Thread: [SOLVED] Trig identities

Given that cos x = e^jx + e^-jx /2 prove that 2cs^2(x)-1 =cos2x

I know that the second part is double angle formula (not that that helps me prove the identitie) But i dont understand what the first part has to do with it. I'm fairly sure its demoivres. But that's the limit of what i understand.

Originally Posted by ally79

Given that cos x = e^jx + e^-jx /2 prove that 2cs^2(x)-1 =cos2x

I know that the second part is double angle formula (not that that helps me prove the identitie) But i dont understand what the first part has to do with it. I'm fairly sure its demoivres. But that's the limit of what i understand.

"2cs^2(x)-1 =cos2x"

Huh?

2 cos^2(x) - 1 = cos(2x)
if that's what you mean.

2 [(e^{jx} + e^{-jx})/2]^2 - 1

= (1/2) (e^{2jx} + e^{-2jx} + 2) - 1

Can you take it from here?

-Dan

yes thank you