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Trig Identity

  1. Jun 18, 2009 #1
    I forget how this one goes.

    A cos(x) + B sin (x) = C sin (x + invtan(?))

    How do you go about condensing both these terms into 1 like the above?
    Last edited: Jun 18, 2009
  2. jcsd
  3. Jun 18, 2009 #2


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    Staff: Mentor

  4. Jun 18, 2009 #3


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    Gold Member

    I'm not sure how much help the wikipedia page will be so I'll provide a relatively simple derivation here:


    Asin(x) + Bcos(x)

    We can define the sine and cosine of an angle y by considering a right triangle with side lengths A and B. The hypotenuse is then given by,

    C = sqrt(A2 + B2)

    Consequently, the sine and cosine of y are given by the following formulas,

    sin(y) = B/sqrt(A2 + B2)

    cos(y) = A/sqrt(A2 + B2)

    Substituting these values into the equation produces,

    C[sin(x)cos(y) + cos(x)sin(y)] = Asin(x) + Bcos(x)


    Csin(x + y) = Asin(x) + Bcos(x)

    Now, we only need determing an expression for y. Using our expressions for sin(y) and cos(y), we know that,

    tan(y) = B/A

    y = arctan(B/A) = tan-1(B/A)

    and consequently,

    Asin(x) + Bcos(x) = Csin(x + tan-1(B/A))

    Hope this helps!
  5. Jun 18, 2009 #4


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    Let A=Ccos(y) and B=Csin(y). So you see immediatedly that:
    and B/A=tany.
  6. Jun 19, 2009 #5


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    Hi HasuChObe! :smile:

    (I think I'm saying the same as other people, but let's just isolate the principle …)

    The object is to get the LHS to look like cos(x)sin(?) + sin(x)cos(?) :wink:
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