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core pure 2

just as: sinx = [(e^ix) - (e^-ix)]/i2

does:

cosx = [(e^ix) + (e^-ix)]/2

?
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harper_
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Yes. Google: exponential representations of trigonometric functions
Original post by FurtherMaths2020
just as: sinx = [(e^ix) - (e^-ix)]/i2

does:

cosx = [(e^ix) + (e^-ix)]/2

?


You should be able to work this out for yourself using:

eix=cos⁡x+isin⁡xe^{ix}=\cos x + i\sin x

and

e−ix=cos⁡(−x)+isin⁡(−x)=cos⁡x−isin⁡x\begin{aligned}e^{-ix} &=\cos (-x )+ i\sin (-x) \\ &=\cos x - i\sin x\end{aligned}
Original post by ghostwalker
You should be able to work this out for yourself using:

eix=cos⁡x+isin⁡xe^{ix}=\cos x + i\sin x

and

e−ix=cos⁡(−x)+isin⁡(−x)=cos⁡x−isin⁡x\begin{aligned}e^{-ix} &=\cos (-x )+ i\sin (-x) \\ &=\cos x - i\sin x\end{aligned}


yes, I know, but is (cosx = [(e^ix) + (e^-ix)]/2) a standard notion?
Original post by FurtherMaths2020
yes, I know, but is (cosx = [(e^ix) + (e^-ix)]/2) a standard notion?


Well, that's not what you asked initially.

Rewriting cos x in that form and knowing it can be written in that form is certainly very useful on occasions; if that's what you mean by a "standard notion", then yes.
Original post by ghostwalker
Well, that's not what you asked initially.

Rewriting cos x in that form and knowing it can be written in that form is certainly very useful on occasions; if that's what you mean by a "standard notion", then yes.


thanks

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