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/
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