已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
问题描述:
已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
答
3sin^2α+2sin^2β=1,
3sin²a=cos2B ①
3sin2α-2sin2β=0
6sinacosa=2sin2B ②
①/②
tana=cot2B
∴ tana=tan(π/2-2B)
∵ a是锐角,
B是锐角,则π/2-2B∈(-π/2,π/2)
∴ a=π/2-2B
即 a+2B=π/2
答
3sin²α+2sin²β=1
===> 3sin²α=1-2sin²β=cos2β…………………………(1)
3sin2α-2sin2β=0
===> 6sinαcosα=2sin2β
===> 3sinαcosα=sin2β…………………………………(2)
(1)/(2)得到:tanα=cot2β
所以,cot[(π/2)-α]=cot2β
已知α、β均为锐角
则,(π/2)-α=2β
所以,α+2β=π/2