MathLabs

Problem 5

Prove that cos⁡π7−cos⁡2π7+cos⁡3π7=12\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac12.
Step 4 of 5: Obtain the three-cosine identity
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2(cos⁡2π7+cos⁡4π7+cos⁡6π7)=−1⟹cos⁡2π7+cos⁡4π7+cos⁡6π7=−122\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)=-1\Longrightarrow\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12
Detailed analysis

Because cos⁡(2π−x)=cos⁡x\cos(2\pi-x)=\cos x, the paired terms yield 2(cos⁡(2π/7)+cos⁡(4π/7)+cos⁡(6π/7))=−12(\cos(2\pi/7)+\cos(4\pi/7)+\cos(6\pi/7))=-1. Dividing by 22 gives the stated three-cosine sum.