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

 Nov 12, 2019

Best Answer 

 #10
avatar+23575 
+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
edited by heureka  Nov 12, 2019
edited by heureka  Nov 12, 2019
 #1
avatar+94 
0

Bro just use the web 2.0 caculator

 

First go to Home located in the top menu

then you should be able to find the calculator in the middle of your screen.

 Nov 12, 2019
 #2
avatar+2505 
0

Yup thats what I did

 

I just copy pasted into the amazing calculator and it cheated homework for me.

 

no jk lol.

But seriously some homework questions are actually dumb. I don't know if the government is trying to steal our brain cells or something.

 Nov 12, 2019
 #3
avatar+94 
0

This is wut I got from the calc: 33.0000000000026774 in degrees 

 Nov 12, 2019
 #4
avatar+2505 
0

yeah same I guess I just round then

CalculatorUser  Nov 12, 2019
 #5
avatar+94 
0

wait a second! R u guest???

VooFIX  Nov 12, 2019
 #6
avatar+2505 
0

no lol

CalculatorUser  Nov 12, 2019
 #7
avatar+2505 
0

if i log out, and post, and log back in. It would make that post mine through the magical use of cookies laugh

 

yes Sherlock Holmes, the tone of my voice sounded as if I was the one asking the question. cheeky

CalculatorUser  Nov 12, 2019
edited by CalculatorUser  Nov 12, 2019
 #8
avatar+94 
0

Oh XDangry

VooFIX  Nov 12, 2019
 #9
avatar+105989 
0

What is squared?   (20)^2    or    (tan20)^2

 Nov 12, 2019
 #10
avatar+23575 
+3
Best Answer

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

heureka Nov 12, 2019
edited by heureka  Nov 12, 2019
edited by heureka  Nov 12, 2019
 #11
avatar+105989 
+2

Thanks Heureka,  cool

 

I have never seen that formula before.  

And if i had I would not have recognised its relevance. 

Melody  Nov 12, 2019
 #12
avatar+23575 
+3

Thank you, Melody !

 

laugh

heureka  Nov 12, 2019
 #13
avatar+105476 
+1

Very nice, Heureka.....not familiar with this "formula," either.....

 

 

cool cool cool

 Nov 12, 2019
 #14
avatar+23575 
+1

Thank you, CPhill !

 

laugh

heureka  Nov 13, 2019

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