Interesting Sum

(tanx)2=(sinx)2+(sinx)4+(sinx)6+(sinx)8+ \large { \left( \tan { x } \right) }^{ 2 }={ \left( \sin { x } \right) }^{ 2 }{ +\left( \sin { x } \right) }^{ 4 }+{ \left( \sin { x } \right) }^{ 6 }+{ \left( \sin { x } \right) }^{ 8 }+\cdots

Prove the trigonometric identity above.

Note by Joel Yip
3 years, 7 months ago

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1 vote

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S=sin2x1sin2x=sin2xcos2x=tan2x\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{S=\dfrac{\sin^2 x}{1-\sin^2 x} = \dfrac{\sin^2x}{\cos^2 x} = \boxed{\tan^2x} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

Nihar Mahajan - 3 years, 7 months ago

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wow it looks like a pyramid. How long did you take to type and count so many boxes :)?

A Former Brilliant Member - 3 years, 7 months ago

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About 5 mins...

Nihar Mahajan - 3 years, 7 months ago

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100% vella! Just like @Mehul Arora

Aditya Kumar - 3 years, 5 months ago

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Haha, I actually copy pasted the latex from some other note :P

It is related to Andrew's recent problem :3

Nihar Mahajan - 3 years, 5 months ago

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The sum of an infinite GP can only be calculated when 1<r<1-1<r<1. However sin2x\sin^2x can be equal to 1. Can you justify?

Rishik Jain - 3 years, 5 months ago

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It should be specified that sinx<1|\sin x|<1

Nihar Mahajan - 3 years, 5 months ago

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@Nihar Mahajan Well it is already in the range!

Joel Yip - 3 years, 5 months ago

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Interesting!

Arnob Roy - 3 years, 7 months ago

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