Numerical solutions help please (iteration)?

Question

The diagram: http://i164.photobucket.com/albums/u32/Codefusionlab/diagram.jpg shows a sector AOB of a circle O and radius r. The angle AOB is α radians, where 0 < α < π. The area of the triangle AOB is half the area of the sector.

(i) Show that α satisfies the equation:
x = 2 sin x.

(ii) Verify by calculation that α lies between π/2 and 2π/3.

(iii) Show that, if a sequence of values given by the iterative formula:
x = 1/3(x + 4 sin x)
converges, then it converges to a root of the equation in part (i).

That's wat im askin you

Thanks for your help

Answer 1

(i) Directly follows from the formulas for the area of the triangle by 2 sides (r) and angle (α) in-between: (1/2)*r^2*sinα, the area of a sector is (1/2)*r^2*α, so
(1/2)*r^2*sinα = (1/2)*(1/2)*r^2*α, or
sinα = (1/2)α or α = 2sinα;
(ii) The function f(x) = x - 2sinx for x belonging to [π/2, 2π/3] is continuous, f(π/2) = π/2 - 2 < 0, f(2π/3) = 2π/3 - sqrt(3) > 0, so there is a root of the equation x - 2sinx = 0 in the given interval /ONLY 1 root, because f'(x) = 1 + 2cos(x) > 0, so f is increasing!/;
(iii) Let L = lim x_{n} when n -> infinity /it exists according the condition!/, where recurrence relationship is:
x_{n+1} = (1/3)(x_{n} + 4 sin x_{n}), then, having n -> infinity we obtain L = (1/3)(L + 4 sin L) or 3L = L + 4 sin L, or
2L = 4 sinL or L = 2sinL, i.e. L = α from (i).

Answer 2

I just got part i.
If you find the area of the sector (as a fraction of the area of the circle), you get Area = (α/2π)(π r^2) = (α/2)r^2.
If you find the area of the triangle (using Area = (1/2)bh):
(where b = length AB)
(1/2) b = r sin (α/2), h = r cos (α/2), so Area = r^2 sin (α/2) cos (α/2).

You know that the area of the triangle is half the area of the sector, so r^2 sin (α/2) cos (α/2) = (1/2) (α/2) r^2,
or 2 sin (α/2) cos (α/2) = α/2.
The left should look like something from the half-angle theorem!
Yes! So sin α = α/2, or α = 2 sin α.
Hope this helps some.

Answer 3

where is the x ?