﻿ Prove the equality \cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{5\pi}{8}+\cos^2\dfrac{7\pi}{8}=2

#### Question

Prove the equality

\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{5\pi}{8}+\cos^2\dfrac{7\pi}{8}=2

Collected in the board: Trigonometry

Steven Zheng posted 8 hours ago

Use the double angle identity to reduce the power

\cos 2\alpha =2 \cos^2 \alpha - 1

\cos^2 \alpha = \dfrac{\cos 2\alpha+1}{2}

Then,

\cos^2\dfrac{\pi}{8}= \dfrac{\cos \dfrac{2\pi}{8}+1}{2}=\dfrac{1}{2} \cos\dfrac{\pi}{4}+\dfrac{1}{2}

\cos^2\dfrac{3\pi}{8}= \dfrac{\cos \dfrac{6\pi}{8}+1}{2}=\dfrac{1}{2} \cos\dfrac{3\pi}{4} +\dfrac{1}{2}=-\dfrac{1}{2} \cos\dfrac{\pi}{4}+\dfrac{1}{2}

\cos^2\dfrac{5\pi}{8}= \dfrac{\cos \dfrac{10\pi}{8}+1}{2}=\dfrac{1}{2} \cos\dfrac{5\pi}{4} +\dfrac{1}{2}=-\dfrac{1}{2} \cos\dfrac{\pi}{4}+\dfrac{1}{2}

\cos^2\dfrac{7\pi}{8}= \dfrac{\cos\dfrac{14\pi}{8}+1}{2}=\dfrac{1}{2} \cos\dfrac{7\pi}{4} +\dfrac{1}{2}=\dfrac{1}{2} \cos\dfrac{\pi}{4}+\dfrac{1}{2}

Sum of four terms gives

\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{5\pi}{8}+\cos^2\dfrac{7\pi}{8}

=\dfrac{1}{2}\cdot 4

=2

Steven Zheng posted 8 hours ago

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