The real part of frac{(cos a+i sin a)^6}{(sin | vedclass

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

Question

(D) Given expression: $Z = \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ Using De Moivre's theorem,the numerator is $(\cos a+i \sin a)^6 = \cos(6a)

Vedclass · Accepted Answer

$\cos (6 a+8 b)$

Vedclass · Answer

(D) Given expression: $Z = \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ Using De Moivre's theorem,the numerator is $(\cos a+i \sin a)^6 = \cos(6a) + i \sin(6a) = e^{i6a}$. For the denominator,note that $\sin b + i \cos b = i(\cos b - i \sin b) = i e^{-ib}$. Thus,$(\sin b + i \cos b)^8 = i^8 (e^{-ib})^8 = 1 \cdot e^{-i8b} = e^{-i8b}$. Substituting these back: $Z = \frac{e^{i6a}}{e^{-i8b}} = e^{i(6a+8b)}$. Expanding using Euler's formula: $Z = \cos(6a+8b) + i \sin(6a+8b)$. The real part is $\cos(6a+8b)$.

The real part of $\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ is

Similar Questions

If $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n^{\text{th}}$ roots of unity,then $\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j =$

$(27)^{1/3} = $

Let $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$,then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

$(\sin \theta - i \cos \theta)^3$ is equal to

If $z+\frac{1}{z}=1$,then $\frac{\left(z^{20}+1\right)\left(z^{40}+1\right)\left(z^{60}+1\right)}{z^{60}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module