ka diketahui bahwa cos 1/2 theta=akar x+1/2x maka x kuadrat-1/x kuadrat

Cos (1/2 Tetha) = akar [{1 + cos tetha)} / 2]

cos (1/2 tetha) = akar [(x+1) / 2x]
akar [{1 + cos tetha)} / 2] = akar [(x+1) / 2x]
 [{1 + cos tetha)} / 2] = [(x+1) / 2x]
(1 + cos tetha) (2x) = (x+1) (2)
2x + 2x cos tetha = 2x + 2
2x cos tetha = 2x - 2x + 2
cos tetha = 2/2x
cos tetha = 1/x
x = 1 / cos tetha

(x^2 -1) / x^2
= [(1/cos tetha)^2 - 1] / [1/cos tetha)^2]
= [(1/cos tetha)^2 / (1/cos tetha)^2 ] - [1/{(1/cos tetha)^2}]
= 1 - cos^2 tetha
= sin^2 tetha