已知tanα=-3，求4sin눀α-3sin눀αcos눀α的值？

解:己知:tan=-3
4sin²α-3sin²acos²α
=4sin²α-3sin²α(1-sin²α)
=4sin²α-3sin²α+3(sinα)^4
=sin²α+3(sinα)^4
=sin²α(1+3sin²α)
=1/cec²α(1+3/cec²α)
=1/(1+ctan²α)[(1+3/(1+ctan²α)]
=1/[1+1/tan²α)][1+3/(1+1/tan²α]
=1/[1+1/(-3)²][1+3/(1+1/(-3)²)]
=1/[1+1/9][1+3/(1+1/9)]
=1/[10/9][1+3/(10/9)]
=9/10X[1+3X9/10]
=9/10X[1+27/10]
=9/10+243/100
=90/100+243/100
=333/100
=3.33

已知tanα=-3，求4sin²α-3sin²αcos²α的值？sin²α+cos²α＝1
sinα＝-3cosα
所以sin²α＝9/10，cos²α＝1/10
4sin²α-3sin²αcos²α
＝4×9/10-3×9/10×1/10
＝36/10-27/100
＝333/100
＝3.33