Riemann Sums:Approximating area with calculator

Using a calculator and *right* Riemann sums, approximate the area of the region described where n = 10,30,60,80 subintervals:
f(x)=2-2sin(x) for [-π/2,π/2]
Now, I know the answer for n=10 is 5.665. I have tried every incorrect combination I know using formula. I just can't seem to get that answer. What is the correct way?

Each subinterval has width: (π/2 − (−π/2)) / n = π/n

Area = subinterval width * ∑[i = 1 to n] f(leftmost point + i*subinterval width)

Area ≈ π/n ∑[i = 1 to n] f(−π/2 + iπ/n)
Area ≈ π/n ∑[i = 1 to n] (2 − 2sin(−π/2 + iπ/n))
Area ≈ π/n { ∑[i = 1 to n] (2) − ∑[i = 1 to n] 2sin(−π/2 + iπ/n) }
Area ≈ π/n { 2n − 2 ∑[i = 1 to n] sin(−π/2 + iπ/n) }
Area ≈ 2π − 2π/n ∑[i = 1 to n] sin(−π/2 + iπ/n)

∑[i = 1 to n] sin(−π/2 + iπ/n)

= sin(−π/2 + π/n) + sin(−π/2 + 2π/n) + sin(−π/2 + 3π/n) + . . .
+ sin(−π/2 + (n-3)π/n) + sin(−π/2 + (n-2)π/n) + sin(−π/2 + (n-1)π/n) + sin(−π/2 + nπ/n)

= sin(−π/2 + π/n) + sin(−π/2 + 2π/n) + sin(−π/2 + 3π/n) + . . .
+ sin(π/2 − 3π/n) + sin(π/2 − 2π/n) + sin(π/2 − π/n) + sin(π/2)

Note that sin(-x) = -sin(x), therefore sin(-x) + sin(x) = 0
Therefore, most terms in this sum get cancelled:
sin(−π/2 + π/n) + sin(π/2 − π/n) = 0
sin(−π/2 + 2π/n) + sin(π/2 − 2π/n) = 0
sin(−π/2 + 3π/n) + sin(π/2 − 3π/n) = 0

When n is odd, the only term that doesn't get cancelled is last term:
sin(π/2) = 1

When n is even, the only 2 terms that don't get cancelled are middle term and last term:
sin(0) + sin(π/2) = 0 + 1 = 1

Therefore: ∑[i = 1 to n] sin(−π/2 + iπ/n) = 1

Area ≈ 2π − 2π/n ∑[i = 1 to n] sin(−π/2 + iπ/n)
Area ≈ 2π − 2π/n (1)
Area ≈ 2π − 2π/n

n = 10
A = 2π − 2π/10
A = 9π/5
A = 5.655

n = 30
A = 2π − 2π/30
A = 29π/15
A = 6.07375

n = 60
A = 2π − 2π/60
A = 59π/30
A = 6.17847

n = 80
A = 2π − 2π/80
A = 79π/40
A = 6.20465