Hi again!

Does anyone here know how to solve arctan(x)=arccos(2x)?

For x of course.

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- Sep 28th 2006, 10:39 AMa4swearcus functions
Hi again!

Does anyone here know how to solve arctan(x)=arccos(2x)?

For x of course. - Sep 28th 2006, 10:51 AMThePerfectHacker
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 - Sep 28th 2006, 10:56 AMa4swe
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.

- Sep 28th 2006, 11:02 AMtopsquark
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 - Sep 28th 2006, 11:18 AMa4swe
Yes I was about to answer you on the incorrect one.

Thanks both of you, now I can continue my work.