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Deriving double-angle formula

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    My notes say:

    Cos(x+a) = CosxCosa - SinxSina


    3. The attempt at a solution

    When I differentiated each term (using the product rule twice on the right side) I got:

    (d/dx) { Sin(x+a) = SinxCosa + CosxSina }

    . Cos(x+a) = -SinxSina + CosaCosx + CosxCosa - SinaSinx

    . Cos(x+a) = 2CosxCosa - 2SinxSina

    Which is not right ... I'm pretty sure I'm just making a stupid mistake with my product rule, but I've tried it a couple times and I keep doing the same thing.

    Does anyone see why I'm getting 2's in front of everything when I shouldn't be?
     
  2. jcsd
  3. Oct 14, 2009 #2
    You are differentiating with respect to x; that means that cos(a) and sin(a) are just constants. Differentiate again, treat them as constants, and you'll get the correct sum formula for cos(x + a). Double-angle formula would be for cos2x.
     
  4. Oct 14, 2009 #3
    hmm ... I don't know if you're right.
    I've seen the double angle formula with two separate variables.
    Plus it shouldn't matter whether it's a constant or a variable, the formula should still work out if the end result is supposed to be in terms of both x and a.
     
  5. Oct 15, 2009 #4

    Office_Shredder

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    Re-word your question. Prove that for any constant a,

    cos(x+a)=cos(x)cos(a)-sin(x)sin(a)

    Do you see how this is equivalent?

    Basically, the error that you made is that even if a is a variable, not a constant, it's independent of x as an input. So when you differentiate with respect to x, you treat the a as constant. This is something covered in Calculus III classes usually (multivariable calc)
     
  6. Oct 15, 2009 #5
    Yes, you do use two different values for the sum formulas. However, you need to differentiate to get the sum formula for cosine, and you are using d/dx which is differentiating with respect to x. Did you notice that when you differentiated the left side, d/dx sin(x + 1), you treated a as a constant? Although a is a constant, you can let it be any value when using the formula, just like when you choose any value for x.
     
  7. Oct 15, 2009 #6
    My friends, first of all "x" & "a" can be both variable or can be a variable "x" and a constant "a" in the expression: Sin(x+a) = SinxCosa + CosxSina. To prove, try solving:
    Sin(30+5) = Sin30Cos5 + Cos30Sin5 and youll find out the formula works. Now, starting from the fact that they can both variable or a constant and a variable, we could think of two possible solutions to this problem which of course will lead to the same result: Cos(x+a) = CosxCosa - SinxSina.
     
  8. Oct 15, 2009 #7
    Lets try the first assumption, i.e., x is a variable and a as a constant:

    Sin(x+a) = SinxCosa + CosxSina--->differentiate both sides with respect to x.
    you'll get: cos(x+a) = [sinx*(0) + cosa*cosx] + [cosx*(0) + sina*(-sinx)]-->notice the derivative of cosa = 0 because the derivative of a constant is 0, likewise the derivative of sina.

    Simplifying, we get: Cos(x+a) = CosxCosa - SinxSina.--->which solves the problem.
     
  9. Oct 15, 2009 #8
    Now assuming that both "x" and "a" are variables, can you try solving the problem? I have the solution already, I just wanna see how you do it.
     
  10. Oct 18, 2009 #9
    Okay, I see what you guys mean.
    ... I tried it again and it worked out fine: I got

    Cos(x+a) = CosaCosx - SinaSinx

    ^_^
    whew
     
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