高一數學(三角函數)急!大師請進Thanks

1.
cos(x) + sin(x/2)
= 1 - 2sin²(x/2) + sin(x/2)
= -2[sin²(x/2) + (1/2)sin(x/2)] + 1
= -2[sin²(x/2) + (1/2)sin(x/2) + (1/4)²] + 2(1/4)² + 1
= -2[sin(x/2) + (1/4)]² + (9/8)

-2[sin(x/2) + (1/4)]² ≤ 0
所求最大值 = -2(0) + (9/8) = 9/8

當 sin(x/2) = 1，-2[sin(x/2) + (1/4)]² 為最小值。
所求的最小值 = -2[1 + (1/4)]² + (9/8) = -2


2.
(3sinx - 2)/(sinx + 2)
= (3sinx + 6 - 8)/(sinx + 2)
= [3(sinx + 2) - 8]/(sinx + 2)
= 3 - [8/(sinx + 2)]

當 sinx = -1，- [8/(sinx + 2)] 為最大值。
所求的取大值 = 3 - [8/(1 + 2)] = 1/3

當 sinx = 1，- [8/(sinx + 2)] 為最小值。
所求的最小值 = 3 - [8/(-1 + 2)] = -5


3.
sinx = 3cosx
sinx/cosx = 3
tanx = 3/1

sinx = 3/√(1² + 3²) = 3/√10
cosx = 1/√(1² + 3²) = 1/√10

sin2x = 2sinx cosx = 2•(3/√10)•(1/√10) = 6/10 = 3/5


4.
恆等式：cos2x = cos²x - sin²x

cos²(π/8) - sin²(π/8)
= cos[2(π/8)]
= cos(π/4)
= (√2)/2


5.
x² + 4αx + (3α + 1) = 0
兩根之和： tan(α/2) + tan(β/2) = 4α
兩根之積： tan(α/2)•tan(β/2) = 3α + 1

tan[(α/2) + (β/2)]
= [tan(α/2) + tan(β/2)] / [1 - tan(α/2)•tan(β/2)]
= 4α/[1 - (3α + 1)]
= 4/3

tan(α + β)
= tan2[(α/2) + (β)/2]
= 2tan[(α/2) + (β/2)] / {1 - tan²[(α/2) + (β/2)]}
= 2(4/3) / [1 - (4/3)²]
= (8/3) / [1 - (16/9)]
= (8/3) / -(7/9)
= -(8/3)•(9/7)
= -24/7

恭喜你答案全對

看起來是這樣