Question

The value of \(\frac{(\cos \theta+i \sin \theta)^4}{(\sin \theta+i \cos \theta)^5}=\) where \(\mathrm{i}=\sqrt{-1}\)

• A. \(\mathrm{cos}\theta -i\mathrm{sin}\theta\)
• B. \(\mathrm{cos}9\theta -\mathrm{isin}9\theta\)
• C. \(\mathrm{sin}\theta -i\mathrm{cos}\theta\)
• D. \(\mathrm{sin}9\theta -\mathrm{icos}9\theta\)

Hint:

Write sin t + i cos t as i times (cos t - i sin t) and apply De Moivre's theorem.

Answer 1

The Correct Option is D

Step 1: Exponential form
Numerator $=e^{4i\theta}$, denominator $=\left(ie^{-i\theta}\right)^5=i\,e^{-5i\theta}$.

Step 2: Divide
$\frac{e^{9i\theta}}{i}=-i\,e^{9i\theta}$.

Step 3: Expand
$-i(\cos9\theta+i\sin9\theta)=\sin9\theta-i\cos9\theta$, the last option.

Final Answer:
The result is $\sin9\theta-i\cos9\theta$. \[ \boxed{\text{(D) }\sin9\theta-i\cos9\theta} \]