Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with this trig identity

  1. Sep 29, 2005 #1
    I ran into this trig identity trying to do my physics hw:


    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. jcsd
  3. Sep 29, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    In order to have a trig identity you must have an equal sign! What is that expression supposed to equal?
  4. Sep 29, 2005 #3
    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: Sep 29, 2005
  5. Sep 29, 2005 #4


    User Avatar
    Science Advisor

    Maybe you're looking for one of these two ?

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


    a cos(t) + b sin(t) = sqrt(a^2 + b^2) cos( t - arctan(b/a))
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook