Trigonometric Sum

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May 22nd 2009, 02:38 PM
magentarita

Trigonometric Sum

I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)
May 22nd 2009, 11:21 PM
Grandad

Trigonometry

Hello magentarita

Quote:

Originally Posted by magentarita

I would like you to show me how to get the amplitude and phase shift of the sum of the trigonometric sum below.

8sin(50pi*x+pi/4)+7sin(50pi*x-pi/3)

$8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)$

= $8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3)$ $-7\cos(50\pi x)\sin(\pi/3)$

$= \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)$

$= \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)$

$= r\sin(50\pi[x-\alpha])$ , where

$r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)$

and $\tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}$

Can you tidy up and simplify now?

Grandad
May 25th 2009, 04:30 AM
magentarita

yes...

Quote:

Originally Posted by Grandad

Hello magentarita $8\sin(50\pi x +\pi/4) +7\sin(50\pi x -\pi/3)$

= $8\sin(50\pi x)\cos(\pi/4) +8\cos(50\pi x)\sin(\pi/4) + 7\sin(50\pi x)\cos(\pi/3)$ $-7\cos(50\pi x)\sin(\pi/3)$

$= \Big(\frac{8}{\sqrt2}+\frac{7}{2}\Big)\sin(50\pi x) + \Big(\frac{8}{\sqrt2}-\frac{7\sqrt3}{2}\Big)\cos(50\pi x)$

$= \frac{1}{2\sqrt2}\Big((16+7\sqrt2)\sin(50\pi x)-(7\sqrt6-16)\cos(50\pi x)\Big)$

$= r\sin(50\pi[x-\alpha])$ , where

$r^2 = \tfrac18\Big((16+7\sqrt2)^2+(7\sqrt6-16)^2\Big)$

and $\tan(50\pi\alpha) = \frac{(7\sqrt6-16)}{(16+7\sqrt2)}$

Can you tidy up and simplify now?

Grandad

Yes, I can. This question is not easy.