有關三角函數的題目有幾題不會，請教教我(我需要計算過程)

1.
cotθ = 1/3
cosθ / sinθ = 1/3
sinθ = 3cosθ

sin²θ + cos²θ = 1
(3cosθ)² + cos²θ = 1
9cos²θ + cos²θ = 1
10cos²θ = 1
cos²θ = 1/10

[1 / (1+sinθ)] + [1 / (1-sinθ)]
= [(1-sinθ) / (1+sinθ)(1-sinθ)] + [(1+sinθ) / (1-sinθ)(1+sinθ)]
= [(1-sinθ)+(1+sinθ)] / (1-sinθ)(1+sinθ)
= 2 / (1-sin²θ)
= 2 / cos²θ
= 2 / (1/10)
= 20


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2.
答案： 2 個
(cosine 和 secant)


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3.
cosθ + sinθ = (√2)sinθ
(cosθ + sinθ) / sinθ = √2
(cosθ / sinθ) + (sinθ/cosθ) = √2
cotθ + 1 = √2
cotθ = -1 + √2

tanθ = 1 / cotθ
tanθ = 1 / (-1 + √2)
tanθ = (1 + √2) / (-1 + √2)(1 + √2)
tanθ = 1 + √2

tanθ + cotθ
= (1 + √2) + (-1 + √2)
= 2√2


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4.
{(1 - tan⁴θ) cos²θ} + tan²θ
= {[(cos⁴θ / cos⁴θ) - (sin⁴θ / cos⁴θ)] cos²θ} + tan²θ
= {[(cos⁴θ - sin⁴θ) / cos⁴θ)] cos²θ} + tan²θ
= [(cos²θ + sin²θ)(cos²θ - sin²θ) / cos²θ] + (sin²θ / cos²θ)
= [(cos²θ - sin²θ) / cos²θ] + (sin²θ / cos²θ)
= [(cos²θ - sin²θ) + sin²θ] / cos²θ
= cos²θ / cos²θ
= 1


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5.
(x - 1)² + (y - 1)²
= (cosθ + sinθ + 1 - 1)² + (cosθ - sinθ + 1 - 1)²
= (cosθ + sinθ)² + (cosθ - sinθ)²
= cos²θ + 2cosθsinθ + sin²θ + cos²θ - 2cosθsinθ + sin²θ
= 2cos²θ + 2sin²θ
= 2(cos²θ + sin²θ)
= 2


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6.
sinA、sinB 為方程式 3x² + x - 2 = 0之二根。
兩根之和： sinA + sinB = -1/3
兩根之積： sinAsinB = -2/3

sin²A + sin²B
= (sin²A + 2sinAsinB + sin²B) - 2sinAsinB
= (sinA + sinB)² - 2sinAsinB
= (-1/3)² - 2(-2/3)
= (1/9) + (4/3)
= (1/9) + (12/9)
= 13/9

2013-11-15 02:00:08 補充：
6. 另解：

3x² + x - 2 = 0
(3x - 2)(x + 1) = 0
x = 2/3 或 x = -1
兩根為 2/3 及 -1。

兩根平方之和
sin²A + sin²B
= (2/3)² + (-1)²
= (4/9) + 1
= 13/9 ...... (答案)