已知函数f(x)=cos^2(x+π/3)-0.5,g(x)=1/2sin(2x+2π/3),求h(x)=f(x)-g(x)的最大
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![已知函数f(x)=cos^2(x+π/3)-0.5,g(x)=1/2sin(2x+2π/3),求h(x)=f(x)-g(x)的最大](/uploads/image/z/6964031-47-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2%28x%2B%CF%80%2F3%29-0.5%2Cg%28x%29%3D1%2F2sin%282x%2B2%CF%80%2F3%29%2C%E6%B1%82h%28x%29%3Df%28x%29-g%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7)
已知函数f(x)=cos^2(x+π/3)-0.5,g(x)=1/2sin(2x+2π/3),求h(x)=f(x)-g(x)的最大
已知函数f(x)=cos^2(x+π/3)-0.5,g(x)=1/2sin(2x+2π/3),求h(x)=f(x)-g(x)的最大
已知函数f(x)=cos^2(x+π/3)-0.5,g(x)=1/2sin(2x+2π/3),求h(x)=f(x)-g(x)的最大
h(x)=f(x)-g(x)
=cos^2(x+π/3)-0.5-1/2sin(2x+2π/3)
=0.5(2cos^2(x+π/3)-1)-1/2sin(2x+2π/3)
=1/2cos(2x+2π/3)-1/2sin(2x+2π/3)
= √2/2[√2/2cos(2x+2π/3)-√2/2sin(2x+2π/3)]
=√2/2[cosπ/4cos(2x+2π/3)-sinπ/4sin(2x+2π/3)]
=√2/2cos(2x+π/4+2π/3)
=√2/2cos(2x+11π/12)
所以
函数最大值为√2/2,最小值为-√2/2