问题
解答题
已知tan(α+
(Ⅰ)求tanα的值; (Ⅱ)求2sin2α-sin(π-α)sin(
|
答案
(Ⅰ)∵tan(α+
)=π 4
=tanα+1 1-tanα
,∴tanα=-1 3
.1 2
(Ⅱ)原式=2sin2α-sinαcosα+cos2α
=
=2sin2α-sinαcosα+cos2α sin2α+cos2α
=2tan2α-tanα+1 tan2α+1
=2×(-
)2-(-1 2
)+11 2 (-
)2+11 2
.8 5