若tanα=3,则(sinα+cosα)²=
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/23 17:52:53
若tanα=3,则(sinα+cosα)²=
若tanα=3,则(sinα+cosα)²=
若tanα=3,则(sinα+cosα)²=
tanα=3>0得sina、cosa同号,所以sina/cosa=3,得sin^2 a=3/4,cos^2 a=1/4,
所以sinacosa=(根号3)/4(sinα+cosα)²=1+(根号3)/2=(2+根号3）/2
答：
tana=3
(sina+cosa)^2
=(sina)^2+2sinacosa+(cosa)^2
=1+2tana*cosa*cosa
=1+2*3*(cosa)^2
=1+6/[1/(cosa)^2]
=1+6/[1+(tana)^2]
=1+6/(1+3^2)
=1+6/10
=8/5
(sinα+cosα)²
=(sin^2α+cos^2α+2sinαcosα)/1
=(sin^2α+cos^2α+2sinαcosα)/(sin^2α+cos^2α)
上下同时除以cos^2a
=(tan^2α+1+2tanα)/(tan^2α+1)
=(9+1+6)/(9+1)
=1.6
sinα/cosα=3，sin²α+cos²α=1.
解得，cosα=√10/10,sinα=3√10/10.
(sinα+cosα)²
=1+2sinαcosα
=1+2×√10/10×3√10/10
=8/5.
tan a=sin a/cos a=3
sin a=3cos a
(sin a)^2+(cos a)^2=1
(cos a)^2=1/10
(sin a + cos a )^2=1+2sin a cos a= 1+6(cos a )^2=8/5=1.6
解
(sina+cosa)²——除以cos²a+sin²a=1，值不变
=(sin²a+2sinacosa+cos²a)/(sin²a+cos²a)——分子分母同时除以cos²a
=(tan²a+2tana+1)/(tan²a+1)
=(9+6+1)/(9+1)
=16/10
=8/5