Thread: arcus functions

Hi again!
Does anyone here know how to solve arctan(x)=arccos(2x)?
For x of course.

Originally Posted by a4swe

Hi again!
Does anyone here know how to solve arctan(x)=arccos(2x)?
For x of course.

Take the tangent of both sides,
tan(arctan(x))=tan(arccos(2x))
x=tan(arccos(2x))

Let,
arccos(2x)=u
Then,
cos u =2x=(2x)/1
Thus,
A right triangle angle is "u" with adjacent side 2x and hypotenuse 1.
By Pythagorean theorem the opposite side is sqrt(1-4x^2)
Then,
tan u
Is opposite over adjacent which is,
sqrt(1-4x^2)/(2x)

Thus we finally have,
x=sqrt(1-4x^2)/(2x)
Multiply by 2x,
2x=sqrt(1-4x^2)
Square,
4x^2=1-4x^2
Thus,
8x^2=1
Thus,
x^2=1/8

To me it seems like that is wrong (sorry) since I can't get that to be sqrt((1/2)(sqrt(2)-1)) witch my lecturer states as the correct answer.

Originally Posted by ThePerfectHacker

Take the tangent of both sides,
tan(arctan(x))=tan(arccos(2x))
x=tan(arccos(2x))

Let,
arccos(2x)=u
Then,
cos u =2x=(2x)/1
Thus,
A right triangle angle is "u" with adjacent side 2x and hypotenuse 1.
By Pythagorean theorem the opposite side is sqrt(1-4x^2)
Then,
tan u
Is opposite over adjacent which is,
sqrt(1-4x^2)/(2x)

Thus we finally have,
x=sqrt(1-4x^2)/(2x)
Multiply by 2x,
2x=sqrt(1-4x^2)

You were correct up to this step. You only multiplied both sides by 2, not 2x.

To continue the last line should have been:
2x^2 = sqrt(1 - 4x^2)
Square both sides:
4x^4 = 1 - 4x^2
4x^4 + 4x^2 - 1 = 0

We may use the quadratic formula to solve this for x^2:
x^2 = (-4 (+/-) sqrt[4^2 - 4*4*-1])/(2*4)

x^2 = -1/2 (+/-) sqrt(2)/2

Note we need to choose the "+" solution since the "-" solution gives a negative x^2.

Thus x = (+/-)sqrt{-1/2 + sqrt(2)/2}

(In case you saw the original, I found a mistake I made. This is the fixed copy.)

I will note that the + solution works, but the - solution doesn't. Perhaps it has to do with the domains on arccos and arctan?

-Dan

Yes I was about to answer you on the incorrect one.
Thanks both of you, now I can continue my work.