Question

已知0＜α＜π/2，且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α+sinαcosα

Answer 1

解答：
因为：3sina=4cosa，所以：tana=4/3
(1)(sin^2α+2sinαcosα)/(3cos^2α-1)=(sin^2α+2sinαcosα)/(2cos^2α-sin^2a)
=(tan^2a+2tana)/(2-tan^2)=(16/9+8/3)/(2-16/9)=20
(2)cos^2α+sinαcosα =(cos^2α+sinαcosα )/(sin^2a+cos^2a)=(1+tana)/(tan^2a+1)
=(1+4/3)/(16/9+1)=21/25

Answer 2

解：
1.(sin^2α+2sinαcosα)/(3cos^2α-1)
=.(sin^2α+2sinαcosα)/(3cos^2α-sin^2a-cos^2a)
=.(sin^2α+2sinαcosα)/(2cos^2α-sin^2a)
=(tan^2a+2tana)/(2-tan^2a)(分子分母同除以cos^2a)
3sinα=4cosα即tana=4/3带入原式
(sin^2α+2sinαcosα)/(3cos^2α-1)=20
2.3sinα=4cosα令sina=4k,cosa=3k则(4k)^2+(3k)^2=1解得k=1/5或-1/5
0＜α＜π/2
所以sina=4/5，cosa=3/5
cos^2α+sinαcosα=(3/5)^2+(4/5)*(3/5)=21/25