Question

Express $(sin\theta +icos\theta {)}^5$ in Eulers form.

• A. i.
• B. i.
• C.
• D.

Answer 1

The correct option is D

$sin\theta +icos\theta$ is not a standard form in which De moivre's theorem can be applied.

So, first convert $sin\theta +icos\theta$ to a standard form of $Cos\theta +isin\theta$ by taking i common.

$(sin\theta +icos\theta {)}^5$

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

i $[cos(-\theta )+isin(-\theta ){]}^5$

i $[cos\theta +isin\theta {]}^{-5}$

Euler's form of this can be written as

= ${ie}^{-i5\theta }$