已知tan²a=2tan²b+1,求证sin²b+1=2sin²a

题目

已知tan²a=2tan²b+1,求证sin²b+1=2sin²a

答案

证明：令 x =(sin a)^2,
y =(sin b)^2,
则 (cos a)^2 =1-x,
(tan a)^2 =x /(1-x).
同理,
(tan b)^2 =y /(1-y).
由已知,
x /(1-x) =2y /(1-y) +1,
所以 x(1-y) =2y(1-x) +(1-x)(1-y),
即 x -xy =2y -2xy +1 -y -x +xy,
所以 y+1 =2x.
即 (sin b)^2 =2 (sin a)^2 +1.
= = = = = = = = =
换元法.
注意次数问题.
x=(sin a)^2,得 (tan a)^2 =x /(1-x) 是关键.