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Thread: Please check that

  1. #1
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    Please check that

    Can someone tell me the solution of following problems:
    1.
    $\displaystyle (sin(pi/2 - x) - cos^3x)/(sin(2x + 4pi))$

    My solution sinx/2


    2.
    I have to get other 2 values
    sinx=2/3
    _______
    sin2x = ?
    tan(pi/4-x) = ?

    3.
    y = x/5 - 1
    y = -5x/4 + 1


    k1 = 1/5
    k2 = -5/4


    $\displaystyle tan()=|(k2-k1)/(1+k1*k2)|$
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  2. #2
    Junior Member
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    1.) Yes, assuming that you are supposed to simplify that expression, I'm also arriving at $\displaystyle \frac{sinx}{2}$.

    2.) Hmm, I think there's some information missing here. In order to evaluate $\displaystyle sin2x$, you could use the identity $\displaystyle sin2x=2sinxcosx$. Since you have the value of $\displaystyle sinx$, $\displaystyle cosx$ can be found using $\displaystyle sin^2x + cos^2x = 1$ - but that gives two values for $\displaystyle cosx$, one positive and another negative. Usually, these type of questions will give a rage for the value of x, or make a comment such as "x is an acute angle", so that you can determine whether $\displaystyle cosx$ is positive or negative.

    3.) Please be clearer. Specifically, $\displaystyle tan()$ makes literally no sense.
    Last edited by Referos; Apr 13th 2009 at 12:06 PM. Reason: grammar
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  3. #3
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    2.) Lets say it is positive. For cosx I get square root of 5 divided by 3. For tanx I get 2 times square root of 5 divided by 5.

    Now I put this values to get sin2x (my solution is 4 times square root of 5 divided by 9) and to get tan(pi/4 - x), my solution to this is 1 - (2 times square root of 5 divided by 5) / 1 + (2 times square root of 5 divided by 5)

    3.) I ment tanφ

    and my solution is: 62 39'

    Thank you very much.
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