Problem with this trig identity

  • Thread starter mcah5
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
Tide
Science Advisor
Homework Helper
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In order to have a trig identity you must have an equal sign! What is that expression supposed to equal?
 
  • #3
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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 [tex]x_0,y_0,\dot {x_0},\dot {y_0}[/tex] 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:
  • #4
uart
Science Advisor
2,797
21
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))
 

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