题目

求函数y=sin²x+3sinxcosx+5cos²x的值域

提问时间：2020-11-10

答案

求函数y=sin²x+3sinxcosx+5cos²x的值域
y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+1
=2(1+cos2x)+(3/2)sin2x+1=2cos2x+(3/2)sin2x+3
=2[cos2x+(3/4)sin2x]+3【令tanφ=3/4,φ∊ (0,π/2)；sinφ=3/5,cosφ=4/5)】
=2[cos2x+tanφsin2x]+3
=(2/cosφ)[cos2xcosφ+sin2xsinφ]+3
=[2/(4/5)]cos(2x-φ)+3
=(5/2)cos(2x-φ）+3
故ymin=-5/2+3=-1/2；ymax=5/2+3=11/2;
即y∊[-1/2,11/2]就是该函数的值域.