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Mathematics Question on Trigonometric Equations

The smallest positive integral value of ′n′'n' such that [1+sin⁡π8+ i cos⁡π81+sin⁡π8− i cos⁡π8]n\left[\frac {1+\sin \frac {\pi}{8}+\,i\,\cos \frac {\pi}{8}}{1+\sin \frac {\pi}{8}-\,i\,\cos \frac {\pi}{8}} \right]^n is purely imaginary is, nn =

A

8

B

4

C

3

D

2

Answer

4

Explanation

Solution

[1+sin⁡π8+icos⁡π81+sin⁡π8−icos⁡π8]n\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{n} =[1+cos⁡α+isin⁡α1+cos⁡α−isin⁡α]n(=\left[\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha}\right]^{n} \quad\left(\right. Put α=π2−π8)\left.\alpha=\frac{\pi}{2}-\frac{\pi}{8}\right) =[2cos⁡2α2+2isin⁡α2cos⁡α22cos⁡2α2−2isin⁡α2cos⁡α2]n=\left[\frac{2 \cos ^{2} \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^{2} \frac{\alpha}{2}-2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right]^{n} =[cos⁡α+isin⁡α2cos⁡α2−isin⁡α2]n=\left[\frac{\cos ^{\alpha}+i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-i \sin \frac{\alpha}{2}}\right]^{n} =(e2iα2)n=einα=\left(e^{2 i \frac{\alpha}{2}}\right)^{n}=e^{i n \alpha} =ein(3π8)=cos⁡3nπ8+isin⁡3nπ8=e^{i n\left(\frac{3 \pi}{8}\right)}=\cos \frac{3 n \pi}{8}+i \sin \frac{3 n \pi}{8} For n=4n=4, we get imaginary part.