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.

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

## Comments

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TopNewestNice observational skill Archit :) – Azhaghu Roopesh M · 2 years, 4 months ago

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– Archit Boobna · 2 years, 4 months ago

Thanks!Log in to reply

This also exists !! – Parth Bhardwaj · 2 years, 4 months ago

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– Archit Boobna · 2 years, 4 months ago

But I thought it myself.Log in to reply

– Parth Bhardwaj · 2 years, 4 months ago

Ok ! Did you try all of my problems - new ones ?Log in to reply