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Math Help - Verify Algebraically

  1. #1
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    Verify Algebraically



    This one really stumped me =[ Does anyone know how to verify it?
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  2. #2
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    Quote Originally Posted by Kitty216 View Post


    This one really stumped me =[ Does anyone know how to verify it?
    I think the (alpha)/3 distracts you. You're used to dealing up to (alpha)/2 only for angles.

    Since the alpha/3 is uniform throughout, then let's say angle alpha/ 3 = angle B.

    Then the original equation/identity will be
    sinBcosB = (1/2)sin(2B)
    ....which is easy.
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  3. #3
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    \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} = \frac{1}{2} \sin \frac{2\alpha}{3}

    <br /> <br />
\begin{aligned}<br />
LHS &= \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \\<br />
  &= \frac{1}{2} \left(2\sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \right)   \\<br />
\end{aligned}<br /> <br />

    Using  \sin 2x = 2 \sin x \cos x

     = \frac{1}{2} \sin \frac{2\alpha}{3}

     = RHS
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  4. #4
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    Wow.. replacing the alpha/3 with a B really helps! So at the end do I just replace the all the B's back with the alpha/3?
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  5. #5
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    Quote Originally Posted by Kitty216 View Post
    Wow.. replacing the alpha/3 with a B really helps! So at the end do I just replace the all the B's back with the alpha/3?
    Yes.

    Later on, in your future Math subjects, you will see that substitutions like that are common.
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  6. #6
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    Thanks for the help! =DD
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