三角學問題

1. 證明sin²θ / 1-cos²θ =1
L.H.S.
=sin²θ / 1-cos²θ
=sin²θ / (sin²θ+cos²θ-cos²θ) [sin²θ+cos²θ=1]
=sin²θ / sin²θ
=1
=R.H.S.

2. 證明sin²θ-cos²θ = 1-2cos²θ
L.H.S.
=sin²θ-cos²θ
=1-cos²θ-cos²θ [sin²θ+cos²θ=1 >sin²θ=1-cos²θ]
=1-2cos²θ
=R.H.S.
3. 證明tanθ√1-sin²θ = sinθ
L.H.S.
=tanθ√(1-sin²θ )
=tanθ√cos²θ
=sinθ/cosθ * (cosθ)
= sinθ
=R.H.S.

4. 證明1+tan²θ =1/cos²θ
L.H.S.
=1+tan²θ
=(cos²θ+sin²θ)/cos²θ
=1/cos²θ
=R.H.S.
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化簡：
1. (1 - cosθ)(1 + cosθ)
=1-cos²θ
=sin²θ

2. (sinθ - cosθ)² + 2sinθcosθ
=sin²θ-2sinθcosθ+cos²θ+ 2sinθcosθ
=1

3. 1 / tan²θ =sin²θ [呢題你打錯]
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已知 tanθ =3/4，計算：
1. 2sinθ + cosθ / sinθ
=2+1/ tanθ
=2+4/3
=10/3

2.3sinθ - 4cosθ / 3sinθ + 4cosθ
=[(3sinθ - 4cosθ )/cosθ]/ [(3sinθ + 4cosθ)/cosθ]
=(3 tanθ -4)/(3 tanθ +4)
=-7/25

証明：
1. sin²θ / 1-cos²θ =1(Tips : sin²θ = 1 - cos²θ)
2. sin²θ-cos²θ = 1-2cos²θ(Tips : sin²θ = 1 - cos²θ)


3. tanθ√1-sin²θ = sinθ (Tips : cos²θ = 1 - sin²θ)
4. 1+tan²θ =1/cos²θ (Tips : 1 = cos²θ/cos²θ)


化簡：
1. (1 - cosθ)(1 + cosθ)
= 1 - cos²θ
= sin²θ

2. (sinθ - cosθ)² + 2sinθcosθ
= 1 - 2sinθcosθ + 2sinθcosθ
= 1

3. 1 / tan²θ = sin²θ
??


已知 tanθ =3/4，計算：
1. 2sinθ + cosθ / sinθ
= 2 + cotθ
= 2 + (4/3)
= 10/3

2.(3sinθ - 4cosθ) / (3sinθ + 4cosθ)
= (3tanθ - 4)/(3tanθ + 4)
= 0