Question

3sinα=-cosα,求1+sinαcosα的值。

Answer 1

由题知tanα=-1/3原式=sinα^2+cosα^2+sinαcosα=（sinα^2+cosα^2+sinαcosα）/(sinα^2+cosα^2)=(tanα^2+1+tanα)/(tanα^2+1)=7/10

Answer 2

3sinα=-cosα
tana=sina/cosa=-1/3
1+sinαcosα
=1+sinαcosα/1
=1+sinαcosα/(sin^2a+cos^2a) (上下同除以cos^2a)
=1+tana/(1+tan^2a)
=1+(-1/3)/(1+1/9)
=7/10

Answer 3

[[[1]]]
3sina=-cosa
(sina)/(cosa)=-1/3
∴tana=-1/3
[[[2]]]
原式
=(sin²a+cos²a+sinacosa)/(sin²a+cos²a)
=(tan²a+tana+1)/(tan²a+1)
=7/10

Answer 4

因为sin²a+cos²a=1，则有9sin²a+sin²a=1，则sina=±根号10/10，而cosa=负正3倍根号10/10，
无论是sina为正还是负，均有sinacosa=3/10
则1+sinacosa=1+3/10=13/10