y=sin(π/3-2x)+cos(2x)的周期怎么求?
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y=sin(π/3-2x)+cos(2x)的周期怎么求?y=sin(π/3-2x)+cos(2x)的周期怎么求?y=sin(π/3-2x)+cos(2x)的周期怎么求?y=sin(π/3-2x)+co
y=sin(π/3-2x)+cos(2x)的周期怎么求?
y=sin(π/3-2x)+cos(2x)的周期怎么求?
y=sin(π/3-2x)+cos(2x)的周期怎么求?
y=sin(π/3-2x)+cos(2x)=—sin(2x)cosπ/3+cos2xsinπ/3+cos(2x)
由于cosπ/3=1/2,sinπ/3=√3/2.则有y=—1/2sin(2x)+(√3+2)/2cos(2x)
令tanb==—1/(√3+2),则有y=sin(2x+b).则有T=π
先化简你的解析式
y=sin(π/3-2x)+cos(2x)
=sinπ/3cos2x-cosπ/3sin2x+cos2x
=√3/2cos2x+1/2sin2x
=sinπ/3cos2x+cosπ/3sin2x
=sin(π/3+2x)
=sin(2x+π/3)
周期就是T=2π/ω=π你算错了吧……不好意思,看错了 就是这样的:y=si...
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先化简你的解析式
y=sin(π/3-2x)+cos(2x)
=sinπ/3cos2x-cosπ/3sin2x+cos2x
=√3/2cos2x+1/2sin2x
=sinπ/3cos2x+cosπ/3sin2x
=sin(π/3+2x)
=sin(2x+π/3)
周期就是T=2π/ω=π
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