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Geometry Level 3

tan ⁡ ( α ) + 2 tan ⁡ ( 2 α ) + 4 tan ⁡ ( 4 α ) + 8 cot ⁡ ( 8 α ) = cot ⁡ ( n α ) \tan(\alpha)+2\tan(2\alpha)+4\tan(4\alpha)+8\cot(8\alpha)=\cot(n\alpha)

If the above trigonomtric equation is true for all α \alpha , what is the value of n n ?

This problem is part of the set Trigonometry .


The answer is 1.

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Omkar Kulkarni by Omkar Kulkarni , 22, India

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3 solutions

Shekhar Prasad
Feb 15, 2015

Take 4 tan 4x + 8 cot 8x together first 8 cot 8x = 8 / tan8x = 4( 1- tan^2 4x)/ tan4x

So 4tan 4x + 8 cot 8 x = 4 cot 4x Take 2 tan2x + 4 cot 4x now Repeat the same exercise. We get 2 tan 2x Now tan x + 2 tan 2x = cot x Hence n= 1

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Noel Lo
Apr 25, 2015

Same as Divij Handa here. I do appreciate a general solution as well. For that, thank you very much Shekhar Prasad.

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Aman Kumar
Feb 14, 2015

Say a=15 tan15= 2-sqrt(3) tan(2 15)=1/sqrt3 tan(4 15)=sqrt3 cot(8*15)=cot120=cot(180-60)=-cot60=-tan30=-1/sqrt3 put these values in L.H.S, we get: 2+sqrt3=cot(15n) or, cot15=2+sqrt3 so. 15n=15, therefore, n=1

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Is there a more general way of doing this. I dont want to just put values and check.

Divij Handa - 6 years, 3 months ago

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