Verified answer

set the derivative to zero

0=1-(1/2) csc^2(t/2)

add (1/2) csc^2(t/2) to both sides

(1/2) csc^2(t/2)=1

multiply both sides by 2

csc^2(t/2)=2

csc(t/2)=plus or minus rad2

csc=1/sin

1/sin(t/2)=plus or minus rad2

sin(t/2)=+-(1/rad2)=+-(rad2/2)

t/2=π/4,3π/4,5π/4,or 7π/4

t=π/2,3π/2,5π/2, and 7π/2

the values that work on the range [π/4, 7π/4]. are t=π/2,3π/2

Well, you're most of the way there.

0 = 1 - (csc^2(t/2))/2

1 = (csc^2(t/2))/2

2 = csc^2(t/2) = 1+ cot^2(t/2)

1 = cot^2(t/2)

let t/2 = x

cot^2x = (cotx)^2

1 = (cotx)^2

take the sqrt of both sides

1 = cotx

arccot(1) = arccot(cotx) = x

cot is a/o (adjacent side over opposite side)

so, the angles x where the adjacent and the opposite are the same are π/4, 5π/4

remember t/2=x, so t/2 = π/4, 5π/4 Then t = π/2, 5π/2. But 5π/2 is > 7π/4.

1 - (1/2) csc^2 (t/2) =0

csc^2 (t/2) =2

1/(sin^2 (t/2)) =2

sin^2 (t/2) = 1/2

sin (t/2) = +/- (root 2)/2

t/2 = pi/4, 3pi/4

t = pi/2, 3pi/2

That's because there are no maxima or minima.