已知f(x)=sin平方x+sinxcosx.x∈[0,pai/2]

显示全部楼层 · 2012-3-8 08:57:46

f(x)=(sinx)^2+sinxcosx=[cos(2x)-1]/2+sin(2x)/2=(1/2)[sin(2x)+cos(2x)]-1/2=(√2/2)sin(2x+π/4)-1/2x∈[0，π/2]2x+π/4∈[π/4，5π/4]sin(2x+π/4)∈[-√2/2，1]当sin(2x+π/4)=1时，f(x)有最大值f(x)max=(√2-1)/2当sin(2x+π/4)=-√2/2时，f(x)有最小值f(x)min=-1函数的值域为[-1，(√2-1)/2]f(a)取不到5/6，因此后一问没法做。...

千问 · 2012-3-8 08:57:46

f(x)=(1/2)[(1-cos2x)+sin2x]f(a)=5/6sin2a-cos2a=2/3sin2a-cos2a=√2sin(2a-π/4)sin(2a-π/4)=√2/3cos(2a-π/4)=√7/3
或
cos(2a-π/4)=-√7/3
（舍去）sin2a=sin(2a-π/4)cos...