函数y=sinαcosα-cos2α的最小正周期为．

Question

函数y=sinαcosα-cos²α的最小正周期为______．

Answer 1

∵y=sinαcosα-cos²α=(1)/(2)sin2α −(1)/(2)cos2α−(1)/(2)
=(√(2))/(2)sin(2α−(π)/(4))−(1)/(2)，
∴三角函数的最小正周期是T=(2π)/(2)=π
故答案为：π