Question

设a＞0为常数，已知函数f(x)=cos平方(x-2π/3)+sin平方(x-5π/6)+asinx/2cosx/2的最大值为2，求a的值。

Answer 1

f(x)=(cos(x-2π/3))^2+(sin(x-5π/6))^2+asin(x/2)cos(x/2)
=(cosx*cos(2π/3)+sinx*sin(2π/3))^2+(sinxcos(5π/6)-cosxsin(5π/6))^2+a/2*sinx
=(-1/2*cosx+√3/2*sinx)^2+(-√3/2*sinx-1/2*cosx)^2+a/2*sinx
=1/4*(cosx)^2-√3/2*sinxcosx+3/4*(sinx)^2++3/4*(sinx)^2+√3/2*sinxcosx+1/4*(cosx)^2+a/2*sinx
=1/2*(cosx)^2+3/2*(sinx)^2+a/2*sinx
=1/2+(sinx)^2+a/2*sinx
=1/2+(sinx+a/4)^2-a^2/16
当 a>=0时 sinx取1时达到最大值，这时 f(x)=1/2+1+a/2=2，解得 a=1
当 a<0时 sinx取-1时达到最大值，这时， f(x)=1/2+1-a/2=a ，解得 a=-1
所以 a=±1