Finding the Value of arctan(1) Without a Calculator

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All I know about the arctan function is that its domain of def. is the whole real line and that the range of values is (-0,5*pi , 0,5*pi), and also that arctan(0)=0.

But is there a way to know that arctan(1)=(1/4)*pi without recognizing the decimal number the calculator gives?
 
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I think I figured out why after some brainwork:

1=tan (arctan (1))

so because tan(pi/4)=1, arctan(1)=pi/4
 
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i don't know if this will help but take the unit circle and reverse it. instead of looking for the angle then finding the value. look for the value then look for the angle that gives you the value. for arcsin(1), you would go along looking a the value of tan(angle) then when you find the value, the x in arcsin(x), you look at the angle that matches up with it.
 
That's easy. Just denote \arctan 1 =x and apply the \tan on both members of the equation. You'll find the eqn \tan x =1 which can easily be solved.

Daniel.
 
The simplest definition of tan(\theta) is "opposite side divided by near side" in a right triangle. If tan(\theta)= 1 then the two legs of the right triangle are the same length- it is an isosceles right triangle. What does that tell you about the angles?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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