Question

超难的二倍角三角函数题.已知3sin²x+2sin²y=1,3sin2x-2sin2y=0,且x,y

Answer 1

3sin²x = 1- 2sin²y = cos(2y)

3sin2x = 2sin(2y)

0 = cos(x+2y)

= cos(x) cos(2y) - sin(x)sin(2y)

= cos(x) 3sin^2(x) -sin(x)3/2sin(2x)

= cos(x)(3sin^2(x) - 3sin^2(x))

= 0

当然实际上可以把x,y算出来,因为

注意到sin²(2y) + cos²(2y) = 1

所以9sin^4(x) + 9/4 * sin^2(2x) = 1

即9sin^4(x) + 9sin^2(x)cos^2(x) = sin^2(x)(9sin^2(x) + 9cos^2(x)) = 9sin^2(x)=1