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Math Help - Proving an identity.

  1. #1
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    Proving an identity.

    Hello! In need of help yet again. I've seemed have gotten stumped :|

    sin(3x)+sin(x) = 2sin(2x)cos(x)

    Took the left side and changed it to:
    sin(2x+x)+sin(x)

    From there I got it equaled to:

    sin(2x)cos(x)+cos(2x)sin(x)+sin(x)

    from then on I get confused. I tried using the double angle formulas to expand the sin(2x) and the cos(2x) but that gets me no where. Any ideas on what I could do next?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dagbayani481 View Post
    Hello! In need of help yet again. I've seemed have gotten stumped :|

    sin(3x)+sin(x) = 2sin(2x)cos(x)

    Took the left side and changed it to:
    sin(2x+x)+sin(x)

    From there I got it equaled to:

    sin(2x)cos(x)+cos(2x)sin(x)+sin(x)

    from then on I get confused. I tried using the double angle formulas to expand the sin(2x) and the cos(2x) but that gets me no where. Any ideas on what I could do next?
    You're on the right track...here's an interesting observation:

    \begin{aligned}\sin(2x)\cos(x) + \cos(2x)\sin(x)+\sin(x) &= \sin(2x)\cos(x)+2\sin x\left(\dfrac{\cos(2x)+1}{2}\right)\\ &=\sin(2x)\cos(x)+2\sin x\cos^2x\\ &=\sin(2x)\cos(x)+(2\sin(x)\cos(x))\cos(x)\\ &= \sin(2x)\cos(x)+\sin(2x)\cos(x)\\ &=2\sin(2x)\cos(x)\end{aligned}



    I hope this makes sense!
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    But it's a direct conclusion by transforming the left side into a product by using sum-product identity.
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