Waste less time on Facebook — follow Brilliant.
×

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
9 months, 1 week ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

\(\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{S=\dfrac{\sin^2 x}{1-\sin^2 x} = \dfrac{\sin^2x}{\cos^2 x} = \boxed{\tan^2x} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} \) Nihar Mahajan · 9 months, 1 week ago

Log in to reply

@Nihar Mahajan 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? Rishik Jain · 7 months ago

Log in to reply

@Rishik Jain It should be specified that \(|\sin x|<1\) Nihar Mahajan · 7 months ago

Log in to reply

@Nihar Mahajan Well it is already in the range! Joel Yip · 7 months ago

Log in to reply

@Nihar Mahajan 100% vella! Just like @Mehul Arora Aditya Kumar · 7 months ago

Log in to reply

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

It is related to Andrew's recent problem :3 Nihar Mahajan · 7 months ago

Log in to reply

@Nihar Mahajan wow it looks like a pyramid. How long did you take to type and count so many boxes :)? Svatejas Shivakumar · 9 months, 1 week ago

Log in to reply

@Svatejas Shivakumar About 5 mins... Nihar Mahajan · 9 months, 1 week ago

Log in to reply

Interesting! Arnob Roy · 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...