If $a=\cos \theta+i \sin \theta, b=\cos \phi+i \sin \phi$ prove that

(i) $\cos (\theta+\phi)=\frac{1}{2}\left[a b+\frac{1}{a b}\right]$

(ii) $\sin (\theta-\phi)=\frac{1}{2 i}\left[\frac{a}{b}-\frac{b}{a}\right]$

Solution:

Given, $\mathrm{a}=\cos \theta+i \sin \theta$

& $\quad \mathrm{~b}=\cos \phi+\mathrm{i} \sin \phi$

(i) $\cos (\theta+\phi)=\frac{1}{2}\left[a b+\frac{1}{a b}\right]$ prove that,

Now, $a b=(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)\\ $

$ \Rightarrow a b=\cos (\theta+\phi)+i \sin (\theta+\phi)\\ $

$also, \frac{1}{a b}=(a b)^{-1}=[\cos (\theta+\phi)+i \sin (\theta+\phi)]\\$ $ \Rightarrow \frac{1}{a b}=\cos (\theta+\phi)-i \sin (\theta+\phi) \ldots \ldots . .(2) $ $$ {\therefore(1)+(2)} $$

$ \begin{aligned} \Rightarrow& a b+\frac{1}{a b}=\cos (\theta+\phi)+i \sin (\theta+\phi)+\cos (\theta+\phi)-i \sin (\theta+\phi) \\ \Rightarrow & a b+\frac{1}{a b}=2 \cos (\theta+\phi) \\ \Rightarrow & \cos (\theta+\phi)=\frac{1}{2}\left[a b+\frac{1}{a b}\right]\\ \end{aligned} $

(ii) $\sin (\theta-\phi)=\frac{1}{2 i}\left[\frac{a}{b}-\frac{b}{a}\right]$ prove that,

$ \begin{array}{l} =(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)^{-1} \\\\ =(\cos \theta+i \sin \theta)(\cos \phi-i \sin \phi)\\\\ \end{array} $

$$ \begin{array}{l} \Rightarrow \frac{a}{b}=\cos (\theta-\phi)+i \sin (\theta-\phi)\\\\ a l s o, \frac{b}{a}=\left(\frac{a}{b}\right)^{-1}=[\cos (\theta-\phi)+i \sin (\theta-\phi)]^{-1}\\ \\ \Rightarrow \frac{b}{a}=\cos (\theta-\phi)-i \sin (\theta-\phi)\\\\ {\therefore(3) - (4)}\\ \frac{a}{b}-\frac{b}{a}=\cos (\theta-\phi)+i \sin (\theta-\phi)-\cos (\theta-\phi)+i \sin (\theta-\phi) \\\\ \Rightarrow \frac{a}{b}-\frac{b}{a}=2 i \sin (\theta-\phi)\\\\ \end{array} $$