中4數廷三角學題

a)sinθ + cosθ = 7/13(sinθ + cosθ)² = 49/169sin²θ + cos²θ + 2sinθ cosθ = 49/1691 + 2sinθ cosθ = 49/169- 2sinθ cosθ = 120/169sin²θ + cos²θ - 2sinθ cosθ = 1 + 120/169(sinθ - cosθ)² = 289/169sinθ - cosθ = 17/13 (捨去因 3π/2 < θ < 2π , 有 sinθ < 0 < cosθ ) 或
sinθ - cosθ = - 17/13
b)

由a) :sinθ + cosθ = 7/13 .....(1)
{
sinθ - cosθ = - 17/13 .....(2)(1) - (2) :2cosθ = 7/13 + 17/13cos θ = - 12/13故sin(3π/2 - θ)= - cos θ= 12/13


2011-06-21 17:21:16 補充：
修正 b):

(1) - (2) :

2cosθ = 7/13 + 17/13

cos θ = 12/13

故

sin(3π/2 - θ)

= - cos θ

= - 12/13

Sinθ+Cosθ = 7/13 其中 3π/2<θ<2π
求Sinθ－Cosθ的值
Sol
3π/2<θ<2π
Sinθ<0，Cosθ>0
Set Sinθ－Cosθ=p<0
Sinθ=p/2+7/26
Cosθ=7/26－p/2
(p/2+7/26)^2+(7/26－p/2)^2=1
(p+7/13)^2+(7/13－p)^2=4
2p^2+98/169=4
2p^2=578/169
p^2=289/169
p=－17/13

a. 17/13
b. -cos(2)