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

Prove the trigonometric identity above.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest\(\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{S=\dfrac{\sin^2 x}{1-\sin^2 x} = \dfrac{\sin^2x}{\cos^2 x} = \boxed{\tan^2x} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} \)

Log in to reply

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

Log in to reply

About 5 mins...

Log in to reply

100% vella! Just like @Mehul Arora

Log in to reply

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

It is related to Andrew's recent problem :3

Log in to reply

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?

Log in to reply

It should be specified that \(|\sin x|<1\)

Log in to reply

Log in to reply

Interesting!

Log in to reply