已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1求f(x)在区间{-π/4,π/4}上的最大值和最小值x∈[-π/4,π/4]①2x∈[-π/2,π/2]②2x+π/4∈[-π/4,3π/4]③-1
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![已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1求f(x)在区间{-π/4,π/4}上的最大值和最小值x∈[-π/4,π/4]①2x∈[-π/2,π/2]②2x+π/4∈[-π/4,3π/4]③-1](/uploads/image/z/14633133-69-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F3%29%2Bsin%282x-%CF%80%2F3%29%2B2cos%5E2x-1%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BD%9B-%CF%80%2F4%2C%CF%80%2F4%EF%BD%9D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BCx%E2%88%88%5B-%CF%80%2F4%2C%CF%80%2F4%5D%E2%91%A02x%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D%E2%91%A12x%2B%CF%80%2F4%E2%88%88%5B-%CF%80%2F4%2C3%CF%80%2F4%5D%E2%91%A2-1)
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已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1求f(x)在区间{-π/4,π/4}上的最大值和最小值x∈[-π/4,π/4]①2x∈[-π/2,π/2]②2x+π/4∈[-π/4,3π/4]③-1
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1
求f(x)在区间{-π/4,π/4}上的最大值和最小值
x∈[-π/4,π/4]①
2x∈[-π/2,π/2]②
2x+π/4∈[-π/4,3π/4]③
-1
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1求f(x)在区间{-π/4,π/4}上的最大值和最小值x∈[-π/4,π/4]①2x∈[-π/2,π/2]②2x+π/4∈[-π/4,3π/4]③-1
2x+π/4∈[-π/4,3π/4] 可以去到正弦值最大的角π/2,可以去到正弦值最小的角-π/4
所以-√2/2=
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