos a \cos b - i \cos a \sin b + i \sin \sin a \cos b - i^2 \sin a \sin b)} {r_2(\cos b \cos b - i \cos b \sin b + i \sin \sin b \cos b - i^2 \sin b \sin b)}$

$\frac{r_1((\cos a \cos b + \sin a \sin b) + i (\sin a \cos b - \cos a \sin b)} {r_2(\cos b i \sin^2 \sin b)}$

Use your trig identities to simplify.