Question

Find the value of $1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + .......{ + ^n}{C_n}\cos n\theta $
A. ${\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos \dfrac{{n\theta }}{2}$
B. $2{\cos ^2}\dfrac{{n\theta }}{2}$
C. $2{\cos ^{2n}}\dfrac{\theta }{2}$
D. ${\left( {2{{\cos }^2}\dfrac{\theta }{2}} \right)^n}$

Answer 1

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.
${\left( {1 + x} \right)^n} = 1{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + ....{ + ^n}{C_n}{x^n}$
 Now, we will substitute $x = \cos \theta + \iota \sin \theta $ in above equation, where $\iota = \sqrt { - 1} $
${\left( {1 + \cos \theta + \iota \sin \theta } \right)^n} = 1{ + ^n}{C_1}\left( {\cos \theta + \iota \sin \theta } \right){ + ^n}{C_2}{\left( {\cos \theta + \iota \sin \theta } \right)^2} + ....{ + ^n}{C_n}{\left( {\cos \theta + \iota \sin \theta } \right)^n}$ … (1)
 Now, we will use de Moivre’s theorem to RHS which states that: ${\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 2\theta = 2\sin \theta \cos \theta $ and
 $
  \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
${\left( {2{{\cos }^2}\dfrac{\theta }{2} + 2\iota \sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right)^n} = 1{ + ^n}{C_1}\left( {\cos \theta + \iota \sin \theta } \right){ + ^n}{C_2}\left( {\cos 2\theta + \iota \sin 2\theta } \right) + .....{ + ^n}{C_n}\left( {\cos n\theta + \iota \sin n\theta } \right)$
Now, by simplifying it we will get,
$
   \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
$
   \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.
$1{ + ^n}{C_1}\cos \theta { + ^n}{C_2}\cos 2\theta + {...^n}{C_n}\cos n\theta = {\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos n\dfrac{\theta }{2}$
Hence, the answer is ${\left( {2\cos \dfrac{\theta }{2}} \right)^n}\cos \dfrac{{n\theta }}{2}$
Hence, the correct option is A.

Note – The binomial theorem tells us how to expand expressions of the form ${\left( {a + b} \right)^n}$ 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.