Question

函数y=（a+sinx)(a+cosx)（a>=0)的最大值为2/3，求a

Answer 1

f(x)=(a+sinx)(a+cosx) =a^2+a(sinx+cosx)+sinxcosx =a^2+√2asin(x+π/4)+1/2sin2x =a^2+√2asin(x+π/4)-1/2cos(2x+π/4) =a^2+√2asinθ-1/2cos2θ ,(θ=x+π/4) =a^2+√2asinθ-1/2[1-2sin^2θ] =t^2+√2at+a^2-1/2 ,(t=sinθ) =(t+√2/2*a)^2+(a^2-1)/2 ，(-1≤t≤1) 当a≥√2时 f(t)≤f(1)=a^2+√2a+1/2 f(t)≥f(-1)=a^2-√2a+1/2 当0≤a≤√2 f(t)≥f(-√2/2*a)=(a^2-1)/2 f(t)≤f(1)=a^2+√2a+1/2 ∴f(x)的最大值为a^2+√2a+1/2 当a≥√2时， f(x)的最小值为a^2-√2a+1/2 当0≤a≤√2时 f(x)的最小值为（a^2-1)/2 因此a^2+√2a+1/2=2/3 解得a=（-3根号2+2根号6）/6，另一负值舍去