# sum of cosines

• Jan 14th 2009, 02:23 AM
qwerty321
sum of cosines

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

• Jan 14th 2009, 04:58 AM
Jester
Quote:

Originally Posted by qwerty321
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?
• Jan 14th 2009, 05:11 AM
qwerty321
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

• Jan 14th 2009, 10:02 PM
Sum of cosines
Hello qwerty321
Quote:

Originally Posted by qwerty321

Quote:

Originally Posted by qwerty321
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 $\displaystyle A$ to stand for $\displaystyle 300\pi t$ and $\displaystyle B$ for $\displaystyle 500 \pi t$:

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

Now make repeated use of the following identities:

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

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

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

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

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

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

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

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

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

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