Question

已知(tanα-3)(sinα+cosα+3)=0 求值(1) 4sinα+2cosα/5cosα+3sinα (2) sin²α+4cosα²

Answer 1

解：∵tanα=3
∴sinα=3cosα..........(1)
==>(3cosα)²+(cosα)²=1
==>cos²α=1/10..........(2)
故（1）(4sinα-2cosα)/(5cosα+3sinα)
=(4(3cosα)-2cosα)/(5cosα+3(3cosα)) (由(1)得)
=(10cosα)/(14cosα)
=5/7；
(2)sin²α+4cosα²
=1-cos^2a+4cos^2a
=3cos^2a+1
=3*1/10+1
=13/10
不懂的欢迎追问，如有帮助请采纳，谢谢!

追问

貌似=(4(3cosα)-2cosα)/(5cosα+3(3cosα)) (由(1)得)
=(10cosα)/(14cosα) 这里到这里的符号不对吧

追答

=(4(3cosα)-2cosα)/(5cosα+3(3cosα))
=(12cosa-2cosa)/(5cosa+9cosa)
=10cosa/14cosa
=5/7
符号哪里不对了??
不懂的欢迎追问，如有帮助请采纳，谢谢!