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Simple trig identity

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data
    I have had a brain malfunction and I need help to understand something simple. It would be great if someone could show the process of attaining the end form.

    How does; ##a\cos{(x)}+b\sin{(x)} = c\sin{(x+\phi)}## where a,b are arbitrary constants, c results from whatever identity is applied to the original expression (so is somehow related to a and b) and phi is a phase constant.

    2. Relevant equations

    sum-to-product identity; ##\sin{(u)}+\sin{(v)} = 2\sin{(\frac{u+v}{2})}\cos{(\frac{u-v}{2})} ##
    3. The attempt at a solution
    I have tried to use the sum-to-product trig identity by changing the cos into a sin. But as far as i can tell, this identity only applies when a=b=1. Thanks for any help.
    Last edited: May 6, 2015
  2. jcsd
  3. May 6, 2015 #2
    where did the theta come from? usually when ure dealing with trig there are formulas u just need to accept and not dwell too much on. if abc are abitrary then = 1 and u have sinx + cosx = sin(x+theta) not seen this formula ever unless u made a mistake somehwere in writing it.
  4. May 6, 2015 #3
    My bad, I was unclear in what i wrote. a and b are arbitrary, but c i suspect would depend on a and b.
  5. May 6, 2015 #4
    Also, the phi is a phase constant, but since i don't know how to derive one side from the other im not sure how phi came into the picture.
  6. May 6, 2015 #5
    Maybe if i add in the context it would be more clear.
    I am Trying to find the fourier series (but not calculating the fourier coefficients) of a trapped wave in a pipe closed at both ends.
    So What im having trouble with is the progression from the following line to the next;
    ##u(x,t)=\sum_{n=1}^\infty [A_n\cos{(n\omega t)}+B_n\sin{(n\omega t)}]sin{(n\lambda x)}##
    ##=\sum_{n=1}^\infty C_n\sin{(n\omega t+\phi_n)}sin{(n\lambda x)}##
    where ##\omega = c\lambda##
  7. May 6, 2015 #6


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    This is equivalent to
    $$\sin(x+\phi)=\frac a c\cos x+\frac b c\sin x,$$ but we have
    $$\sin(x+\phi)=\sin x\cos\phi+\cos x\sin\phi.$$ So if the formula in the quote above holds, we have
    $$\sin x\cos\phi+\cos x\sin\phi=\frac a c\cos x+\frac b c\sin x.$$ Since sin and cos are linearly independent functions, this would imply that if ##a,b,c## are independent of ##x##, we have
    $$\frac a c =\sin\phi,\qquad \frac b c =\cos\phi.$$ You called ##a,b,c## "constants". If you meant that they're also independent of ##\phi##, then the claim that you want to prove is false, because if ##c## is independent of ##\phi##, then ##a## isn't.
  8. May 6, 2015 #7
    Thanks you for your reply Fredrik.
    In the OP I called a and b constants and said that c was dependent on a and b. Also i said phi was a phase constant (By which i meant is also dependent on a and b)
    With regards to the original question, i have come to the conclusion (with no proof, but by inspection) that ##c=\sqrt{a^2+b^2}\text{ and }\phi = \arctan{(\frac{a}{b})}##
  9. May 6, 2015 #8


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    What you want to prove, [itex]a cos( x)+ b sin(x)= c sin(x+ \phi)[/itex], is very nearly the "sum rule", [itex]sin(x+ \phi)= sin(\phi)cos(x)+N cos(\phi)sin(x)[/itex]. Are you allowed to use that? If so then [itex]\phi[/itex] must be such that [itex]c sin(\phi)= a[/itex] and [itex]c cos(\phi)= b[/itex]. Then [tex]c^2 sin^2(\phi)+ c^2 cos(\phi)= c^2= a^2+ b^2[/tex]. And, of course, [itex]\frac{a sin(\phi)}{b cos(\phi)}= tan(\phi)= \frac{a}{b}[/itex].
  10. May 6, 2015 #9


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    OK, then you can ignore the last two sentences in my previous post, but the rest of what I said still holds. In particular the results ##a=c\sin\phi## and ##b=c\cos\phi##. It's very easy to use these two equalities to prove the two equalities you just mentioned in post #7.

    I think it's easier to go from the second line to the first (in post #5), than from the first to the second. You can rewrite ##c\sin(x+\phi)## as
    $$c(\sin x\cos\phi+\cos x\sin\phi)$$ and then define ##a=c\sin\phi## and ##b=c\cos\phi##.
  11. May 6, 2015 #10
    Yes it is indeed very easy, thanks for your guidance. I am now able to understand the progression from the first line to the second line in the post #5.
  12. May 6, 2015 #11
    This is how I ended up proving it. I have been wondering about it for a while now because my lecturer consistently uses it in our applied course so he can half the amount of terms (and typing) needed, but he never bothered showing us how he arrived at the conclusion. Now I see why, its quite trivial and I feel stupid. Lol.
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