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Math Help - Deriving a compound angle formula..

  1. #1
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    Deriving a compound angle formula..

    The compound angle formulas can be used to derive many other formulas. One of these is the formula


    a) Show that the formula is valid for x = 2pi/3 and y = pi/3. (I did this, both sides were equal to square root of 3)

    b) Use appropriate equivalent trigonometric expressions to derive a similar formula for sin x - sin y.

    Answer for b:


    How would I show my work as to get to that answer for b? I don't really know what b is even asking.
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  2. #2
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    You will use the fact that \sin y is an odd function, i.e.,  - \sin y = \sin (-y).
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  3. #3
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    Simple question... are these two the same thing?:

    sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)

    and

    sin x - sin y = 2 sin ((x-y) / 2) cos ((x+y) / 2)

    if they are I think I got it
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  4. #4
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    You are asking is  u \cdot v = v \cdot u. You understand that it doesn't matter in which order I multiply numbers yes?
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  5. #5
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    Just making sure

    b) Use appropriate equivalent trigonometric expressions to derive a similar formula for sin x - sin y.

    So is all I have to do for that is come up with sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)? Do I have to do any calculations to prove that it works?
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  6. #6
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    No, you just need to use trig identities to show it is equal.
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  7. #7
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    Hello, kmjt!

    If they want us to derive "from scratch", it requires some acrobatics.


    b) Use appropriate equivalent trigonometric expressions to derive
    a similar formula for: . \sin x - \sin y

    Answer: . \sin x - \sin y \:=\:2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)
    Note that: . \begin{array}{ccc}x &=& \dfrac{x+y}{2} + \dfrac{x-y}{2} \\ \\[-3mm] y &=& \dfrac{x+y}{2} - \dfrac{x-y}{2} \end{array}


    Then:

    . . \sin x \;=\; \sin\left(\frac{x+y}{2} + \frac{x-y}{2}\right) . =\; \sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\;\;[1]

    . . \sin y \;=\; \sin\left(\frac{x+y}{2} - \frac{x-y}{2}\right) . =\; \sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\;\;[2]


    Subtract [2] from [1]: . \sin x - \sin y \;=\;2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)

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  8. #8
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    You guys have both been very helpful. Thanks!
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  9. #9
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    I accept with information:sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)
    and sin x - sin y = 2 sin ((x-y) / 2) cos ((x+y) / 2)
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