Question

If $a=cis2\alpha ,b=cis2\beta$, then $\cos (\alpha -\beta )$ is
where $cis\theta =\cos \theta +i\sin \theta$

• A. $\frac{a+b}{2\sqrt{ab}}$
• B. $\frac{a-b}{\sqrt{ab}}$
• C. $\frac{a-b}{4\sqrt{ab}}$
• D. $\frac{ab}{2\sqrt{a-b}}$

Answer 1

The correct option is A $\frac{a+b}{2\sqrt{ab}}$

$a=cis2\alpha =\cos 2\alpha +i\sin 2\alpha$
$\Rightarrow a=e^{i2\alpha }$
$a={\left(e^{i\alpha }\right)}^2=(cis\alpha {)}^2$
$\Rightarrow \sqrt{a}=cis\alpha =e^{i\alpha }$
Similarly,
$\sqrt{b}=cis\beta =e^{i\beta }$

Hence,
$\frac{\sqrt{a}}{\sqrt{b}}=e^{i(\alpha -\beta )}=cis(\alpha -\beta )\frac{\sqrt{b}}{\sqrt{a}}=e^{i(\beta -\alpha )}=cis(\beta -\alpha )\therefore 2\cos (\alpha -\beta )=\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{a}}\Rightarrow \cos (\alpha -\beta )=\frac{a+b}{2\sqrt{ab}}$