Question

How do you find an expression for `sin(x)` in terms of `e^(ix)` and `e^(ix)`?

Answer 1

`sinx = (e^(ix) - e^(-ix))/(2i)`

Explanation:

Start from the MacLaurin series of the exponential function:

`e^x = sum_(n=0)^oo x^n/(n!)`

so:

`e^(ix) = sum_(n=0)^oo (ix)^n/(n!) = sum_(n=0)^oo i^nx^n/(n!)`

Separate now the terms for `n` even and `n` odd, and let `n=2k` in the first case, `n= 2k+1` in the second:

`e^(ix) = sum_(k=0)^oo i^(2k) x^(2k)/((2k)!) + sum_(k=0)^oo i^(2k+1)x^(2k+1)/((2k+1)!)`

Note now that:

`i^(2k) = (i^2)^k = (-1)^k`

`i^(2k+1) = i*i^(2k) = i*(-1)^k`

so:

`e^(ix) = sum_(k=0)^oo (-1)^k x^(2k)/((2k)!) + isum_(k=0)^oo (-1)^k x^(2k+1)/((2k+1)!)`

and we can recognize the MacLaurin expansions of `cosx` and `sinx`:

`e^(ix) = cosx +i sinx`

which is Euler's formula.

Considering that `cosx` is an even function and `sinx` and odd function then we have:

`e^(-ix) = cos(-x) + i sin(-x) = cosx-i sinx`

then:

`e^(ix) - e^(-ix) = 2i sinx`

and finally:

`sinx = (e^(ix) - e^(-ix))/(2i)`

Answer 2

Other approach to problem. See below

Explanation:

We know that `e^(ix)=cosx+isinx` (Euler)

Similarly, `e^(-ix)=cos(-x)+isin(-x)`

But we know that `cos(-x)=cosx` and `sin(-x)=-sinx`

Then we have

`e^(ix)=cosx+isinx`
`e^(-ix)=cosx-isinx`

Adding both identities

`e^(ix)+e^(-ix)=2cosx` and finally `cosx=(e^(ix)+e^(-ix))/2`

Subtarcting both, we have

`e^(ix)-e^(-ix)=2isinx` and then `sinx=(e^(ix)-e^(-ix))/(2i)`

Answer 3

Compare the Maclaurin series of `sinx` and `e^x` and construct the relation from that:
`sinx=1/(2i)(e^(ix)-e^(-ix))`

Explanation:

I assume the final formula in the question should read `e^(-ix)`?

Compare the Maclaurin series of `sinx` and `e^x` and construct the relation from that. We'll take as given the series for these functions. Deriving these is a pleasure in itself, one easily found elsewhere on the web, e.g.
http://blogs.ubc.ca/infiniteseriesmodule/units/unit-3-power-series/taylor-series/maclaurin-expansion-of-sinx/
http://www.songho.ca/math/taylor/taylor_exp.html
Note that both of these series are convergent over the whole range of `x`.

`sinx=x-(x^3)/(3!)+(x^5)/(5!)-...+(-1)^nx^(2n+1)/((2n+1)!)+...`
`e^x=1+x+x^2/(2!)+x^3/(3!)+...+x^n/(n!)+...`

We can immediately see that the terms in the sine series are very similar to those in the exponential series - they're the same size where they exist, but often have the opposite sign, and half of them are missing.

Recalling that the powers of `i` change in a periodic four-step pattern that has two successive plus signs and two successive minus signs, we wonder if changing `x` to `ix` in the exponential series might help our sign problem. We substitute:

`e^(ix)=1+ix+(ix)^2/(2!)+(ix)^3/(3!)+...+(ix)^n/(n!)+...`
which becomes
`e^(ix)=1+ix-x^2/(2!)-ix^3/(3!)+x^4/(4!)+...+(ix)^n/(n!)+...`

To remove every second term, we combine it with the series for `e^(-ix)`:
`e^(-ix)=1+(-ix)+(-ix)^2/(2!)+(-ix)^3/(3!)+...+(-ix)^n/(n!)+...`
which becomes (be careful combining minus signs and `i^2`s!)
`e^(-ix)=1-ix-x^2/(2!)-ix^3/(3!)+x^4/(2!)-...+(-ix)^n/(n!)+...`

When we take the difference of these series term by term, we get closer to what we want (NB taking the sum of them gives us a relation for `cosx` instead - give it a try). Note that the terms of even powers of `x` are identical in the two series, so their difference is 0.

`e^(ix)-e^(-ix)=2ix-2ix^3/(3!)+2ix^5/(5!)-...+2i(-1)^nx^(2n+1)/((2n+1)!)+...`

which is just the series above for `sinx` multiplied by `2i`. So we have our desired relation:

`sinx=1/(2i)(e^(ix)-e^(-ix))`

Compare at this point the hyperbolic functions, which you may have been introduced to already. In particular, note the definition of `sinhx` ("hyperbolic sine"; "sinh" is pronounced in one of several ways - "shine", "sinch", etc.):

`sinhx=1/2(e^x-e^(-x))`

The hyperbolic functions are a set of functions closely related to the trig functions via these formulae. As you progress with differential equations, you'll encounter situations where a simple change of sign to a coefficient makes the difference between finding trig function and hyperbolic function solutions. The relation between the two sets of functions is an important one.