Trig3.6

From Exampleproblems

Jump to: navigation, search

Prove that [1+\cos \frac{\pi}{8}][1+\cos \frac{3\pi}{8}][1+\cos \frac {5\pi}{8}][1+\cos \frac{7\pi}{8}]=\frac{1}{8}\,

We have \cos \frac{3\pi}{8}=\cos [\frac{\pi}{2}-\frac{\pi}{8}]=\sin \frac{\pi}{8}\,

\cos \frac{5\pi}{8}=\cos [\frac{\pi}{2}+\frac{3\pi}{8}]=-\sin \frac{\pi}{8}\,

\cos \frac{7\pi}{8}=\cos [\pi-\frac{\pi}{8}]=-\cos \frac{\pi}{8}\,

Therefore,LHS=[1+\cos \frac{\pi}{8}][1+\sin \frac{\pi}{8}][1-\sin \frac {\pi}{8}][1-\cos \frac{\pi}{8}]\,

[1-\cos^ \frac{\pi}{8}][1-\sin^2 \frac{\pi}{8}]=\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\,

\frac{1}{4}[2\sin \frac{\pi}{8} \cos \frac{\pi}{8}]^2=\frac{1}{4} \sin^2 \frac{\pi}{4}=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}\,=RHS

Main Page:Trigonometry

Retrieved from "http://www.exampleproblems.com/wiki/index.php/Trig3.6"
Argan Oil
Natural Skin Care
Organic Skin Care
visitor stats