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Mathematics Question on Trigonometric Functions

The value of sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8\sin^{4} \frac{\pi}{8} + \sin^{4} \frac{3 \pi}{8} + \sin^{4} \frac{5\pi}{8} + \sin^{4} \frac{7\pi}{8} is

A

34\frac{\sqrt{3}}{4}

B

34\frac{3}{4}

C

32\frac{\sqrt{3}}{2}

D

32\frac{3}{2}

Answer

32\frac{3}{2}

Explanation

Solution

We have, 7π8=π−π8\frac{7\pi}{8}=\pi-\frac{\pi}{8}
and 5π8=π−3π8\frac{5\pi}{8}=\pi-\frac{3\pi}{8}
⇒sin⁡7π8=sin⁡(π−π8)\Rightarrow \sin \frac{7\pi}{8}=\sin\left(\pi-\frac{\pi}{8}\right)
and sin⁡5π8=sin⁡(π−3π8)\sin \frac{5\pi}{8}=\sin\left(\pi-\frac{3\pi}{8}\right)
⇒sin⁡7π8=sin⁡π8\Rightarrow \sin \frac{7\pi}{8}=\sin \frac{\pi}{8} and sin⁡5π8=sin⁡3π8\sin \frac{5\pi}{8}=\sin \frac{3\pi}{8}
⇒sin⁡4(7π8)=sin⁡4(π8)\Rightarrow \sin^{4}\left(\frac{7\pi}{8}\right)=\sin^{4}\left(\frac{\pi}{8}\right)
and sin⁡4(5π8)=sin⁡4(3π8)\sin^{4}\left(\frac{5\pi}{8}\right)=\sin^{4}\left(\frac{3\pi}{8}\right)
Now, sin⁡4(π8)+sin⁡4(3π8)+sin⁡4(5π8)+sin⁡4(7π8)\sin^{4}\left(\frac{\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{5\pi}{8}\right)+\sin^{4}\left(\frac{7\pi}{8}\right)
=sin⁡4(π8)+sin⁡4(3π8)+sin⁡4(3π8)+sin⁡4(π8)=\sin^{4}\left(\frac{\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{3\pi}{8}\right)+\sin^{4}\left(\frac{\pi}{8}\right)
=2sin⁡4(π8)+2sin⁡4(3π8)=2\sin^{4}\left(\frac{\pi}{8}\right)+2\sin^{4}\left(\frac{3\pi}{8}\right)
=2[(sin⁡2π8)2+(sin⁡2(3π8))2]=2\left[\left(\sin^{2} \frac{\pi}{8}\right)^{2}+\left(\sin^{2}\left(\frac{3\pi}{8}\right)\right)^{2}\right]
=2[(1−cos⁡ 2(π8)2)2+(1−cos⁡ 2(3π8)2)2]=2\left[\left(\frac{1-\cos\,2\left(\frac{\pi}{8}\right)}{2}\right)^{2}+\left(\frac{1-\cos\,2\left(\frac{3\pi}{8}\right)}{2}\right)^{2}\right]
=2[(1−cos⁡π4)24+(1−cos⁡3π4)24]=2\left[\frac{\left(1-\cos \frac{\pi}{4}\right)^{2}}{4}+\frac{\left(1-\cos \frac{3\pi}{4}\right)^{2}}{4}\right]
=12[(1−12)2+(1+12)2]=\frac{1}{2}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]
[∵cos⁡π4=12 and cos⁡3π4=−12]\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \,\text{and} \,\cos \frac{3\pi}{4}=\frac{-1}{\sqrt{2}}\right]
=12[1+12−22+1+12+22]=\frac{1}{2}\left[1+\frac{1}{2}-\frac{2}{\sqrt{2}}+1+\frac{1}{2}+\frac{2}{\sqrt{2}}\right]
=12[2+22]=12[2+1]=32=\frac{1}{2}\left[2+\frac{2}{2}\right]=\frac{1}{2}\left[2+1\right]=\frac{3}{2}