Question about Euler's formula

Question

I have a question about Euler's formula

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

I want to show

$$\sin(ax)\sin(bx) = \frac{1}{2}(\cos((a-b)x)-\cos((a+b)x))$$

and

$$ \cos(ax)\cos(bx) = \frac{1}{2}(\cos((a-b)x)+\cos((a+b)x))$$

I'm not really sure how to get started here.

Can someone help me?

score 2 · Answer 1 · answered Aug 07 '15 at 20:25

2

HINT:

$$\begin{align} \cos (x\pm y)+i\sin(x\pm y)&=e^{i(x\pm y)}\\\\ &=e^{ix}e^{\pm iy}\\\\ &=\left(\cos x+i \sin x\right)\left(\cos y\pm i \sin y\right)\\\\ &=(\cos x \cos y\mp \sin x \sin y)+i(\sin x\cos y\pm \sin y\cos x) \end{align}$$

answered Aug 07 '15 at 20:25

Mark Viola

179,405

I'm confused on your pluses and minuses? How am I suppose to know which one applies to which? YOu have a plus/minus on everything. – Jackson Hart Aug 07 '15 at 20:27
1

The convention is to follow the upper sign consistently to get one result. Follow the lower sign consistently to get the other result. – Mark Viola Aug 07 '15 at 20:29
@JacksonHart I am happy to delete the answer if this was not useful. – Mark Viola Aug 07 '15 at 21:48
Do not delete it, please ! It is a useful answer. Cheers. – Claude Leibovici Aug 08 '15 at 04:57
@ClaudeLeibovici Hello my friend! How are you? – Mark Viola Aug 08 '15 at 04:59

haqnatural · Accepted Answer · 2015-08-12T07:42:18.833

2

$$\sin { \left( ax \right) } \sin { \left( bx \right) =\left( \frac { { e }^{ aix }-{ e }^{ -aix } }{ 2i } \right) \left( \frac { { e }^{ bix }-{ e }^{ -bix } }{ 2i } \right) } =\frac { { e }^{ \left( a+b \right) ix }-e^{ \left( a-b \right) ix }-{ e }^{ \left( b-a \right) ix }+{ e }^{ -\left( a+b \right) ix } }{ -4 } \\ =-\frac { 1 }{ 2 } \left( \frac { { e }^{ \left( a+b \right) ix }+{ e }^{ -\left( a+b \right) ix } }{ 2 } -\frac { { e }^{ \left( a-b \right) ix }+{ e }^{ -\left( a-b \right) ix } }{ 2 } \right) =\frac { 1 }{ 2 } \left( \cos { \left( a-b \right) x-\cos { \left( a+b \right) x } } \right) $$

same method you can do with $\cos { \left( ax \right) \cos { \left( bx \right) } } $

Edit: $$\int { \sin { \left( ax \right) \sin { \left( bx \right) } } dx=\frac { 1 }{ 2 } \int { \left[ \cos { \left( a-b \right) x-\cos { \left( a+b \right) x } } \right] dx=\quad } } $$$$\frac { 1 }{ 2 } \int { \cos { \left( a-b \right) xdx } } -\frac { 1 }{ 2 } \int { \cos { \left( a+b \right) xdx= } } $$

now to order calculate $\int { \cos { \left( a+b \right) xdx } } $ write $$t=\left( a+b \right) x\quad \Rightarrow \quad x=\frac { t }{ a+b } \quad \Rightarrow dx=\frac { 1 }{ a+b } dt\\ \int { \cos { \left( a+b \right) xdx=\frac { 1 }{ a+b } \int { \cos { \left( t \right) } dt=\frac { 1 }{ a+b } \sin { \left( t \right) = } } \frac { 1 }{ a+b } \sin { \left( a+b \right) x } } } +C\\ $$

edited Aug 12 '15 at 07:42

answered Aug 07 '15 at 20:28

haqnatural

21,578

Do you know how this result could be used to calculate the integrals of sin(ax)sin(bx)dx? and integral of cos(ax)cos(bx)dx? – Jackson Hart Aug 07 '15 at 20:33
$\int { \sin { \left( ax \right) \sin { \left( bx \right) } } dx=\frac { 1 }{ 2 } \int { \left( \cos { \left( a+b \right) x-\cos { \left( a-b \right) x } } \right) dx=\frac { 1 }{ 2\left( a+b \right) } \int { \cos { \left( a+b \right) xd\left( a+b \right) x } } -\frac { 1 }{ 2\left( a-b \right) } \int { \cos { \left( a-b \right) xd\left( a-b \right) x= } } } } \ =\frac { 1 }{ 2\left( a+b \right) } \sin { \left( a+b \right) x-\frac { 1 }{ 2\left( a-b \right) } \sin { \left( a-b \right) x } } +C$ – haqnatural Aug 07 '15 at 20:40
could you add the integral to your answeR? it is overflowing on my page and i cannot read all of it. – Jackson Hart Aug 07 '15 at 21:22
Also, it seems like you got a (a+b) factor outside the integral before integrating, where did this come from – Jackson Hart Aug 07 '15 at 21:23
1

ok,i will expand my post with integral – haqnatural Aug 07 '15 at 21:26
ok thanks! ill look for the edit – Jackson Hart Aug 07 '15 at 21:36
ask ,if it is not clear – haqnatural Aug 07 '15 at 21:45
Shouldnt the sin definition have an "i" in the denominator? – Jackson Hart Aug 12 '15 at 03:11

score 0 · Answer 3 · edited Aug 07 '15 at 20:53

0

Try using the identity

$$\sin A \cos B \equiv \tfrac{1}{2}\left(\sin(A + B) + \sin(A - B)\right)$$

edited Aug 07 '15 at 20:53

Fly by Night

32,272

answered Aug 07 '15 at 20:10

pg18

1

1

how would this be using eulers formula? – Jackson Hart Aug 07 '15 at 20:20

Alex R. · Answer 4 · 2015-08-07T20:17:09.800

0

If you want to just use Euler's equation and no trig identities (except the fact that $\sin(x),\cos(x)$ are odd/even respectively), write $\sin(x)=\frac{\exp(ix)-\exp(-ix)}{2i}$ and $\cos(x)=\frac{\exp(ix)+\exp(-ix)}{2}$ and go to town simplifying the product $\sin(ax)\sin(bx)$.

edited Aug 07 '15 at 20:17

answered Aug 07 '15 at 20:11

Alex R.

32,771

How does this specifically use exp(ix) = cos(x)+isin(x)? Sorry the methodology here isn't entirely clear to me. – Jackson Hart Aug 07 '15 at 20:13
@JacksonHart: Write down $\exp(ix)-\exp(-ix)$ in terms of $\sin(x),\cos(x)$ and use the fact that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$. This will prove the formula for $\sin(x)$. Do the same for $\cos(x)$. – Alex R. Aug 07 '15 at 20:15
I have sin(ax), not sin(x). Would that just make it exp(iax)? – Jackson Hart Aug 07 '15 at 20:18

Question about Euler's formula

4 Answers4

Linked