FurtherMaths2020
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#1
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just as: sinx = [(e^ix) - (e^-ix)]/i2

does:

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

?
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harper_
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#2
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Yes. Google: exponential representations of trigonometric functions
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ghostwalker
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(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:

e^{ix}=\cos x + i\sin x

and

\begin{aligned}e^{-ix} &=\cos (-x )+ i\sin (-x) \\ &=\cos x - i\sin x\end{aligned}
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FurtherMaths2020
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(Original post by ghostwalker)
You should be able to work this out for yourself using:

e^{ix}=\cos x + i\sin x

and

\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?
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ghostwalker
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(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.
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FurtherMaths2020
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(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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