The value of frac{(cos theta + i sin theta)^4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given expression: $E = \frac{(\cos 	heta + i \sin 	heta)^4}{(\sin 	heta + i \cos 	heta)^5}$ Using De Moivre's Theorem,the numerator is $(\cos

Vedclass · Accepted Answer

$\sin 9 	heta - i \cos 9 	heta$

Vedclass · Answer

(D) Given expression: $E = \frac{(\cos 	heta + i \sin 	heta)^4}{(\sin 	heta + i \cos 	heta)^5}$ Using De Moivre's Theorem,the numerator is $(\cos 	heta + i \sin 	heta)^4 = \cos 4 	heta + i \sin 4 	heta$. For the denominator,rewrite $\sin 	heta + i \cos 	heta$ as $i(\cos 	heta - i \sin 	heta) = i(\cos(-	heta) + i \sin(-	heta))$. So,$(\sin 	heta + i \cos 	heta)^5 = i^5 (\cos(-	heta) + i \sin(-	heta))^5 = i (\cos(-5 	heta) + i \sin(-5 	heta)) = i(\cos 5 	heta - i \sin 5 	heta)$. Thus,$E = \frac{\cos 4 	heta + i \sin 4 	heta}{i(\cos 5 	heta - i \sin 5 	heta)} = \frac{1}{i} \cdot \frac{\cos 4 	heta + i \sin 4 	heta}{\cos 5 	heta - i \sin 5 	heta}$. Using the property $\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = \cos(\alpha - \beta) + i \sin(\alpha - \beta)$,we have: $E = -i \cdot \frac{\cos 4 	heta + i \sin 4 	heta}{\cos(-5 	heta) + i \sin(-5 	heta)} = -i (\cos(4 	heta - (-5 	heta)) + i \sin(4 	heta - (-5 	heta)))$ $E = -i (\cos 9 	heta + i \sin 9 	heta) = -i \cos 9 	heta - i^2 \sin 9 	heta = \sin 9 	heta - i \cos 9 	heta$.

The value of $\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^5}$ is equal to,where $i = \sqrt{-1}$:

Similar Questions

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

If $n, K \in N$ such that $n \neq 3K$,then $(\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = $

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0$. The equation whose roots are $\alpha^{19}$ and $\beta^7$ is

The cube roots of unity when represented on the Argand plane form the vertices of an

Two values of $(-8-8 \sqrt{3} i)^{1/4}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module