# Trigonometric Sum

• May 22nd 2009, 01:38 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, 10:21 PM
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)

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

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

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

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

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

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

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

Can you tidy up and simplify now?

• May 25th 2009, 03:30 AM
magentarita
yes...
Quote:

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

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

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

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

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

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

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

Can you tidy up and simplify now?