Question

Express $\frac{(cos\theta +isin\theta {)}^4}{(sin\theta +icos\theta {)}^5}$ in a+ib form where i=$\sqrt{-1}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Denominator $(sin\theta +icos\theta {)}^5$ is not a standard form in which De moivre's theorem can be used.

So, first convert $(sin\theta +icos\theta {)}^5$ to a standard form of $cos\theta +isin\theta$ .

This can be done by taking i common.

=$(i{)}^5$ $(cos\theta -isin\theta {)}^5$

=$\frac{{(cos\theta +isin\theta )}^4}{i^5{(cos\theta +isin\theta )}^5}$

$\frac{(cos\theta +isin\theta {)}^4}{i(cos\theta +isin\theta {)}^{-5}}$

=$\frac{(cos\theta +isin\theta {)}^{4+5}}{i}$

=$\frac{(cos\theta +isin\theta {)}^9}{i}$

Now use De moivre's theorem in numerator

$\frac{1}{i}$ $\left((cos9\theta +isin9\theta )\right)$

=$icos9\theta +sin9\theta$ = $sin9\theta -icos9\theta$