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Question: Find the value of \(1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + .......{ + ^n}{C_n}\cos n\th...

Find the value of 1+nC1cos⁡θ+nC2cos⁡2θ+.......+nCncos⁡nθ1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + .......{ + ^n}{C_n}\cos n\theta
A. (2cos⁡θ2)ncos⁡nθ2{\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos \dfrac{{n\theta }}{2}
B. 2cos⁡2nθ22{\cos ^2}\dfrac{{n\theta }}{2}
C. 2cos⁡2nθ22{\cos ^{2n}}\dfrac{\theta }{2}
D. (2cos⁡2θ2)n{\left( {2{{\cos }^2}\dfrac{\theta }{2}} \right)^n}

Explanation

Solution

Hint – We will use binomial theorem and de Moivre’s theorem to get the given equation in question and its result, also with the little help of trigonometric identities.

Complete step-by-step answer:
We will solve the above equation by using the binomial theorem i.e.
(1+x)n=1+nC1x+nC2x2+....+nCnxn{\left( {1 + x} \right)^n} = 1{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + ....{ + ^n}{C_n}{x^n}
Now, we will substitute x=cos⁡θ+ιsin⁡θx = \cos \theta + \iota \sin \theta in above equation, where ι=−1\iota = \sqrt { - 1}
(1+cos⁡θ+ιsin⁡θ)n=1+nC1(cos⁡θ+ιsin⁡θ)+nC2(cos⁡θ+ιsin⁡θ)2+....+nCn(cos⁡θ+ιsin⁡θ)n{\left( {1 + \cos \theta + \iota \sin \theta } \right)^n} = 1{ + ^n}{C_1}\left( {\cos \theta + \iota \sin \theta } \right){ + ^n}{C_2}{\left( {\cos \theta + \iota \sin \theta } \right)^2} + ....{ + ^n}{C_n}{\left( {\cos \theta + \iota \sin \theta } \right)^n} … (1)
Now, we will use de Moivre’s theorem to RHS which states that: (cos⁡θ+ιsin⁡θ)n=cos⁡nθ+ιsin⁡nθ{\left( {\cos \theta + \iota \sin \theta } \right)^n} = \cos n\theta + \iota \sin n\theta
And to LHS we will use trigonometric ‘half angle’ and ‘double angle’ identities i.e.
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta and
cos⁡θ2=1+cos⁡θ2 ⇒2cos⁡2θ2=1+cos⁡θ  \cos \dfrac{\theta }{2} = \sqrt {\dfrac{{1 + \cos \theta }}{2}} \\\ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = 1 + \cos \theta \\\
So, we will get the equation (1) as
(2cos⁡2θ2+2ιsin⁡θ2cos⁡θ2)n=1+nC1(cos⁡θ+ιsin⁡θ)+nC2(cos⁡2θ+ιsin⁡2θ)+.....+nCn(cos⁡nθ+ιsin⁡nθ){\left( {2{{\cos }^2}\dfrac{\theta }{2} + 2\iota \sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right)^n} = 1{ + ^n}{C_1}\left( {\cos \theta + \iota \sin \theta } \right){ + ^n}{C_2}\left( {\cos 2\theta + \iota \sin 2\theta } \right) + .....{ + ^n}{C_n}\left( {\cos n\theta + \iota \sin n\theta } \right)
Now, by simplifying it we will get,
⇒[(2cos⁡θ2)(cos⁡θ2+ιsin⁡θ2)]n=(1+nC1cos⁡θ+nC2cos⁡2θ+...nCncos⁡nθ) \+ι(nC1sin⁡θ+nC2sin⁡2θ+...+nCnsin⁡nθ)  \Rightarrow {\left[ {\left( {2\cos \dfrac{\theta }{2}} \right)\left( {\cos \dfrac{\theta }{2} + \iota \sin \dfrac{\theta }{2}} \right)} \right]^n} = \left( {1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + {{...}^n}{C_n}\cos n\theta } \right) \\\ \+ \iota \left( {^n{C_1}\sin \theta { + ^n}{C_2}\sin 2\theta + ...{ + ^n}{C_n}\sin n\theta } \right) \\\
We will again use de Moivre’s theorem on LHS
⇒(2cos⁡θ2)n(cos⁡nθ2+ιsin⁡nθ2)=(1+nC1cos⁡θ+nC2cos⁡2θ+...nCncos⁡nθ) \+ι(nC1sin⁡θ+nC2sin⁡2θ+...+nCnsin⁡nθ)  \Rightarrow {\left( {2\cos \dfrac{\theta }{2}} \right)^n}\left( {\cos n\dfrac{\theta }{2} + \iota \sin n\dfrac{\theta }{2}} \right) = \left( {1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + {{...}^n}{C_n}\cos n\theta } \right) \\\ \+ \iota \left( {^n{C_1}\sin \theta { + ^n}{C_2}\sin 2\theta + ...{ + ^n}{C_n}\sin n\theta } \right) \\\
Now, equate the real part from both sides.
1+nC1cos⁡θ+nC2cos⁡2θ+...nCncos⁡nθ=(2cos⁡θ2)ncos⁡nθ21{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + {...^n}{C_n}\cos n\theta = {\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos n\dfrac{\theta }{2}
Hence, the answer is (2cos⁡θ2)ncos⁡nθ2{\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos \dfrac{{n\theta }}{2}
Hence, the correct option is A.

Note – The binomial theorem tells us how to expand expressions of the form (a+b)n{\left( {a + b} \right)^n} which we explained above. Whereas, the de Moivre’s theorem gives us a formula for computing powers of complex numbers. It is to be noted that this is the only way possible to solve this question with the help of these theorems and trigonometric identities.