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Thread: trig help....

  1. #1
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    trig help....

    i have two problems that i can't figure out how to do...


    1. Give the exact real number value for Tan inverse [tan 7pi/12]





    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ilaura View Post
    i have two problems that i can't figure out how to do...


    1. Give the exact real number value for Tan inverse [tan 7pi/12]





    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
    On a restriced interval $\displaystyle \arctan(x)$ is an inverse function of $\displaystyle \tan(x)$ so $\displaystyle \tan(\arctan(x))=x$ and for the second one use the fact that $\displaystyle \sin(\arccos(x))=\sqrt{1-x^2}$ and $\displaystyle \sin(\arcsin(x))=x$ after using the identity $\displaystyle \sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$
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  3. #3
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    i'm still really confused.....


    i tried doing something with the second one,


    i don't think its right but i got:


    6(square root of 2) + 3(square root of 21) -1
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ilaura View Post
    i'm still really confused.....


    i tried doing something with the second one,


    i don't think its right but i got:


    6(square root of 2) + 3(square root of 21) -1
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    $\displaystyle \begin{aligned}\sin\left(\arccos\left(\frac{1}{3}\ right)\right)+\arcsin\left(\frac{-2}{5}\right)&=\sin\left(\arccos\left(\frac{1}{3}\r ight)\right)\cos\left(\arcsin\left(\frac{-2}{5}\right)\right)\\
    &+\sin\left(\arcsin\left(\frac{-2}{5}\right)\right)\cos\left(\arccos\left(\frac{1} {3}\right)\right)\\
    &=\sqrt{1-\left(\frac{1}{3}\right)^2}\sqrt{1-\left(\frac{-2}{5}\right)^2}+\frac{1}{2}\cdot\frac{-2}{5}\\
    &=\frac{2\sqrt{42}-2}{15}
    \end{aligned}$
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    Quote Originally Posted by Mathstud28 View Post
    ......and for the second one use the fact that $\displaystyle \sin(\arccos(x))=\sqrt{1-x^2}$ and $\displaystyle \sin(\arcsin(x))=x$



    i don't understand where these come from....
    are they some kind of identities?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ilaura View Post
    i don't understand where these come from....
    are they some kind of identities?
    Yes List of trigonometric identities - Wikipedia, the free encyclopedia
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  7. #7
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    Quote Originally Posted by ilaura View Post
    [snip]
    and 2. Evaluate: (show your work. do not give a decimal approximation)

    Sin[cos inverse(1/3) + sin inverse(-2/5)]
    Let $\displaystyle \cos^{-1} \left( \frac{1}{3} \right)= \alpha \Rightarrow \cos \alpha = \frac{1}{3}$.

    Let $\displaystyle \sin^{-1} \left(-\frac{2}{5}\right) = \beta \Rightarrow \sin \beta = - \frac{2}{5}$.

    Then $\displaystyle \sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \, .... $
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    $\displaystyle \begin{aligned}\sin\left(\arccos\left(\frac{1}{3}\ right)\right)+\arcsin\left(\frac{-2}{5}\right)&=\sin\left(\arccos\left(\frac{1}{3}\r ight)\right)\cos\left(\arcsin\left(\frac{-2}{5}\right)\right)\\
    &+\sin\left(\arcsin\left(\frac{-2}{5}\right)\right)\cos\left(\arccos\left(\frac{1} {3}\right)\right)\\
    &=\sqrt{1-\left(\frac{1}{3}\right)^2}\sqrt{1-\left(\frac{-2}{5}\right)^2}+\frac{1}{2}\cdot\frac{-2}{5}\\
    &=\frac{2\sqrt{42}-2}{15}
    \end{aligned}$
    where did you get the 1/2? is it part of the identity?
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  9. #9
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    Quote Originally Posted by ilaura View Post
    where did you get the 1/2? is it part of the identity?
    Clearly it's a typo and should be 1/3.
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ilaura View Post
    My homework says:


    1. Give the exact real number value for arctan [tan (7pi/12)]

    steps would be appreciated. thanks
    Note that arctangent and tangent are inverses of each other.

    Thus, if we have $\displaystyle \tan^{-1}\left(\tan(\varphi)\right)$, it just reduces to $\displaystyle \varphi$. The same thing applies to $\displaystyle \tan\left(\tan^{-1}(\varphi)\right)$. It also equals $\displaystyle \varphi$.

    Knowing this, what do you think the answer is?
    Last edited by Chris L T521; Nov 30th 2008 at 11:22 AM.
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