fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期请给一个完整的过程,
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![fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期请给一个完整的过程,](/uploads/image/z/6870288-48-8.jpg?t=fx%E7%AD%89%E4%BA%8Esin%282x-3%2F%CF%80%29%2Bcos%282x-6%2F%CF%80%29%2B2cos%26%23178%3B%2B1%2C%E6%B1%82%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E8%AF%B7%E7%BB%99%E4%B8%80%E4%B8%AA%E5%AE%8C%E6%95%B4%E7%9A%84%E8%BF%87%E7%A8%8B%2C)
fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期请给一个完整的过程,
fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期
请给一个完整的过程,
fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期请给一个完整的过程,
f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos²x+1 /顺带说一句,π/3、π/6你写反了.
=sin(2x)cos(π/3)-cos(2x)sin(π/3)+cos(2x)cos(π/6)+sin(2x)sin(π/6)+cos(2x)+2
=(1/2)sin(2x)-(√3/2)cos(2x)+(√3/2)cos(2x)+(1/2)sin(2x)+cos(2x)+2
=sin(2x)+cos(2x)+2
=√2sin(2x+π/4)+2
最小正周期=2π/2=π
答:
f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos²x+1
=sin2xcosπ/3-cos2xsinπ/3+cos2xcosπ/6+sin2xsinπ/6+1+cos2x+1
=sin2x+cos2x+2
=√2*[(√2/2)sin2x+(√2/2)cos2x]+2
=√2sin(2x+π/4)+2
所以:f(x)的最新正周期T=2π/2=π
.f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos²x+1,
=sin2x*1/2-cos2x*根号3/2+cos2x*根号3/2+sin2x*1/2+1+cos2x+1
=sin2x+cos2x+2
=根号2sin(2x+π/4)+2
T=2π/2=π
你好这个题 是个化简题 你可以展开
f(x)=(sin2xcos3/π-cos2xsin3/π)+(cos2xcos6/π+sin2xsin6/π)+cos2x+2
=2sin2xsin6/π+cos2x+2
周期=2π/2=π