How do you find an expression for #sin(x)# in terms of #e^(ix)# and #e^(ix)#?
3 Answers
Explanation:
Start from the MacLaurin series of the exponential function:
so:
Separate now the terms for
Note now that:
so:
and we can recognize the MacLaurin expansions of
which is Euler's formula.
Considering that
then:
and finally:
Other approach to problem. See below
Explanation:
We know that
Similarly,
But we know that
Then we have
Adding both identities
Subtarcting both, we have
Compare the Maclaurin series of
Explanation:
I assume the final formula in the question should read
Compare the Maclaurin series of
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
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
which becomes
To remove every second term, we combine it with the series for
which becomes (be careful combining minus signs and
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
which is just the series above for
Compare at this point the hyperbolic functions, which you may have been introduced to already. In particular, note the definition of
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.