高三数学题,大神速来详细解答.跪谢已知tanα=负三分之四.求(1)sinα-4cosα除以5sinα+2cosα (2)sin²α+2sinαcosα 的值
2019-05-07
高三数学题,大神速来详细解答.跪谢
已知tanα=负三分之四.求(1)sinα-4cosα除以5sinα+2cosα (2)sin²α+2sinαcosα 的值
优质解答
(1)
(sinα-4cosα)/(5sinα+2cosα )
=(tanα-4)/(5tanα+2 ) (分子分母同时除以cosα而得)
=(-4/3-4)/(-5x4/3+2 )
=(-16/3)/(-14/3 )
=16/14
=8/7 (7分之8)
(2)
sin²α+2sinαcosα
=(sin²α+2sinαcosα)/(sin²α+cos²α)
=(tan²α+2tanα)/(tan²α+1) (分子分母同时除以cos²α而得)
=[(-4/3)²+2(-4/3)]/[(-4/3)²+1]
=(16/9-8/3)/(16/9+1)
=(-8/9)/(25/9)
=-8/25 (-8分之25)
(1)
(sinα-4cosα)/(5sinα+2cosα )
=(tanα-4)/(5tanα+2 ) (分子分母同时除以cosα而得)
=(-4/3-4)/(-5x4/3+2 )
=(-16/3)/(-14/3 )
=16/14
=8/7 (7分之8)
(2)
sin²α+2sinαcosα
=(sin²α+2sinαcosα)/(sin²α+cos²α)
=(tan²α+2tanα)/(tan²α+1) (分子分母同时除以cos²α而得)
=[(-4/3)²+2(-4/3)]/[(-4/3)²+1]
=(16/9-8/3)/(16/9+1)
=(-8/9)/(25/9)
=-8/25 (-8分之25)