已知sin2 X-2cos2 x=0
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y=√2×sin2x×cos2x化简得y=√2/2*sin4x所以函数的振幅为√2/2周期为π/2当x∈(kπ/2-π/8,kπ/2+π/8)时为增函数当x∈(kπ/2+π/8,kπ/2+3π/8)时
sin2x=2sinxcosxcos2x=(cosx)^2-(sinx)^2sin2x+sinxcosx-cos2x=3sinxcosx-(cosx)^2+(sinx)^2由于sinx-2cosx=0
(sinx)^2表示sinx的平方(sinx)^2+2(sinx)^2cosx+(cosx)^2-(sinx)^2=12(sinx)^2cosx+(cosx)^2=(sinx)^2+(cosx)^22
只要cos2x≠0即可,所以定义域为{x|x∈R,且x≠k∏+∏/2,k∈Z}f(x)=1/2(tan2x+1/cos2x+1)tan2x和cos2x在第一象限时的值域为(0,+∞)tan2x和cos
分式有意义,cosx≠0f(x)=[sin(2x)+cos(2x)+1]/(2cosx)=(2sinxcosx+cos²x-sin²x+cos²x+sin²x)
tanx=tan[(x-π/4)+π/4]=[tan(x-π/4)+tan(π/4)]/[1-tan(x-π/4)tan(π/4)]=(2+1)/(1-2*1)=-3(sin2x+cos2x)/(2c
sin2x=cos2xsin2x^2+cos2x^2=1∴sin2x=cos2x=根号2/2∴2x=n*pi+pi/4(n为整数)∴x=n*pi/2+pi/8
已知sin2x=2sinxcosxcos2x=(cosx)^2-(sinx)^2所以1-cos2x=2(sinx)^21+cos2x=2(cosx)^2所以(1+cos2x)/2cosx=sin2x/
解题思路:灵活利用三角函数的公式进行化简,最后套“周期公式”。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prced
(3sin2x+4cos2x)/(cos2x-3sin2x)用万能公式sin2x=2tanx/(1+tan²x)=4/5cos2x=(1-tan²x)/(1+tan²x)
已知函数f(x)=根号3sin2x+cos2x+21求f(x)的最大值及f(x)取得最大值时自变量x集合f(x)=根号3sin2x+cos2x+2=2[(根号3/2)sin2x+(1/2)cos2x]
数学之美团为你解答(1+tanx)/(1-tanx)=3解得tanx=1/2(sin²x+2sinxcosx-cos²x)/(sin²x+2cos²x)=(ta
因为tanx=2所以tan2x=2tanx/[1-(tanx)^2]=2*2/(1-2^2)=-4/3所以(sin2x+cos2x)/(cos2x-sin2x)=[(sin2x/cos2x)+(cos
分两部分求2sin2x=4sinxcosx注:sin2x=2sinxcosx=4sinxcosx/{(cosx)^2+(sinx)^2}注:{(cosx)^2+(sinx)^2=1=4tanx/{1+
(1)∵y=sin2x+sin2x+3cos2x=sin2x+cos2x+2=2sin(2x+π4)+2,∴当2x+π4=2kπ-π2(k∈Z),即x=kπ-3π8(k∈Z)时,f(x)取得最小值2-
(1-2sinx×cosx)/cos²x-sin²x=(cosx-sinx)²/[(cosx-sinx)(cosx+sinx)]=(cosx-sinx)/(cosx+si
sin²2x+sin2x+cos2x=1sn2x+cos2x=1-sin²2x=cos²2xsin2x=cos²2x-cos2x----------------
(1)a×b=√3sin2xcos2x-(cos2x)^2=√3/2sin4x-1/2cos4x-1/2=-cos(4x+60°)-1/2,所以cos(4x+60°)=3/5,因为52.5°
(cos2x-sin2x)/[(1-cos2x)(1-tan2x)]=cos2x[1-(sin2x/cos2x)]/[(1-cos2x)(1-tan2x)](分母部分提出cos2x)=cos2x(1-
sin2x+cos2x=√2sin(2x+45°),他的递增区间是[-(3/8)π+kπ,(1/8)π+kπ]k取整数.值域[-√2,√2]y=log1/2(x)在定义域(0,+∞)内递减,求交集;[