Waste less time on Facebook — follow Brilliant.
×

Trigonometric Identity

We know that \(\cos { 2x } =\cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } \)

This is just a proof I have made. I don't know if it already exists but I just wanted to share it with you.

\({ \left( { e }^{ ix } \right) }^{ 2 }={ e }^{ 2ix }\\ \\ { \left( { e }^{ ix } \right) }^{ 2 }={ e }^{ i\times 2x }\\ \\ { \left( \cos { x } +i\sin { x } \right) }^{ 2 }=cos2x+isin2x\\ \\ \cos ^{ 2 }{ x } +{ i }^{ 2 }\sin ^{ 2 }{ x } +2i\cos { x } \sin { x } =cos2x+isin2x\\ \\ \cos ^{ 2 }{ x } +\left( -1 \right) \sin ^{ 2 }{ x } +i\times 2\cos { x } \sin { x } =cos2x+isin2x\\ \\ \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } +i\sin { 2x } =cos2x+isin2x\\ \\ So,\quad \cos { 2x } =\cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } \)

Note by Archit Boobna
1 year, 10 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Nice observational skill Archit :) Azhaghu Roopesh M · 1 year, 10 months ago

Log in to reply

@Azhaghu Roopesh M Thanks! Archit Boobna · 1 year, 10 months ago

Log in to reply

This also exists !! Parth Bhardwaj · 1 year, 10 months ago

Log in to reply

@Parth Bhardwaj But I thought it myself. Archit Boobna · 1 year, 10 months ago

Log in to reply

@Archit Boobna Ok ! Did you try all of my problems - new ones ? Parth Bhardwaj · 1 year, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...