Factoring exponents from trig functions

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June 29th 2012, 07:57 AM
daigo

Factoring exponents from trig functions

I tend to forget some of the trigonometric functions and someone showed me how to derive the double angle identities from what I think is Euler's formula:

$e^{ix} = \cos x + i\sin x$
=
$e^{i2x} = \cos 2x + i\sin 2x$
=
$(e^{ix})^{2} = (\cos x + i\sin x)^{2}$

I have a question about this step...I understand how the '2' from the e^(i2x) was pulled out, but is it okay to pull the '2' out from the '2x' of the trigonometric functions and to square the right hand expression? How does that work?
June 29th 2012, 08:19 AM
Prove It

Re: Factoring exponents from trig functions

Quote:

Originally Posted by daigo

I tend to forget some of the trigonometric functions and someone showed me how to derive the double angle identities from what I think is Euler's formula:

$e^{ix} = \cos x + i\sin x$
=
$e^{i2x} = \cos 2x + i\sin 2x$
=
$(e^{ix})^{2} = (\cos x + i\sin x)^{2}$

I have a question about this step...I understand how the '2' from the e^(i2x) was pulled out, but is it okay to pull the '2' out from the '2x' of the trigonometric functions and to square the right hand expression? How does that work?

DeMoivre's Theorem states that $\displaystyle \begin{align*} \left[ r \left( \cos{x} + i\sin{x} \right) \right]^n = r^n\left[\cos{(nx)} + i\sin{(nx)}\right] \end{align*}$ , so yes, for Complex Numbers, you can do as it looks, which is to "pull the 2 out of the 2x in the trigonometric function".
June 29th 2012, 08:31 AM
emakarov

Re: Factoring exponents from trig functions

In fact, what is written in OP does not require DeMoivre's Theorem. From Euler's formula, we have

$e^{ix} = \cos x + i\sin x$ (1)

and

$e^{i2x} = \cos 2x + i\sin 2x$ (2)

Squaring both sides of (1), we get

$e^{i2x} = (\cos x + i\sin x)^2$ (3)

Now we compare the right-hand side of (2) and (3) and equate the real and imaginary parts to derive the identities for double-angle sine and cosine.
June 29th 2012, 08:31 PM
Prove It

Re: Factoring exponents from trig functions

Quote:

Originally Posted by emakarov

In fact, what is written in OP does not require DeMoivre's Theorem. From Euler's formula, we have

$e^{ix} = \cos x + i\sin x$ (1)

and

$e^{i2x} = \cos 2x + i\sin 2x$ (2)

Squaring both sides of (1), we get

$e^{i2x} = (\cos x + i\sin x)^2$ (3)

Now we compare the right-hand side of (2) and (3) and equate the real and imaginary parts to derive the identities for double-angle sine and cosine.

I was specifically answering the OP's question about "pulling out" the angle multiple as a power though.