Question

已知sin^2α/sin^2β+cos^2αcos^2θ=1 若α＝9π／4, β＝13π／3 求|sinθ|的值

Answer 1

1. cos2θ=(cosθ)^2-(sinθ)^2=1-2(sinθ)^2=3/5 可以算出(sinθ)^2=1/5 同理(cosθ)^2=4/5
(sinθ)^4+(cosθ)^4=17/25
2。因为tan(α+β)=(tanα+tanβ)/(1-tanα*tanβ)=-1;
两边同时乘以分母，
所以tanα*tanβ-(tanα+tanβ)＝1
故所求=1-(tanα+tanβ)+tanα*tanβ＝1+1＝2

Answer 2

α＝9π/4=2π+π/4
sinα=sin(π/4)=√2/2,cosα=cos(π/4)=√2/2
sin²α=cos²α=1/2
β＝13π/3=4π+π/3
sinβ=sin(π/3)=√3/2
sin²β=3/4

sin²α/sin²β+cos²αcos²θ=1
(1/2)/(3/4)+1/2*cos²θ=1
cos²θ=2/3
|sinθ|=√(1-cos²θ)=√(1-2/3)=√(1/3)=√3/3