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
1 year, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

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 - 1 year, 8 months ago

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?

Rishik Jain - 1 year, 6 months ago

Log in to reply

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

Nihar Mahajan - 1 year, 6 months ago

Log in to reply

@Nihar Mahajan Well it is already in the range!

Joel Yip - 1 year, 6 months ago

Log in to reply

100% vella! Just like @Mehul Arora

Aditya Kumar - 1 year, 6 months ago

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

Nihar Mahajan - 1 year, 6 months ago

Log in to reply

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

Brilliant Member - 1 year, 8 months ago

Log in to reply

About 5 mins...

Nihar Mahajan - 1 year, 8 months ago

Log in to reply

Interesting!

Arnob Roy - 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...