solving equations

[deep basin]

#1020426321261756536 #/analysis

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[raw basin]

1. Bring arctg(x) to the left side, then take tangent of both sides.
2. Take sine of both sides.
3. This is a quadratic equation in arcsin(x).
   If you get several solutions in 1 and 2, make sure to check them.

[deep basin]

For the first one I mean what after going into tan

[raw basin]

Do you remember the fomula for tan(a + b)?

[deep basin]

No

[raw basin]

tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b))

And, of course, tan(arctan(x)) = x.

[deep basin]

Got it thanks

[raw basin]

You're welcome!

[deep basin]

Sorry to bother again

[deep basin]

What about this one?

@raw basin

[raw basin]

Hmm...

[deep basin]

Interesting one, isn't it?

[raw basin]

Well, I think I have the general idea of why that restriction is needed, but not sure how to arrive to it.

[deep basin]

Okay go ahead sharing it up, maybe I can work with it

[raw basin]

Well, basically, we know that the range of arctan(x) is (-π/2, π/2). So, when we add arctangents of two independent variables, the range of the result is (-π, π).
However, we also have an arctangent on the other side, which must still have range (-π/2, π/2).

So, there needs to be some sort of restriction on x and y.

[deep basin]

So are we going to kinda of restrict the arctan(x) + arctan(y) between π/2 and -π/2

?

[raw basin]

Yes.

Not 100% sure how to do that, though.

[deep basin]

Ok tysm

[deep basin]

Got an idea about that ?

[raw basin]

For now, no.

[deep basin]

What about the last question here

@raw basin

[raw basin]

Hm...
Judging by its derivative f'(x) = 2sgn(x)/√(1 - x^2), I'd say f(x) = 2arcsin(|x|).

[deep basin]

How did you find that?

[raw basin]

The derivative? Well, just by differentiating f(x).

[deep basin]

Got it thank ya Man, I just asked bout the absolute value

Why has it to be done there ?

[raw basin]

Oh.
Well, suppose we have an even function f(x). Then sgn(x)f(x) is odd, as sgn(x) is odd.
Moreover, the derivative of an even function will be odd, and vice versa.
Using that, we can conclude the following:
Suppose we have an even function f(x) that has an odd antiderivative F(x). Then the antiderivative of f(x)sgn(x) will be F(|x|).
We can see that by differentiating the latter:
(d/dx)F(|x|) = f(|x|)(d/dx)|x| = f(|x|)sgn(x) = f(x)sgn(x)

[deep basin]

Brooooo, thank you sooooo much