Question

已知sin(π-x)=4/5,x∈(0,π/2),（1）求sin2x-cos²x/2的值。

（2）求函数f（x）=5/6cosxsin2x-1/2cos2x的单调递增区间。

Answer 1

sin(π-x)=4/5,x∈(0,π/2),
sinx=4/5,cosx= -3/5
sin2x-cos²x/2=(sinx+cosx)²-1/2(2cos²x/2-1)-3/2
=(sinx+cosx)²-1/2cosx-3/2
=(4/5 -3/5)²+1/2*3/5-3/2
= -29/25
（2）f（x）=5/6cosxsin2x-1/2cos2x
= -5/6*3/5sin2x-1/2cos2x
=-1/2sin2x-1/2cos2x
=-√2/2sin(2x-π/4)
2Kπ-π/2<=2x-π/4<=2Kπ+π/2
2Kπ-π/8<=x-<=2Kπ+3π/8
f（x）=5/6cosxsin2x-1/2cos2x的单调递增区间为:
2Kπ-π/8<=x-<=2Kπ+3π/8