3sin^2A +2sin^2B=1,3sin2A- 2sin2B=0,求证A+2B=90度

Question

3sin^2A +2sin^2B=1,3sin2A- 2sin2B=0,求证A+2B=90度

Answer 1

因为3(sinA)^2=1-2(sinB)^2=cos2B
3sin2A/2=sin2B
(cos2B)^2+(sin2B)^2=1
所以[3(sinA)^2]^2+(3sin2A/2)^2=1
-->9(sinA)^4+9(sinAcosA)^2=1
-->9(sinA)^4+9(sinA)^2[1-(sinA)^2]=1
设(sinA)^2=t,则9t^2+9t(1-t)=1,
解得t=1/9,从而sinA=1/3,cosA=2*根号2/3.
因为2B∈(0,2π)且cos2B=3(sinA)^2=1/3>0,所以2B∈(0,π).
因为A∈(0,π/2),所以(π/2-A)∈(0,π/2).
由cos2B=sinA=cos(π/2-A)且余弦函数在(0,π/2)有单调性,
可得2B=π/2-A
即A+2B=π/2.证毕.