已知函数fx=3sin(x 2 派 6) 3
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f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,
f(x)=sin(2x+a),|a|>π/2,图像经过点p(π/6,1)所以sin(π/3+a)=1所以a=π/6f(x)=sin(2x+π/6)最小正周期T=2π/2=πf(π/4)=sin(π/2
再答:亲,满意请采纳再问:谢谢亲再问:再问:可以帮我解决第二问吗再答:恩,我看下再问:麻烦了。O(∩_∩)O再答:再答:再问:谢谢亲再答:😊😊再问:美丽的姑凉谢谢你再答
当x=11π/4时,1/3x-π/6=3π/4所以,f(11π/4)=sin3π/4=二分之根号二希望我的答案能让您满意,如有不明白的地方,请继续发问
求最大值时候要看整个定义域内的最大值因为在定义域内函数新增后减所以在取得的最大值应该是在函数波形的峰值处即x=pi/3时取得最大值fx=1再问:哦,明白了,谢谢
f(x)=2sin(2x+π/3)最小正周期:2π/ω=2π/2=π最小值:f(x)=2*(-1)=-2最大值:f(x)=2*1=2当sin(2x+π/3)=-1时,取得最小值;2x+π/3=2kπ-
f(x)=√3asinx+bcos(x-π/3)f(x)图像过点(派/3,1/2),(7派/6,0)所以3/2*a+b=1/2-√3/2a-√3/2b=0解得:a=2+√3,b=-2-√3∴f(x)=
f(x)=sin(x+3π/2)sin(x-2π)=-cosxsinx=-1/2sin2x最大值1/2最小值-1/2最小正周期2π/2=πf(π/6)=-1/2sinπ/3=-√3/4f(π/12)=
fx=2cosxsin(x+π/3)-√3sin^2x+sinxcosx+1=2cosx(√3/2cosx+1/2sinx)-√3sin^2x+sinxcosx+1=√3cos^2x-√3sin^2x
答:f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x
f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0
f(x)=√3sin(2x-π/6)-2cos²(x-π/12)+1=√3sin(2x-π/6)-cos(2x-π/6)=2{sin(2x-π/6)cosπ/6-cos(2x-π/6)sin
所以T=6π=2π/ww=1/3fx=2sin(1/3x-Ψ)将(π/2,0)代入0=2sin(1/3*π/2-Ψ)所以Ψ=π/6
别灰心.(1)f(x)=sin(x+π/4)+√2cos(x+π/2)(改题了)=(1/√2)(sinx+cosx)-√2sinx=(1/√2)(cosx-sinx)=cos(x+π/4),x∈[0,
f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co
f(x)=sinx-cosx=√2sin(x-4/π)(1).T=2π(2).f(x)max=√2f(x)min=-√2(3).sina+cosa=√2cos(a-π/4)cos(a-π/4)=√[1
第一题A.第二题B
解答;f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)
解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数