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Asinx + bcosy into single trig function

  1. Mar 5, 2015 #1
    I know that asinx + bcosx can be put into a single trig fom, but can also the above sum be put into a single trig function?
     
  2. jcsd
  3. Mar 5, 2015 #2

    jedishrfu

    Staff: Mentor

    If you could convert it to this format:

    A(cosy*sinx + siny*cosx)

    then yes it could be converted to Asin(y+x) as an example.

    However, that requires that a and b must have the following relationships:

    a=Acosy and b=Asiny

    which will not be true in the general case for any a and b.
     
  4. Mar 5, 2015 #3
    Thanks Jedi
    I am looking for a general solution for any a and b. Is this possible? I have tried the above method but currently limited by the above restrictions.
     
  5. Mar 5, 2015 #4

    jedishrfu

    Staff: Mentor

    I don't think there is any hope here. Usually this kind of problem might push someone to use Fourier methods but again your initial solution is probably optimal.

    I am not one of the Math mentors but perhaps @Mark44 or @HallsofIvy would have a better answer to your problem.
     
  6. Mar 5, 2015 #5

    Mark44

    Staff: Mentor

    Not that I'm aware of...
     
  7. Mar 9, 2015 #6

    Svein

    User Avatar
    Science Advisor

    Off the top of my head: Let [itex] A=\sqrt{a^{2}+b^{2}}[/itex] and [itex]a_{1}=\frac{a}{A}, b_{1}=\frac{b}{A} [/itex]. Then [itex]a_{1}^{2}+b_{1}^{2}=1 [/itex], therefore you can find an angle φ such that a1=cos(φ) and b1=sin(φ).
    ∴a⋅sin(x)+b⋅cos(x) = A(cos(φ)sin(x)+sin(φ)cos(x))=A⋅sin(φ+x).
     
  8. Mar 9, 2015 #7

    pasmith

    User Avatar
    Homework Helper

    Actually it is: [itex]\tan y = b/a[/itex] (there exists exactly one such [itex]y \in (-\pi/2, \pi/2)[/itex] for each possible choice of [itex]a[/itex] and [itex]b \neq 0[/itex]) and [itex]A = \sqrt{a^2 + b^2}[/itex].

    See also here.
     
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