若5/2π

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若5/2π
若5/2π

若5/2π
∵ (5 / 2)π < a < (11 / 4)π
∴ 5 π < 2a < (11 / 2)π
∵ sin 2a = - 4 / 5
∴ cos 2a = - 3 / 5
∴ cos ² a - sin ² a = - 3 / 5
(cos ² a - sin ² a)/ (sin ² a + cos ² a)= - 3 / 5
(1 - tan ² a)/(1 + tan ² a)= - 3 / 5
∴ 5(1 - tan ² a)= - 3 (1 + tan ² a)
5 - 5 tan ² a = - 3 - 3 tan ² a
2 tan ² a = 8
tan ² a = 4
tan a = ± 2
∵ (5 / 2)π < a < (11 / 4)π
∴ tan a = 2
∴ 2 tan(a / 2)/ (1 - tan ² a / 2)= 2
设 tan a / 2 = m ,则有:
2 m / (1 - m ²)= 2
2 m = 2 - 2 m ²
m ² + m = 1
m ² + m + (1 / 2)² = 1 + (1 / 2)²
(m + 1 / 2)² = 5 / 4
m + 1 / 2 = ±√5 / 2
m = ±√5 / 2 - 1 / 2
m1 = (√5 - 1)/ 2
m2 = - (√5 +1)/ 2
∵ (5 / 2)π < a < (11 / 4)π
∴ (5 / 4)π < a / 2 < (11 / 8)π
∴ tan a / 2 > 0
∴ tan a / 2 = (√5 - 1)/ 2