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Math Help - Do not understand this question

  1. #1
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    Do not understand this question

    Find an algebraic expression equivalent to the given expression. (hint : use right triangles)

    a. sin(tan inverse(^-1 it says in paper) x)
    b. cos(tan inverse(^-1 it says in paper) x)

    what does it want me to do? How can i approach this problem?

    Thank You
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  2. #2
    Eater of Worlds
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    They want you to use a right triangle to express the inverse tangent functions in terms of x.

    Like so

    sin(tan^{-1}(x))=\frac{x}{\sqrt{x^{2}+1}}

    It can also be done algebraically.
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  3. #3
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    Quote Originally Posted by galactus View Post
    They want you to use a right triangle to express the inverse tangent functions in terms of x.

    Like so

    sin(tan^{-1}(x))=\frac{x}{\sqrt{x^{2}+1}}

    It can also be done algebraically.
    How did you get that?
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  4. #4
    o_O
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    Let \theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}

    So construct a right triangle with angle \theta such that the side opposite it has length x and the side adjacent it is 1. By the Pythagorean theorem, the hypotenuse must be \sqrt{x^2 + 1}

    So what does \sin \theta equal to? What was \theta in the first place?
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  5. #5
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    Quote Originally Posted by o_O View Post
    Let \theta = \tan^{-1} x \ \Leftrightarrow \ \tan \theta = \frac{x}{1} = \text{"} \frac{\text{opp}}{\text{adj}} \text{"}

    So construct a right triangle with angle \theta such that the side opposite it has length x and the side adjacent it is 1. By the Pythagorean theorem, the hypotenuse must be \sqrt{x^2 + 1}

    So what does \sin \theta equal to? What was \theta in the first place?
    nice, got it...
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