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 #10
avatar+26367 
+3

Find \(\tan^2(20) + \tan^2(40) + \tan^2(80) \) .  All angles are in degrees.

 

Formula:
\(\begin{array}{rcll} \boxed{ \sum \limits_{l=1}^{n} \tan^2\left( \dfrac{180^\circ \cdot l}{2n+1} \right) = n(2n+1) } \\ \end{array}\)

Source (7): https://math.stackexchange.com/questions/173447/proving-sum-limits-l-1n-sum-limits-k-1n-1-tan-frac-lk-pi-2n1-t/173858#173858

 

\(\begin{array}{|lrcll|} \hline n=1: & \sum \limits_{l=1}^{1} \tan^2\left( \dfrac{180^\circ \cdot l}{2\cdot 1+1} \right) &=& 1\cdot(2\cdot 1 +1) \\ & \sum \limits_{l=1}^{1} \tan^2\left( \dfrac{180^\circ \cdot l}{3} \right) &=& 1\cdot(3) \\ & \mathbf{ \tan^2\left(60^\circ\right)} &=& \mathbf{ 3 } \\ \hline n=4: & \sum \limits_{l=1}^{4} \tan^2\left( \dfrac{180^\circ \cdot l}{2\cdot 4+1} \right) &=& 4\cdot(2\cdot 4 +1) \\ & \sum \limits_{l=1}^{4} \tan^2\left( \dfrac{180^\circ \cdot l}{9} \right) &=& 4\cdot(9) \\ & \mathbf{\tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(60^\circ\right)+\tan^2\left(80^\circ\right) } &=& \mathbf{ 36 } \\ \hline & \tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(80^\circ\right) &=& 36 - 3 \\ & \mathbf{\tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(80^\circ\right)} &=& \mathbf{33} \\ \hline \end{array}\)

 

laugh

Nov 12, 2019
 #14
avatar+142 
0

THANK YOU

Nov 12, 2019
 #8
avatar+159 
0

Oh XDangry

Nov 12, 2019
 #6
avatar+2862 
0

no lol

Nov 12, 2019

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