﻿ Show that \sin^2\dfrac{\pi}{14}+\sin^2\dfrac{3\pi}{14}+\sin^2\dfrac{5\pi}{14}=\dfrac{4}{5}

#### Question

Show that \sin^2\dfrac{\pi}{14}+\sin^2\dfrac{3\pi}{14}+\sin^2\dfrac{5\pi}{14}=\dfrac{4}{5}

Collected in the board: Trigonometry

Steven Zheng posted 2 weeks ago

Apply double angle identity to reduce the power of trig terms.

\sin^2\dfrac{\pi}{14}+\sin^2\dfrac{3\pi}{14}+\sin^2\dfrac{5\pi}{14}

= \dfrac{1-\cos\dfrac{2\pi}{7} }{14} +\dfrac{1-\cos\dfrac{6\pi}{7} }{14} +\dfrac{1-\cos\dfrac{10\pi}{7} }{14}

=\dfrac{3-\Big( \cos \dfrac{π}{7}+ \cos\dfrac{3π}{7}+ \cos\dfrac{5π}{7}\Big)}{2}

Now what we will do is to find the value of

\cos \dfrac{π}{7}+ \cos\dfrac{3π}{7}+ \cos\dfrac{5π}{7}

Multiply and divide each term by 2\sin\dfrac{π}{7}

= \dfrac{1}{2\sin\dfrac{π}{7}} [ 2\sin\dfrac{π}{7}\cos \dfrac{π}{7}+ 2\sin\dfrac{π}{7}\cos\dfrac{3π}{7}+ 2\sin\dfrac{π}{7}\cos\dfrac{π}{7}]

Use double angle identity for sine function and product identity

\cos \dfrac{π}{7}+ \cos\dfrac{3π}{7}+ \cos\dfrac{5π}{7}

= \dfrac{1}{2\sin\dfrac{π}{7}} [ \sin\dfrac{2π}{7} + \sin(\dfrac{π}{7}+\dfrac{3π}{7}) + \sin(\dfrac{π}{7}-\dfrac{3π}{7}) + \sin(\dfrac{π}{7}+\dfrac{5π}{7}) + \sin(\dfrac{π}{7}-\dfrac{5π}{7}) ]

= \dfrac{1}{2\sin\dfrac{π}{7}} ( \sin\dfrac{2π}{7} + \sin\dfrac{4π}{7} + \sin\dfrac{-2π}{7} + \sin\dfrac{6π}{7} + \sin(\dfrac{-4π}{7})

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin\dfrac{6π}{7}

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin(π-\dfrac{π}{7})

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin\dfrac{π}{7}

=\dfrac{1}{2}

Substitution results in the final value

\sin^2\dfrac{\pi}{14}+\sin^2\dfrac{3\pi}{14}+\sin^2\dfrac{5\pi}{14}

=\dfrac{3-\dfrac{1}{2} }{2} ==\dfrac{5}{4}

Steven Zheng posted 2 weeks ago

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