f.4 maths

(a)
(cosθ)^2/[1+(sinθ)^2]=2/3
3(cosθ)^2=2[1+(sinθ)^2]
3(1-(sinθ)^2)=2+2(sinθ)^2
3-3(sinθ)^2=2+2(sinθ)^2
(sinθ)^2=1/5
(b)
90≦θ≦180
sinθ=1/√5
cosθ=-2/√5 [since (cosθ)^2+(sinθ)^2=1]
cosθ/(1+3sinθ)
=-2/√5/(1+3/√5)
=-2/[√5(1+3/√5)]
=-2/(√5+3)
=-2(√5-3)/(5-9)
=(√5-3)/2


2007-02-02 00:35:24 補充：
講到(b) part做得數多後﹐你一見到sinθ 有確定值θ的象限已知你就應該即時聯想到其他三角函數都可以計到再發現到cosθ/(1+3sinθ)冇明顯三角函數關係式可用就應該即時知道要找cosθ的值再代入去式子中計出答案記住最後要有理化呀我講真架

(cos)^2= 1-(sinθ)^2
so (cosθ)^2/[1+(sinθ)^2]=2/3
[1-(sinθ)^2]/[1+(sinθ)^2]=2/3
1=5(sinθ)^2
1/5 = (sinθ)^2
part b, since 90≦θ≦180
sinθ must be positive....so sinθ is 1/root5

Let me try...

(a)

(cosθ)^2/[1+(sinθ)^2]=2/3
[1-(sinθ)^2]/[1+(sinθ)^2]=2/3
3-3(sinθ)^2]=2+2(sinθ)^2
1=5(sinθ)^2
(sinθ)^2=1/5

2007-02-02 00:24:54 補充：
(b) sinθ=1/SQRT(5); cosθ=-2/SQRT(5) for 90≦θ≦180cosθ/(1+3sinθ)=[-2/SQRT(5)]/(1+3[1/SQRT(5)])=[-2/SQRT(5)]/(1+3[1/SQRT(5)])=(-2)/(SQRT(5)+3)=(-2)[SQRT(5)-3]/(SQRT(5)+3)(SQRT(5)-3)=(-2)[SQRT(5)-3]/(5-9)=(-2)[SQRT(5)-3]/(-4)=[SQRT(5)-3]/2