证明：
sin²x
=sin²x/1
=sin²x/(sin²x+cos²x)
=(sin²x/cos²x)/[(sin²x/cos²x)+(cos²x/cos²x)]
=tan²x/(tan²x+1)
即：sin²x=tan²x/(tan²x+1)
同理：sin²y=tan²y/(tan²y+1)
而：tan²x=2tan²y+1
∴sin²x
=tan²x/(tan²x+1)
=(2tan²y+1)/(2tan²y+1+1)
=(2tan²y+1)/2(tan²y+1)
则：2sin²x=(2tan²y+1)/(tan²y+1)
那么：2sin²x-1
=(2tan²y+1)/(tan²y+1)-1
=(2tan²y+1-tan²y-1)/(tan²y+1)
=tan²y/(tan²y+1)
=sin²y
即：sin²y=2sin²x-1