三角學(完全唔明,幫幫手)

1.sinθ/cosθ= tanθ
2.sin²θ+cos²θ= 1
3.sin(90°- θ)= cos θ
4.cos(90°-θ)= sinθ
5.tan(90°-θ)=1/tanθ

6.(1/cos²θ)﹣1
=(1/cos²θ)﹣(cos²θ/cos²θ)
=(1﹣cos²θ)/cos²θ
=sin²θ/cos²θ
=tan²θ

7.(1/cosθ)﹣sinθtanθ
=(1/cosθ)﹣sinθ‧sinθ/cosθ
=(1/cosθ)﹣sin²θ/cosθ
=(1﹣sin²θ)/cosθ
=cos²θ/cosθ
=cosθ

8.tanθ√(1- sin2θ) = sinθ = ?
你是不是想證明恆等式?
左方
= tanθ√(1 - sin2θ)
= tanθ√cos2θ
= tanθ‧cosθ
= sinθ/cosθ‧cosθ
= sinθ
右方
= sinθ
=左方
∴tanθ√(1- sin2θ) ≡ sinθ

9.1+tan²θ=
=1 + sin2θ/cos2θ
=cos2θ/cos2θ + sin2θ/cos2θ
=(cos2θ+sin2θ)/cos2θ
=1/cos2θ

10.6sinθ/sin(90°-θ)
=6sinθ/cosθ
=6tanθ

11.4cos(90°-θ) - cosθ/tan(90°- θ)
=4sinθ - cosθ÷sin(90°- θ)/cos(90°- θ)
=4sinθ - cosθ÷cosθ/sin θ
=4sinθ - cosθ×sin θ/cosθ
=4sinθ - sin θ
=3sinθ



2007-08-27 23:24:37 補充：
1至5題是公式

Ans :
1.sinθ/cosθ=tanθ
2.sin²θ+cos²θ=1
3.sin(90°-θ)=cosθ
4.cos(90°-θ)=sinθ
呢d都係公式黎