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Math Help - Taking 1/2 of arcsin/arccos/arctan

  1. #1
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    Taking 1/2 of arcsin/arccos/arctan

    Could I get some help with the following:

    sin{1/2acrsin(-7/25)}
    cos(1/2arctan 8/15)
    tan(1/2acrcos 3/5)


    I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are:

    1. -1/10 * 2^(1/2)
    2.
    4/17 * 17^(1/2)
    3.
    1/2

    Thanks in advance.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by kjv6675 View Post
    Could I get some help with the following:

    sin{1/2acrsin(-7/25)}
    cos(1/2arctan 8/15)
    tan(1/2acrcos 3/5)


    I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are:

    1. -1/10 * 2^(1/2)
    2.
    4/17 * 17^(1/2)
    3.
    1/2

    Thanks in advance.
    List of trigonometric identities - Wikipedia, the free encyclopedia

    This will be helpful here is the first

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

    This gives

    \displaystyle \sin\left( \frac{1}{2}\sin^{-1}(\frac{-7}{25})\right)=\pm \sqrt{\frac{1-\cos(\sin^{-1}(\frac{-7}{25}))}{2}}

    -\sqrt{\frac{1-\frac{24}{25}}{2}}=-\sqrt{\frac{1}{50}}=-\frac{\sqrt{2}}{10}
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  3. #3
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    Let \frac{1}{2}arcsin(\frac{-7}{25}) = \theta

    arcsin(\frac{-7}{25}) = 2\theta

    (\frac{-7}{25}) = \sin{2 \theta}

    (\frac{-7}{25}) =  \frac{2\tan{\theta}}{1 + \tan^2{\theta}}

    Simplify and solve the quadratic to find tanθ. from that find sinθ.

    Similarly try the other problems.
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  4. #4
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    Let \frac{1}{2}arcsin(\frac{-7}{25}) = \theta

    arcsin(\frac{-7}{25}) = 2\theta

    (\frac{-7}{25}) = \sin{2 \theta}

    (\frac{-7}{25}) =  \frac{2\tan{\theta}}{1 + \tan^2{\theta}}

    Simplify and solve the quadratic to find tanθ. from that find sinθ.

    Similarly try the other problems.
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  5. #5
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    Half angle formula...thanks! I don't know why I didn't think of that.
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