Trigo Teasing Again

Geometry Level 4

∑ n = 1 n = 7 tan ⁡ 2 ( n π 16 ) \large \displaystyle \sum_{n=1}^{n=7} \tan^2 \left(\frac{n\pi}{16}\right)

Evaluate the above sum.


The answer is 35.

Vilakshan Gupta by Vilakshan Gupta , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Zico Quintina
Jun 26, 2018

Consider the sum of the first and last terms of the series. Making use of the identity tan ⁡ θ = cot ⁡ ( π 2 − θ ) \tan \theta = \cot \left( \frac{\pi}{2} - \theta \right) we get that tan ⁡ ( 7 π 16 ) = cot ⁡ ( π 16 ) \tan \left( \frac{7 \pi}{16} \right) = \cot \left( \frac{\pi}{16} \right) .

Then letting α = π 16 \alpha = \frac{\pi}{16} , we have

tan ⁡ 2 ( π 16 ) + tan ⁡ 2 ( 7 π 16 ) = tan ⁡ 2 α + cot ⁡ 2 α = ( tan ⁡ 2 α + 2 tan ⁡ α cot ⁡ α + cot ⁡ 2 α ) − 2 tan ⁡ α cot ⁡ α = ( tan ⁡ α + cot ⁡ α ) 2 − 2 = ( sin ⁡ α cos ⁡ α + cos ⁡ α sin ⁡ α ) 2 − 2 = ( 1 sin ⁡ α cos ⁡ α ) 2 − 2 = ( 2 sin ⁡ 2 α ) 2 − 2 = 4 csc ⁡ 2 π 8 − 2 \begin{array}{rl} \tan^2 \left( \dfrac{\pi}{16} \right) + \tan^2 \left( \dfrac{7 \pi}{16} \right) &= \ \ \tan^2 \alpha + \cot^2 \alpha \\ \\ &= \ \ ( \tan^2 \alpha + 2 \tan \alpha \cot \alpha + \cot^2 \alpha ) - 2 \tan \alpha \cot \alpha \\ \\ &= \ \ ( \tan \alpha + \cot \alpha )^2 - 2 \\ \\ &= \ \ \left( \dfrac{\sin \alpha}{\cos \alpha} + \dfrac{\cos \alpha}{\sin \alpha} \right)^2 - 2 \\ \\ &= \ \ \left( \dfrac{1}{\sin \alpha \cos \alpha} \right)^2 - 2 \\ \\ &= \ \ \left( \dfrac{2}{\sin 2\alpha} \right)^2 - 2 \\ \\ &= \ \ 4 \csc^2 \dfrac{\pi}{8} - 2 \end{array}

Similarly, we find that tan ⁡ 2 ( 2 π 16 ) + tan ⁡ 2 ( 6 π 16 ) = 4 csc ⁡ 2 π 4 − 2 = 6 AND tan ⁡ 2 ( 3 π 16 ) + tan ⁡ 2 ( 5 π 16 ) = 4 csc ⁡ 2 3 π 8 − 2 \begin{array}{rcccl} \tan^2 \left( \dfrac{2 \pi}{16} \right) + \tan^2 \left( \dfrac{6 \pi}{16} \right) = 4 \csc^2 \dfrac{\pi}{4} - 2 = 6 & & \text{AND} & & \tan^2 \left( \dfrac{3 \pi}{16} \right) + \tan^2 \left( \dfrac{5 \pi}{16} \right) = 4 \csc^2 \dfrac{3 \pi}{8} - 2 \end{array}

Then

∑ n = 1 n = 7 tan ⁡ 2 ( n π 16 ) = [ tan ⁡ 2 ( π 16 ) + tan ⁡ 2 ( 7 π 16 ) ] + [ tan ⁡ 2 ( 2 π 16 ) + tan ⁡ 2 ( 6 π 16 ) ] + [ tan ⁡ 2 ( 3 π 16 ) + tan ⁡ 2 ( 5 π 16 ) ] + tan ⁡ 2 ( 4 π 16 ) \begin{array}{rl} \displaystyle \sum_{n=1}^{n=7} \tan^2 \left( \dfrac{n \pi}{16} \right) &= \ \ \left[ \tan^2 \left( \dfrac{\pi}{16} \right) + \tan^2 \left( \dfrac{7 \pi}{16} \right) \right] + \left[ \tan^2 \left( \dfrac{2 \pi}{16} \right) + \tan^2 \left( \dfrac{6 \pi}{16} \right) \right] + \left[ \tan^2 \left( \dfrac{3 \pi}{16} \right) + \tan^2 \left( \dfrac{5 \pi}{16} \right) \right] + \tan^2 \left( \dfrac{4 \pi}{16} \right) \\ \\ \end{array} = ( 4 csc ⁡ 2 π 8 − 2 ) + 6 + ( 4 csc ⁡ 2 3 π 8 − 2 ) + 1 = 4 ( csc ⁡ 2 π 8 + sec ⁡ 2 π 8 ) + 3 [Here we used csc ⁡ θ = sec ⁡ ( π 2 − θ ) ] = 4 ( cos ⁡ 2 π 8 + sin ⁡ 2 π 8 sin ⁡ 2 π 8 cos ⁡ 2 π 8 ) + 3 = 4 ( 4 sin ⁡ 2 π 4 ) + 3 [Here we used sin ⁡ θ cos ⁡ θ = 1 2 sin ⁡ 2 θ ] = 4 ( 8 ) + 3 = 35 \begin{array}{rlcccl} &= \ \ \left( 4 \csc^2 \dfrac{\pi}{8} - 2 \right) + 6 + \left( 4 \csc^2 \dfrac{3 \pi}{8} - 2 \right) + 1 \\ \\ &= \ \ 4 \left( \csc^2 \dfrac{\pi}{8} + \sec^2 \dfrac{\pi}{8} \right) + 3 & & & &\small \text{[Here we used } \csc \theta = \sec \left( \frac{\pi}{2} - \theta \right) \ ]\\ \\ &= \ \ 4 \left( \dfrac{\cos^2 \dfrac{\pi}{8} + \sin^2 \dfrac{\pi}{8}}{\sin^2 \dfrac{\pi}{8} \ \cos^2 \dfrac{\pi}{8}} \right) + 3 \\ \\ &= \ \ 4 \left( \dfrac{4}{\sin^2 \dfrac{\pi}{4}} \right) + 3 & & & &\small \text{[Here we used } \sin \theta \ \cos \theta = \frac{1}{2} \sin 2 \theta \ ] \\ \\ &= \ \ 4 (8) + 3 = \boxed{35} \end{array}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...