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Math Help - Stuck on a problem in Homework (Trigonometric Identities and Formulas)

  1. #1
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    Stuck on a problem in Homework (Trigonometric Identities and Formulas)

    I've been trying to figure it out for the past 30 minutes but can't seem to get anywhere.

    I was able to get somewhere but I can't seem to get it balanced:

     \frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2

    I took the left side and I started by combining the two terms:

    \frac {sin(3x)cos(x) - cos(3x)sin(x)}{sin(x)cos(x)}

    From there I recognized that the numerator is an expansion of the sine subtraction formula so I rewrote it to:

    \frac {sin(3x-x)}{sinxcosx}

    which is basically

    \frac {sin(2x)}{sinxcosx}

    That's where I am stuck.
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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
    I've been trying to figure it out for the past 30 minutes but can't seem to get anywhere.

    I was able to get somewhere but I can't seem to get it balanced:

     \frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2

    I took the left side and I started by combining the two terms:

    \frac {sin(3x)cos(x) - cos(3x)sin(x)}{sin(x)cos(x)}

    From there I recognized that the numerator is an expansion of the sine subtraction formula so I rewrote it to:

    \frac {sin(3x-x)}{sinxcosx}

    which is basically

    \frac {sin(2x)}{sinxcosx}

    That's where I am stuck.
    Recall that \sin(2x)=2\sin x\cos x.

    You should be able to finish things off now.
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  3. #3
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    Ha! I had totally forgotten about the double angle formula. I should have known. Thank you!
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