Transforming Trig Functions

Date: 05/08/2001 at 06:11:46
From: Alex
Subject: Transforming trig functions

How would you express 3sin(x) + 5cos(x) in terms of sin(x) so that the 
graph is the same as the first equation?

Date: 05/08/2001 at 08:10:19
From: Doctor Jerry
Subject: Re: Transforming trig functions

Hi Alex,

You have p*sin(a*t)+q*cos(a*t). This can be written as the sine 
function L*sin(a*t+c).

Here's how: let J = sqrt(p^2+q^2).

     p*sin(a*t)+q*cos(a*t) = J((p/J)sin(a*t)+(q/J)cos(a*t))

Since (p/J)^2+(q/J)^2 = 1, the point (p/J,q/J) is on the unit circle. 
Hence, there is an angle c, in radians, for which cos(c) = p/J and 
sin(c) = q/J. So

     p*sin(a*t)+q*cos(a*t) = J(cos(c)*sin(a*t)+sin(c)*cos(a*t))

                           = J*sin(a*t+c)

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/18/2001 at 03:46:26
From: Alex
Subject: Transforming trig functions

Another equation: find an expression for y = a sin(x) + cos(x) in 
terms of cos(x).

Thank you.

Date: 05/18/2001 at 07:46:51
From: Doctor Jerry
Subject: Re: transforming trig functions

Hi Alex,

In the expression 

     y = a sin x + cos x

the factor a affects the amplitude of the graph. When you do the 
algebra I suggested in my earlier note, you will obtain

     y = sqrt(a^2+1) sin(x+c)

So we have a graph with period 2*pi/1 = 2*pi, amplitude sqrt(a^2+1), 
and horizontal displacement (sometimes  called  phase shift) of -c.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/