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

3 Answers
Jun 4, 2018

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)

Jun 4, 2018

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)

Jun 4, 2018

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^2s!)
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.