Question

The value of $(1+\cos \theta +i\sin \theta {)}^n+(1+\cos \theta -i\sin \theta {)}^n=$?

• A. $2^n\cdot {\cos }^n\frac{\theta }{2}\cdot \cos \frac{n\theta }{2}$
• B. $2^{n+1}\cdot {\cos }^n\frac{\theta }{2}\cdot sin\frac{\theta }{2}$
• C. $2^{n+1}\cdot {\cos }^n\frac{\theta }{2}\cdot \cos \frac{n\theta }{2}$
• D. $2^n\cdot {\cos }^n\frac{\theta }{2}\cdot \cos \frac{\theta }{2}$

Answer 1

The correct option is C $2^{n+1}\cdot {\cos }^n\frac{\theta }{2}\cdot \cos \frac{n\theta }{2}$

$1+\cos \theta +i\sin \theta =2{\cos }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})=2\cos \frac{\theta }{2}{e}^{i\frac{\theta }{2}}$

$1+\cos \theta -i\sin \theta =2{\cos }^2\frac{\theta }{2}-i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}-i\sin \frac{\theta }{2})=2\cos \frac{\theta }{2}{e}^{-i\frac{\theta }{2}}$

$(1+\cos \theta +i\sin \theta {)}^n+(1+\cos \theta -i\sin \theta {)}^n=(2\cos \frac{\theta }{2}{e}^{i\frac{\theta }{2}}{)}^n+(2\cos \frac{\theta }{2}{e}^{-i\frac{\theta }{2}}{)}^n$

$=2^n{\cos }^n\frac{\theta }{2}(e^{i\frac{n\theta }{2}}+e^{-i\frac{n\theta }{2}})$

$=2^{n+1}{\cos }^n\frac{\theta }{2}\cos \frac{n\theta }{2}$