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Math Help - Double and Half Angle Trouble...

  1. #1
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    Double and Half Angle Trouble...

    There are two equations in my HW that im having trouble with. They happen to be next to each other so I took a picture.

    They are separated by a white line, the answer I got is in blue and the answer the book gives is in red.

    Really appreciate the assistance, as I have a test tomorrow and this stuff will be included.

    First one is half angle, second is double..


    What am I missing here? The formulas are correct, and I believe I'm doing them right.
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  2. #2
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    Your answer to the second problem is correct, you just have to simplify it further. \sqrt{168}=2\sqrt{42}.

    I don't understand what you did on the first problem. You need to find tan(\frac{1}{2}\theta) right?
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  3. #3
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    Thanks for the response, I feel stupid now

    For the first one, yes we're given tan(theta) = sqrt(7)/3
    Then you need to find tan(theta/2) or as you said tan(\frac{1}{2}\theta).

    I made the assumption that since sinx/cosx = tanx that sinx=sqrt(7) and cosx=3.

    So i substituted those into a tangent formula and ended up with the answer which I circled in blue.
    Last edited by topdnbass; November 10th 2009 at 06:19 PM.
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  4. #4
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    Quote Originally Posted by topdnbass View Post
    Thanks for the response, I feel stupid now

    For the first one, yes we're given tan(theta) = sqrt(7)/3
    Then you need to find tan(theta/2) or as you said tan(\frac{1}{2}\theta).

    I made the assumption that since sinx/cosx = tanx that sinx=sqrt(7) and cosx=3.

    So i substituted those into a tangent formula and ended up with the circled blue one.
    Ok, I see where you went wrong. You can't make that assumption. Sine can't equal the \sqrt{7} because sine is always less than 1. To solve this one, I would just use the formula for the tangent of twice an angle:

    tan(2A)=\frac{2tan(A)}{1-tan^2(A)}

    If we let \theta=2A then A=\frac{1}{2}\theta. Therefore:

    tan(\theta)=\frac{2tan(\frac{1}{2}\theta)}{1-tan^2(\frac{1}{2}\theta)}

    \frac{\sqrt{7}}{3}=\frac{2tan(\frac{1}{2}\theta)}{  1-tan^2(\frac{1}{2}\theta)}

    If you do the algebra you should obtain a quadratic equation where you can solve for tan(\frac{1}{2}\theta)
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