“根号2/2”是什么??
若sinx>cosx则2kπ+π/4<x<2kπ+5π/4此时-1<=cosx<√2/2f(x)=1/2(sinx+cosx)-1/2(sinx-cosx)=cosx所以-1<=f(x)<√2/2若sinx<cosx则2kπ-5π/4<x<2kπ+π/4此时-1<=sinx<√2/2f(x)=1/2(sinx+cosx)-1/2(cosx-sinx)=sinx所以-1<=f(x)<√2/2若sinx=cosx,则 sinx=cosx=±√2/2此时f(x)=1/2(sinx+cosx)=±√2/2综上值域[-1,√2/2]
f(x)=cosx•sinx+1/2cos²x∵sina=√2/2,a∈(0,π/2)∴cosa=√2/2即f(a)=√2/2×√2/2+1/2(√2/2)²=1/2