三角學的應用

13.已知tanθ=3/4。計算3sinθ-4cosθ/3sinθ+4cosθ
tanθ=3/4
sinθ/cosθ=3k/4k (where k is a constant)
so sinθ=3k and cos =4k
3sinθ-4cosθ/3sinθ+4cosθ
=(9k-16k)/(9k+16k)
=-7k/25
=-7/25//

14.化簡
(a)1/tan^2θ*sin^2 θ
=sin^2 θ/(sin^2θ/cos^2θ)
=sin^2 θ*cos^2θ / sin^2θ
=cos^2θ//
(b)(sinθ-cos θ)^2+2sinθcosθ
=sin^2θ-2cosθsinθ+2sinθcosθ+cos^2θ
=sin^2 θ+cos^2 θ
=1//

tanθ√(1-sin^2 θ)
=tanθ√(cos^2θ)
=sinθ/cosθ * cosθ
=inθ//

12a)tanθ= y
sinθ= y / √(y^2 + 1) = y√(y^2 + 1) / (y^2 + 1)
cosθ= 1 / √(y^2 + 1) = √(y^2 + 1) / (y^2 + 1)
b)tanθ= x/y
sinθ= x / √(x^2 + y^2) = x√(x^2 + y^2) / (x^2 + y^2)
cosθ= y / √(x^2 + y^2) = y√(x^2 + y^2) / (x^2 + y^2)

1 - tan^2 a = 1 - (sin^2 a/ cos^2 a) = 1 / cos^2 a