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Math Help - inverse trig

  1. #1
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    inverse trig

    I have a question on a test, but have no clue how to work the problem. My instructions are to evaluate the expression exactly:

    sin (arctan 2)

    My possible answers are:

    A=5sq rt 2/2
    b=2 sq rt5/5
    c=2sq rt 5
    d 5 sq rt 2
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  2. #2
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    Quote Originally Posted by scottsw1 View Post
    ...evaluate the expression exactly: sin (arctan 2)
    Draw a right triangle; the particular dimensions don't matter.

    Label the non-right base angle as \theta.

    By definition, \arctan(2)\, =\, \theta means that \tan(\theta)\, =\, 2\, =\, \frac{2}{1}

    So label the "opposite" and "adjacent" sides of the triangle to reflect this.

    Use the Pythagorean Theorem to find the value of the hypotenuse.

    Since \arctan(2)\, =\, \theta, then \sin(\arctan(2)) means \sin(\theta).

    Read the value of the sine from your picture.
    Last edited by stapel; April 5th 2009 at 04:40 PM. Reason: Forgot a "close paren".
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  3. #3
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    It looks like to me hypontenous is sq rt 5. Please help me with the equation, I didn't understand what you were trying to describe to me. Thanks for taking the time to help!!
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  4. #4
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    Quote Originally Posted by scottsw1 View Post
    I have a question on a test, but have no clue how to work the problem. My instructions are to evaluate the expression exactly:

    sin (arctan 2)

    My possible answers are:

    A=5sq rt 2/2
    b=2 sq rt5/5
    c=2sq rt 5
    d 5 sq rt 2
    Let \alpha = \arctan 2 \Rightarrow \tan \alpha = 2.

    You need to find the value of \sin \alpha. Do you know how to use a right-angle triangle to do this (stapel has described how)?
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  5. #5
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    I came up with the answer that sin=sq rt 5. Is this correct? If it is, it does not equal the multiple choice answers on the paper.
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  6. #6
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    I solved the problem!! It was 2 sq rt 5.

    Now here is another:

    6 arcsin x= pi

    Any suggestions on how to solve?
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  7. #7
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    Quote Originally Posted by scottsw1 View Post
    I solved the problem!! It was 2 sq rt 5. Mr F says: No doubt you meant to include a vinculum ....

    Now here is another:

    6 arcsin x= pi

    Any suggestions on how to solve?
    \Rightarrow \arcsin x = \frac{\pi}{6} \Rightarrow x = \sin \frac{\pi}{6} \, ....
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