Program to calculate sin(0.75) close, but not working? - fortran

Question

I have already posted a question regarding this problem, and have implemented what i've learned from the answers. I'm now at a point where the answers that are printed out on the screen are very close, but incorrect. Here is the code I now have:
program taylor
implicit none
integer :: k = 0
real :: y = 0.75
real :: x = 0.75
do while (abs(y - sin(0.75)) > 1E-6)
k = k + 1
y = y + ((y * (-x * x)) /( 2 * k * (2 * k + 1 )))
print *, y
end do
end program taylor
I can't seem to spot the error here, why is this not working? The first answer it prints is correct, but then it seems to get progressively lower, instead of closer to the true value. (The do while loop is to ensure the program stops when the absolute value between the calculation and the intrinsic sin function is less than 1E-6). I can see that the program is constantly reducing the final output, and the Taylor series is suppose to alternate between - and +, so how can I write that in my program?
Thanks.

Answer 1

The taylor series uses factorials and power of x.
taylor_sin(x) = sum[0->n] ( [(-1 ^ n) / (2n + 1)!] * x**(2n + 1) )
(Sorry for my poor keyboard math skills...)
program taylor
implicit none
integer :: n = 0
real :: fac = 1
real :: y = 0.75
real :: x = 0.75
real :: px = 1
integer :: sgn = -1
integer :: s = 1
do while (abs(y - sin(0.75)) > 1E-6)
n = n + 1
fac = fac * (2 * n) * ((2 * n) + 1)
px = px * (x * x)
s = s * sgn
y = y + ((s /fac) * (px * x))
print *, y
end do
end program taylor
Computing the factorial fac in a loop is rather trivial.
To compute x**(2n + 1), compute and keep x**(2n) and add an extra multiplication by x.
I use an integer to compute -1**n since using a real may lose precision over time.
[EDIT] silly me... you can save that extra multiplication by x by setting px to 0.75 before the loop.