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Thread: Am I doing this right?

  1. #1
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    Am I doing this right?

    It's a simplification. I'm supposed to simplify so that only one variable is present at the end.

    x = [sin(A+B)/2]/sin(A/2)

    Here's what I've done so far...

    x = 2[sin(A+B)] / 2[sin(A)

    x = sin(A + B) / sin(A)

    am I totally off....?

    Thank you!
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  2. #2
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    Quote Originally Posted by Tall Jessica View Post
    It's a simplification. I'm supposed to simplify so that only one variable is present at the end.

    x = [sin(A+B)/2]/sin(A/2)

    Here's what I've done so far...

    x = 2[sin(A+B)] / 2[sin(A)

    x = sin(A + B) / sin(A)

    am I totally off....?

    Thank you!
    is the equation ...

    $\displaystyle x = \frac{\sin\left(\frac{A+B}{2}\right)}{\sin\left(\f rac{A}{2}\right)}$

    or

    $\displaystyle x = \frac{\frac{\sin(A+B)}{2}}{\sin\left(\frac{A}{2}\r ight)}$

    or something else ?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    is the equation ...

    $\displaystyle x = \frac{\sin\left(\frac{A+B}{2}\right)}{\sin\left(\f rac{A}{2}\right)}$

    or

    $\displaystyle x = \frac{\frac{\sin(A+B)}{2}}{\sin\left(\frac{A}{2}\r ight)}$

    or something else ?
    The first one. Thanks :-)
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  4. #4
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    I don't know if the original expression can be "simplified" any more than it is already. You're not going to get an expression with only A or B alone unless there is a known relationship between A and B.

    I tried using the half-angle formula for sine ...

    $\displaystyle \sin\left(\frac{A+B}{2}\right) = \pm \sqrt{\frac{1 - \cos(A+B)}{2}}$

    $\displaystyle \sin\left(\frac{A}{2}\right) = \pm \sqrt{\frac{1 - \cos{A}}{2}}$


    doing the division ...

    $\displaystyle \pm \sqrt{\frac{1 - \cos(A+B)}{1 - \cos{A}}}$


    I also did the sum expansion for $\displaystyle \sin\left(\frac{A}{2} + \frac{B}{2}\right) $ , divided by $\displaystyle \sin\left(\frac{A}{2}\right)$ and got the following expression ...

    $\displaystyle \cos\left(\frac{B}{2}\right) + \cot\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right)$

    not much there as far as "simplification" goes, either.

    Is there more to this problem that you haven't posted?
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  5. #5
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    Well... I probably copied it wrong. I'll check tomorrow.

    Thank you and sorry!
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