The value of left[ frac{1 - cos frac{pi}{10} | vedclass

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

Question

(B) Let $	heta = \frac{\pi}{10}$. The expression is $\left[ \frac{1 - \cos 	heta + i\sin 	heta}{1 - \cos 	heta - i\sin 	heta} ight]^{10}$.Using

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $	heta = \frac{\pi}{10}$. The expression is $\left[ \frac{1 - \cos 	heta + i\sin 	heta}{1 - \cos 	heta - i\sin 	heta} ight]^{10}$.Using the identities $1 - \cos 	heta = 2\sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2\sin \frac{	heta}{2} \cos \frac{	heta}{2}$:Numerator $= 2\sin^2 \frac{	heta}{2} + i(2\sin \frac{	heta}{2} \cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} (\sin \frac{	heta}{2} + i\cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} [i(\cos \frac{	heta}{2} - i\sin \frac{	heta}{2})] = 2i\sin \frac{	heta}{2} e^{-i	heta/2}$.Denominator $= 2\sin^2 \frac{	heta}{2} - i(2\sin \frac{	heta}{2} \cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} (\sin \frac{	heta}{2} - i\cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} [-i(\cos \frac{	heta}{2} + i\sin \frac{	heta}{2})] = -2i\sin \frac{	heta}{2} e^{i	heta/2}$.Dividing the two: $\frac{2i\sin \frac{	heta}{2} e^{-i	heta/2}}{-2i\sin \frac{	heta}{2} e^{i	heta/2}} = -e^{-i	heta} = -(\cos 	heta - i\sin 	heta)$.Raising to the power of $10$: $(-1)^{10} (\cos 	heta - i\sin 	heta)^{10} = 1 \cdot (\cos 10	heta - i\sin 10	heta) = \cos \pi - i\sin \pi = -1 - 0 = -1$.

The value of $\left[ \frac{1 - \cos \frac{\pi}{10} + i\sin \frac{\pi}{10}}{1 - \cos \frac{\pi}{10} - i\sin \frac{\pi}{10}} \right]^{10} = $

Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then the value of $(x+y)^2+(x \omega+y \omega^2)^2+(x \omega^2+y \omega)^2$ is

If $n$ is an integer which leaves a remainder of $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals

$\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} \right) - i \cos \left( \frac{2n\pi}{21} \right) \right] = $

Let $1, \omega$ and $\omega^2$ be the cube roots of unity. What is the value of $(1-\omega+\omega^{-1})^5-2(1+\omega-\omega^{-1})^4$?

If ${z_1}, {z_2}, {z_3}, ......, {z_n}$ are the $n^{th}$ roots of unity,then for $k = 1, 2, ....., n-1$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module