Question

$1+{}^n{C}_1\cos \theta +{}^n{C}_2\cos 2\theta +........+{}^n{C}_n\cos n\theta$ equals

• A. ${\left(\begin{array}{c}2\cos \frac{\theta }{2}\end{array}\right)}^n\cos \frac{n\theta }{2}$
• B. $2{\cos }^2\frac{n\theta }{2}$
• C. $2{\cos }^{2n}\frac{\theta }{2}$
• D. ${\left(\begin{array}{c}2{\cos }^2\frac{\theta }{2}\end{array}\right)}^n$

Answer 1

The correct option is A ${\left(\begin{array}{c}2\cos \frac{\theta }{2}\end{array}\right)}^n\cos \frac{n\theta }{2}$

We know that, $(1+x){}^n=1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+....+{}^n{C}_{n}x{}^n$

Substitute $x=\cos \theta +i\sin \theta$

$(1+\cos \theta +i\sin \theta ){}^n=1+{}^n{C}_1(\cos \theta +i\sin \theta )+{}^n{C}_2(\cos \theta +i\sin \theta {)}^2+....+{}^n{C}_n(\cos \theta +i\sin \theta ){}^n$

Now use Demoivre's theorem to RHS,

$(2{\cos }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}){}^n=$$1+{}^n{C}_1(\cos \theta +i\sin \theta )+{}^n{C}_2(\cos 2\theta +i\sin 2\theta )+....+{}^n{C}_n(\cos n\theta +i\sin n\theta )$

$\Rightarrow \left(2\cos \frac{\theta }{2}\right)(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}){}^n=$$(1+{}^n{C}_1\cos \theta +{}^n{C}_2\cos 2\theta +...+{}^n{C}_n\cos n\theta )+i({}^n{C}_1\sin \theta +{}^n{C}_2\sin 2\theta +...+{}^n{C}_n\sin n\theta )$

$\Rightarrow \left(2\cos \frac{\theta }{2}\right)(\cos \frac{n\theta }{2}+i\sin \frac{n\theta }{2})=$$(1+{}^n{C}_1\cos \theta +{}^n{C}_2\cos 2\theta +...+{}^n{C}_n\cos n\theta )+i({}^n{C}_1\sin \theta +{}^n{C}_2\sin 2\theta +...+{}^n{C}_n\sin n\theta )$

Now equate the real parts from both sides,

$1+{}^n{C}_1\cos \theta +{}^n{C}_2\cos 2\theta +........+{}^n{C}_n\cos n\theta ={\left(2\cos \frac{\theta }{2}\right)}^n\left(\cos \frac{n\theta }{2}\right)$