Why does sinx=(1/2i)[e^(ix)-e^(-ix)]?

[asdf1]

why does sinx=(1/2i)[e^(ix)-e^(-ix)]?

 

[siddharth]

This can be shown from Euler's formula.

http://mathworld.wolfram.com/EulerFormula.html"

 

[asdf1]

e^ix=cosx+isinx
e^(-ix)=?
do you really add a negative sign and it becomes cos(-x)+isin(-x)?

 

[LeonhardEuler]

e^ix=cosx+isinx
e^(-ix)=?
do you really add a negative sign and it becomes cos(-x)+isin(-x)?

Yes, and cos(-x)=cos(x), while sin(-x)=-sin(x), because they are even and odd functions, respectively.

 

[benorin]

e^ix=cosx+isinx
e^(-ix)=?
do you really add a negative sign and it becomes cos(-x)+isin(-x)?

yes, and recall that sine is an odd function and cosine is an even function so we have

$$e^{-ix} = \cos(-x) + i\sin(-x)=\cos(x)-i\sin(x)$$

and from Euler's formula we have $e^{ix}=\cos(x)+i\sin(x)$

and so subtracting the first formula from the second gives

$$e^{ix}-e^{-ix} = 2i\sin(x)$$

hence

$$\sin(x)=\frac{1}{2i} \left( e^{ix}-e^{-ix}\right)$$

 

[asdf1]

wow~
amazing...
thank you very much!