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# Interesting Sum

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

Prove the trigonometric identity above.

Note by Joel Yip
11 months ago

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$$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{S=\dfrac{\sin^2 x}{1-\sin^2 x} = \dfrac{\sin^2x}{\cos^2 x} = \boxed{\tan^2x} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$$ · 11 months ago

The sum of an infinite GP can only be calculated when $$-1<r<1$$. However $$\sin^2x$$ can be equal to 1. Can you justify? · 8 months, 3 weeks ago

It should be specified that $$|\sin x|<1$$ · 8 months, 3 weeks ago

Well it is already in the range! · 8 months, 3 weeks ago

100% vella! Just like @Mehul Arora · 8 months, 3 weeks ago

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

It is related to Andrew's recent problem :3 · 8 months, 3 weeks ago

wow it looks like a pyramid. How long did you take to type and count so many boxes :)? · 10 months, 4 weeks ago

About 5 mins... · 10 months, 4 weeks ago