f(x)=cos(x +圆周率/2)cos(x +圆周率/6)的最小正周期是多少?
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![f(x)=cos(x +圆周率/2)cos(x +圆周率/6)的最小正周期是多少?](/uploads/image/z/11969313-33-3.jpg?t=f%28x%29%3Dcos%28x+%EF%BC%8B%E5%9C%86%E5%91%A8%E7%8E%87%2F2%29cos%28x+%EF%BC%8B%E5%9C%86%E5%91%A8%E7%8E%87%2F6%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E6%98%AF%E5%A4%9A%E5%B0%91%3F)
f(x)=cos(x +圆周率/2)cos(x +圆周率/6)的最小正周期是多少?
f(x)=cos(x +圆周率/2)cos(x +圆周率/6)的最小正周期是多少?
f(x)=cos(x +圆周率/2)cos(x +圆周率/6)的最小正周期是多少?
f(x) =cos(x+π/2)cos(x+π/6)
= (cos( 2x+2π/3)+ cos(π/3))/2
= cos( 2x+2π/3) /2 + 1/4
最小正周期 = π
f(x)=cos(x +圆周率/2)cos(x +圆周率/6)
=-sinx*(cosx*根号3/2-sinx*1/2)
=1/2(sinx)^2-根号3/2 sinxcosx
=1/2*(1-cos2x)/2-根号3/2*1/2sin2x
=1/4-1/2(1/2cos2x-根号3/2sin2x)
=1/4-1/2cos(2x+Pai/3)
故最小正周期T=2Pai/2=Pai.