1+{ }^{n} C_{1} cos theta+{ }^{n} C_{2} cos 2 | vedclass

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(A) The given expression is the real part of the binomial expansion of $(1+e^{i	heta})^n$.Let $S = 1+{ }^{n} C_{1} \cos 	heta+{ }^{n} C_{2} \cos 2 \

Vedclass · Accepted Answer

$\left(2 \cos \frac{	heta}{2}ight)^{n} \cos \frac{n 	heta}{2}$

Vedclass · Answer

(A) The given expression is the real part of the binomial expansion of $(1+e^{i	heta})^n$.Let $S = 1+{ }^{n} C_{1} \cos 	heta+{ }^{n} C_{2} \cos 2 	heta+\ldots+{ }^{n} C_{n} \cos n 	heta$.Then $S = \operatorname{Re}\left(\sum_{k=0}^{n} { }^{n} C_{k} e^{ik	heta}ight) = \operatorname{Re}((1+e^{i	heta})^n)$.Using $1+e^{i	heta} = 1+\cos 	heta + i \sin 	heta = 2 \cos^2 \frac{	heta}{2} + i 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$.$1+e^{i	heta} = 2 \cos \frac{	heta}{2} \left(\cos \frac{	heta}{2} + i \sin \frac{	heta}{2}ight) = 2 \cos \frac{	heta}{2} e^{i	heta/2}$.Thus, $(1+e^{i	heta})^n = (2 \cos \frac{	heta}{2})^n e^{in	heta/2} = (2 \cos \frac{	heta}{2})^n \left(\cos \frac{n	heta}{2} + i \sin \frac{n	heta}{2}ight)$.Taking the real part, we get $S = (2 \cos \frac{	heta}{2})^n \cos \frac{n	heta}{2}$.

$1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$ equals

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