已知f(x)=sin平方x+sinxcosx.x∈[0,pai/2]

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)-1/2
x∈[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.

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