sum of cosines

qwerty321 · #1 $Old$ January 14th, 2009, 04:23 AM

I have the function: y(t) = (cos(300πt) + sin(500πt))^3

I need too expand the expression of y to get a sum of cosines with positive frequencies. Using
trigonometric identities find the frequencies of resulting sine waves.Can someone help me with that?
Thank you

Danny · #2 $Old$ January 14th, 2009, 06:58 AM

Quote:

Originally Posted by qwerty321 $View Post$

I have the function: y(t) = (cos(300πt) + sin(500πt))^3

I need too expand the expression of y to get a sum of cosines with positive frequencies. Using
trigonometric identities find the frequencies of resulting sine waves.Can someone help me with that?
Thank you

Why only cosine terms? Why not sine and cosine term?

qwerty321 · #3 $Old$ January 14th, 2009, 07:11 AM

because what I have to do is:

Expand the expression of

y(t) to get a sum of cosines with positive frequencies. Using trigonometric identities find the frequencies of resulting sine waves and compare with the frequency components obtained using MATLAB

Grandad · #4 $Old$ January 15th, 2009, 12:02 AM

Hello qwerty321

Quote:

Originally Posted by qwerty321 $View Post$

Quote:

Originally Posted by qwerty321 $View Post$

I have the function: y(t) = (cos(300πt) + sin(500πt))^3

I need too expand the expression of y to get a sum of cosines with positive frequencies. Using
trigonometric identities find the frequencies of resulting sine waves.Can someone help me with that?
Thank you

Using $A$ to stand for $300\pi t$ and $B$ for $500 \pi t$ :

$(\cos A + \sin B)^3 = \cos^3A+3\cos^2A\sin B + 3\cos A \sin^2B+\sin^3B$

Now make repeated use of the following identities:

$\cos^2x = \frac{1}{2}(\cos 2x -1)$

$\sin^2x = \frac{1}{2}(1-\cos 2x)$

$\cos x \cos y = \frac{1}{2}(\cos(x+y) + \cos(x-y))$

$\sin x \cos y = \frac{1}{2}(\sin(x+y) + sin(x-y))$

So, for example: $\cos^3A = \frac{1}{2}(\cos 2A -1)\cos A$

$= \frac{1}{2}\left(\frac{1}{2}(\cos3A + \cos A) -\cos A\right)$

$= \frac{1}{4}(\cos 3A - \cos A)$

And the second term is: $3 \cos^2 A \sin B = \frac{3}{2}(\cos 2A -1) \sin B$

$= \frac{3}{2}\left(\frac{1}{2}(\sin(B+2A)+\sin(B-2A)) -\sin B \right)$

Similarly with the last two terms. Finally, if you need to get an expression in terms of cosine only, you could use $\sin x = \cos(\pi /2 - x)$ .

Grandad