I want to know who can answer this.?

Question

I just want to see people's solutions to the problem, but I already know the answer.

If for a certain positive real number x, cos (arctan x) = x, find the value of x^2.

Answer 1

Call arctan(x) = t.

Then the equation implies that cos(t) = x. Draw a right triangle whose side adjacent to angle t is x, and whose hypotenuse is 1 (so that cos(t) = adj/hyp = x/1 = x).

Then the side opposite of angle t is sqrt(1-x^2) by the pythagorean theorem.

Now, we have that arctan(x) = t => tan(t) = x.

But tan(t) = opp/adj = sqrt(1-x^2)/x

Thus, we have

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

=>

1-x^2 = x^4

=>

x^4 + x^2 - 1 = 0.

Let A = x^2. Then we have

A^2 + A - 1 = 0,

whose solution is A = x^2 = 1/2 * (-1 + sqrt(5))

Answer 2

Let's just make us a little triangle there. You have a tan that is equal to x so, for now, make a side x and a side of 1. That about does it for tanx. Now you want the cos of that there angle to equal x so let's just make the 1 an x and the x an x^2 and the long side there a 1. So what we got here:

tanA = x^2 / x = x
cosA = x/1 = x

So we got a triangle with sides x, x^2 and a longest side of 1. Then using that squared thing we get:

1^2 = x^2 + (x^2)^2
1 = x^2 + x^4

using y = x^2 we have y^2 + y - 1 = 0
and y = (-1 +/- SQRT(1 - 4(1)(-1))/2
y = -1/2 +/- SQRT(5)/2
x = SQRT(-1/2 + SQRT(5)/2) since you want a real solution
x = 0.7862

Let's just check that thing out now:
A = 38.1727 degrees
cos(a) = 0.7862 Yup that looks like a good answer

Answer 3

Since Cos(z) can't exceed 1, this puts a limit on the arctan(x), which is less than 1 for x

Answer 4

Let arctan x = y, so cos y = x. We know that tan y = x, so
(sin y)/(cos y) = x. Therefore, (sin y)/x = x, so x^2 = sin y.