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Math Help - Inverse trig functions

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Inverse trig functions

    one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

    how is arctan square root of 3 divided by 3 = pi/6?

    how is (arctan x = pi/4) equal to 1?

    how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

    and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?

    thank you
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

    how is arctan square root of 3 divided by 3 = pi/6?

    how is (arctan x = pi/4) equal to 1?

    how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

    and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?

    thank you
    first part cos(x)=\frac{e^{ix}+e^{-ix}}{2}\Rightarrow{cos(0)=\frac{e^{0i}+e^{-0i}}{2}}\Rightarrow{\frac{1+1}{2}=1} and for the other sin(x)=\frac{e^{ix}-e^{-ix}}{2}\Rightarrow{ \frac{1-1}{2}=0} we jumped the steps using the same logic as above...for the second thing because tan\bigg(\frac{\pi}{4}\bigg)=1...for the next one follow same logic and for the next one...for the last one it is because tan(30)=\frac{\sqrt{3}}{3}
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