高二三角函數的的一些題目不會寫?

1.
tan(x + 45°) = k
[tan x + tan(45°)] / [1 - tan x tan(45°)] = k
[tan x + 1] / [1 - tan x] = k
tan x + 1 = k - k tan x
k tan x + tan x = k - 1
(k + 1) tan x = k - 1
tan x = (k - 1)/(k+ 1)
tan²x = [(k - 1)/(k + 1)]²
tan²x = (k² - 2k + 1)/(k² + 2k + 1)
1 + tan²x = 1 + (k² - 2k + 1)/(k² + 2k + 1)
sec²x = [(k² + 2k + 1) + (k² - 2k + 1)]/(k² + 2k + 1)
1/cos²x = (2k² + 2)/(k² + 2k + 1)
cos²x = (k² + 2k + 1)/(2k² + 2)
2 cos²x - 1 = [2 (k² + 2k + 1)/(2k² + 2)] - 1
cos(2x) = [(2k² + 4k + 2) - (2k² + 2)]/(2k² + 2)
cos(2x) = 4k/[(2k² + 2)]
cos(2x) = 2k/(k² + 1)

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2.
90° < θ < 180°
450° < 5θ < 900°

sin(2θ) = cos(3θ)
sin(2θ) = sin(90° - 3θ)
sin(2θ) = sin(360°×2 + 90° - 3θ)
sin(2θ) = sin(810° - 3θ)
2θ = 810° - 3θ
5θ = 810°
θ = 162°


另解：
當90° < θ < 180° 時：
sinθ > 0  及  cosθ < 0

sin(2θ) = cos(3θ)
2 sinθ cosθ = 4 cos³θ - 3 cosθ
2 sinθ cosθ - 4 cos³θ + 3 cosθ = 0
cosθ (2 sinθ - 4 cos²θ + 3) = 0
cosθ = 0 (拾棄，因 cosθ < 0)  或  2 sinθ - 4 cos²θ + 3 = 0
2 sinθ - 4 (1 - sin²θ) + 3 = 0
4 sin²θ + 2 sinθ - 1 = 0
sinθ = [-2 ± √(2² + 4*2*1)] /(2*4)
sinθ = (-1 + √5)/4  或 sinθ = (-1 - √5)/4 (捨棄，因 sinθ > 0)
θ = 180° - 18°
θ = 162°

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3. (1)
因為 180° < θ < 270°，所以 90° < θ < 135°
因此 sinθ > 0  及   cosθ < 0

√[1 + cosθ] + √[1 - cosθ]
= √[2 (1 + cosθ)/2] + √[2 (1 - cosθ)/2]
= √[2 cos²(θ/2)] + √[2 sin²(θ/2)]
= -√2 cos(θ/2) + √2 sin(θ/2)
= 2 [sin(θ/2) (1/√2) - cos(θ/2) (1/√2)]
= 2 [sin(θ/2) cos45° - cos(θ/2) sin45°]
= 2 sin[(θ/2) - 45°]


3. (2)
由(1): √(1 + cosθ) + √(1 - cosθ) = 2 sin[(θ/2) - 45°]
當 θ 轉換成 (90° - θ) 時：
[1 + cos(90° - θ)] + √[1 - cos(90° - θ)] = 2 sin{[(90° - θ)/2) - 45°]
√(1 + sinθ) + √(1 - sinθ) = 2 sin[45° - (θ/2) - 45°]
√(1 + sinθ) + √(1 - sinθ) = 2 sin(θ/2)

1.
tan(x+45°)=k
    = (tan(x)+tan(45°))/(1-tan(x)tan(45°)) 
    = (1+tan(x))/(1-tan(x))
∴ tan(x) = (k-1)/(k+1)

cos(2x) = 2 cos^2(x) - 1
       = 2/sec^2(x) - 1
       = 2/(1+tan^2(x)) - 1
       = 2/[1+(k-1)^2/(k+1)^2] - 1
       = (k+1)^2/(k^2+1) - 1
       = 2k/(k^2+1)

2.
sin(2θ) = cos(3θ)
   <==> 2sin(θ)cos(θ) = cos^3(θ) - 3sin^2(θ)cos(θ)
                                  = (1-4sin^2(θ))cos(θ)
∴ cos(θ) = 0 或 2sin(θ) = 1-4sin^2(θ)

∵ 90°<θ<180°
∴ cos(θ) < 0 , sin(θ) > 0
∴ 4sin^2(θ) + 2sin(θ) - 1  = 0
∴ sin(θ) = (√5 - 1)/4
    θ = 180° - asin((√5 -1)/4) = 162°

3.
180°<θ<270° 則 
    sin(θ) < 0, cos(θ) < 0.
又: 90° < θ/2 < 135°, sin(θ/2) > 0, cos(θ/2) < 0.

因問題寫的不清楚, 暫以我所認為的問題回答:

(1)
 √(1+cos(θ)) + √(1-cos(θ))
    = √(2cos^2(θ/2)) +  √(2sin^2(θ/2))
    = -√2 cos(θ/2) + √2 sin(θ/2)
    = 2(sin(θ/2)cos(45°)-cos(θ/2)sin(45°))
    = 2 sin(θ/2-45°)

(2)
√(1+sin(θ)) + √(1-sin(θ))
   = √(1+cos(90°-θ)) + √(1-cos(90°-θ))
   = 2 sin((90°-θ)/2 - 45°)                      (利用 (1))
   = 2 sin(-θ/2)
   = -2sin(θ/2)