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Math Help - Proof that two trig functions = Pi/2

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    Proof that two trig functions = Pi/2

    The problem: prove that

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

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

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

    I am not sure how you "prove" this.
    Define f(x) = \arctan(x) + arccot(x). Show that f'(x) = 0. Therefore .... Now note that f(1) = .... Therefore ....
    Last edited by mr fantastic; February 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

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

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

    \displaystyle \arctan{x} + \textrm{arccot}\,{x} = \frac{\pi}{2}, not \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 \theta and \phi are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), sin(\theta)= cos(\phi), [tex], and sec(\theta)= csc(\phi). And, of course, those two angles add to \frac{\pi}{2} (did I get it right, Prove It?).
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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 \theta and \phi are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), sin(\theta)= cos(\phi), [tex], and sec(\theta)= csc(\phi). And, of course, those two angles add to \frac{\pi}{2} (did I get it right, Prove It?).
    I just meant because \displaystyle \pi is the ratio of the circumference to the diameter of a circle, while \displaystyle \Pi is the symbol for product :P
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