How do you prove $arccos (\frac{x}{2}) + arctan (x) = \frac{π}{2}$ $arccos(x/2) + arctan(x) = pi/2$ ?

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How do you prove $arccos (\frac{x}{2}) + arctan (x) = \frac{π}{2}$ $arccos(x/2) + arctan(x) = pi/2$ ?

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Answer 1 · 2016-05-24T21:31:24

You can't prove it. It is not always true.

Explanation:

For example, with $x = 1$ $x=1$ , we get

$arccos (\frac{1}{2}) + arctan (1) = \frac{π}{3} + \frac{π}{4} = \frac{7 π}{12} \neq \frac{π}{2}$ $arccos(1/2)+arctan(1) = pi/3+pi/4 = (7pi)/12 != pi/2$

It is true only for $x = 0$ $x=0$ , $x = \sqrt{3}$ $x=sqrt3$ , and $x = - \sqrt{3}$ $x=-sqrt3$ .

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How do you prove $arccos (\frac{x}{2}) + arctan (x) = \frac{π}{2}$ $arccos(x/2) + arctan(x) = pi/2$ ?

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How do you prove arccos(x/2) + arctan(x) = pi/2arccos(x2)+arctan(x)=π2arccos(x/2) + arctan(x) = pi/2?

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How do you prove $arccos (\frac{x}{2}) + arctan (x) = \frac{π}{2}$ $arccos(x/2) + arctan(x) = pi/2$ ?