作业帮已知f（x）=2cos(π/2-x)cosx-√3cos2x,x属于R.问1.求f(π/6)的值.2.当x∈[0,π/2]时,求f(x)的值
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![已知f（x）=2cos(π/2-x)cosx-√3cos2x,x属于R.问1.求f(π/6)的值.2.当x∈[0,π/2]时,求f(x)的值](/uploads/image/z/3779664-24-4.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3D2cos%28%CF%80%2F2-x%29cosx-%E2%88%9A3cos2x%2Cx%E5%B1%9E%E4%BA%8ER.%E9%97%AE1.%E6%B1%82f%28%CF%80%2F6%29%E7%9A%84%E5%80%BC.2.%E5%BD%93x%E2%88%88%5B0%2C%CF%80%2F2%5D%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC)
已知f（x）=2cos(π/2-x)cosx-√3cos2x,x属于R.问1.求f(π/6)的值.2.当x∈[0,π/2]时,求f(x)的值
已知f（x）=2cos(π/2-x)cosx-√3cos2x,x属于R.问1.求f(π/6)的值.2.当x∈[0,π/2]时,求f(x)的值
已知f（x）=2cos(π/2-x)cosx-√3cos2x,x属于R.问1.求f(π/6)的值.2.当x∈[0,π/2]时,求f(x)的值
1、当x=π/6=30度时,f(x)=2cos60度cos30度-√3cos60度
=2×1/2×√3/2-√3×1/2
=0
2、当x∈[0,π/2]时,可化简,但数值仍不固定
可化简为 f(x)=2sinx cosx-√3cos2x
=sin2x-√3cos2x
=2×(1/2sin2x-√3/2 cos2x)
=2×(cosπ/3sin2x-sinπ/3cos2x)
=2×sin(2x-π/3)