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Thread: Proof that two trig functions = Pi/2

  1. #1
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    Proof that two trig functions = Pi/2

    The problem: prove that

    $\displaystyle ArcTan(x)+ArcCot(x) = \Pi/2$

    I am not sure how you "prove" this.
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  2. #2
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    Quote Originally Posted by skyd171 View Post
    The problem: prove that

    $\displaystyle ArcTan(x)+ArcCot(x) = \Pi/2$

    I am not sure how you "prove" this.
    Define $\displaystyle f(x) = \arctan(x) + arccot(x)$. Show that f'(x) = 0. Therefore .... Now note that f(1) = .... Therefore ....
    Last edited by mr fantastic; Feb 19th 2011 at 07:32 PM.
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    Thanks.
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    Quote Originally Posted by skyd171 View Post
    The problem: prove that

    $\displaystyle ArcTan(x)+ArcCot(x) = \Pi/2$

    I am not sure how you "prove" this.
    Actually, this is not true...

    $\displaystyle \displaystyle \arctan{x} + \textrm{arccot}\,{x} = \frac{\pi}{2}$, not $\displaystyle \displaystyle \frac{\Pi}{2}$...
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    Now that's going to confuse a lot of people, Prove It!

    More generally, tan and cotan, sin and cos, sec and csc "swap" complementary angles. That is, if $\displaystyle \theta$ and $\displaystyle \phi$ are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), $\displaystyle sin(\theta)= cos(\phi)$, [tex], and $\displaystyle sec(\theta)= csc(\phi)$. And, of course, those two angles add to $\displaystyle \frac{\pi}{2}$ (did I get it right, Prove It?).
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Now that's going to confuse a lot of people, Prove It!

    More generally, tan and cotan, sin and cos, sec and csc "swap" complementary angles. That is, if $\displaystyle \theta$ and $\displaystyle \phi$ are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), $\displaystyle sin(\theta)= cos(\phi)$, [tex], and $\displaystyle sec(\theta)= csc(\phi)$. And, of course, those two angles add to $\displaystyle \frac{\pi}{2}$ (did I get it right, Prove It?).
    I just meant because $\displaystyle \displaystyle \pi$ is the ratio of the circumference to the diameter of a circle, while $\displaystyle \displaystyle \Pi$ is the symbol for product :P
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