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Thread: 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. $\displaystyle \sqrt{168}=2\sqrt{42}$.

    I don't understand what you did on the first problem. You need to find $\displaystyle tan(\frac{1}{2}\theta)$ 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 $\displaystyle tan(theta) = sqrt(7)/3$
    Then you need to find $\displaystyle tan(theta/2)$ or as you said $\displaystyle tan(\frac{1}{2}\theta)$.

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

    So i substituted those into a tangent formula and ended up with the answer which I circled in blue.
    Last edited by topdnbass; Nov 10th 2009 at 05:19 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 $\displaystyle tan(theta) = sqrt(7)/3$
    Then you need to find $\displaystyle tan(theta/2)$ or as you said $\displaystyle tan(\frac{1}{2}\theta)$.

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

    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 $\displaystyle \sqrt{7}$ because sine is always less than 1. To solve this one, I would just use the formula for the tangent of twice an angle:

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

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

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

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

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