Problem with this trig identity

1. Sep 29, 2005

mcah5

I ran into this trig identity trying to do my physics hw:

a*cos(s)+b*cos(t)

I tried deriving it using an analogous approach to deriving the product-sum trig identity, but ran into problems. I was wondering if this trig identity exists, or if I should just try to find some clever way to solve the physics problem without the trig identity.

2. Sep 29, 2005

Tide

In order to have a trig identity you must have an equal sign! What is that expression supposed to equal?

3. Sep 29, 2005

mcah5

Edit2: Never mind, found another way without the blasted trig identity.

That's the problem, I don't know the identity, I was trying to derive it but ran into problems. I was wondering if there was a common name for the identity so I could google for it.

Edit: The problem I'm trying to solve is "Prove that superimposing initial conditions will lead to the superposition of of the corresponding motion of coupled oscillators." So basically, I'm trying to show that for the coupled equations

\frac {d^2x} {dt^2] = ax+by
\frac {d^2y} {dt^2} = cx+dy

that for the a solution x and y with initial conditions $$x_0,y_0,\dot {x_0},\dot {y_0}$$ is the superposition of two other solutions where the initial conditions add up to the original initial conditions. After a bunch of math, I need to show that:

a*cos(s+t) = b*cos(s+u)+c*cos(s+v) where a,s,t,b,s,u,c,v are a bunch of random constants

A trig identity for a*cos(s)+b*cos(t) would help solve this problem

Last edited: Sep 29, 2005
4. Sep 29, 2005

uart

Maybe you're looking for one of these two ?

cos(a-b) + cos(a+b) = 2 cos(a) cos(b)

or

a cos(t) + b sin(t) = sqrt(a^2 + b^2) cos( t - arctan(b/a))