sinx cosx=2 3
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(I)∵sinxcosx=12sin2x,cos2x=12(1+cos2x)∴f(x)=23sinxcosx+2cos2x-t=3sin2x+cos2x+1-t=2(sin2xcosπ6+cos2xs
sinx/cosx=3sinx=3cosx(sinx)^2+(cosx)^2=110(cosx)^2=1(cosx)^2=1/10sinxcosx=3(cosx)^2=3/10再问:10(cosx)^
(1)∵f(x)=23sinxcosx-(sinx+cosx)(sinx-cosx)=3sin2x+cos2x=2sin(2x+π6)=1,∴sin(2x+π6)=12,∴2x+π6=2kπ+π6或2
1、由题得tanx=sinx/cosx=3sinx=3cosxsin^2x+cos^2x=110cos^2x=1cosx=√10/10sinx=3cosx=3√10/10所以sinx·cosx=3/1
sinxcosx=(sinx+cosx)的平方减1再除以2,然后把sinx+cosx看成整体,再根据平方法化简可得最后结果是(sinx+cosx+1)^2/2-1.
sinxcosx=(1/2)*sin(2x)=(1/2)*{2tanx/[1+(tanx)^2]}=2/5
正解.引自吉米多维奇著《数学分析习题集》
sinx-cosx=sinxcosx两边平方(sinx)^2-2sinxcosx+(cosx)^2=(sinxcosx)^21-2sinxcosx=(sinxcosx)^2(sinxcosx)^2+2
解题思路:利用三角函数正弦的和公式sin(x+x)可得结果解题过程:解:因为sin2x=sin(x+x)=sinxcosx+cosxsinx=2sinxcosx,所以y=2sinxcosx=sin2x
怎么感觉cosx应该是平方啊再问:嗯的,打错了再答:(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c
(Ⅰ)∵f(x)=sin2x+23sinxcosx+3cos2x,=3sin2x+2-1−cos2x2+1=3sin2x+cos2x+2=2sin(2x+π6)+2.所以最小正周期为:T=2π2=π当
(1-2sinxcosx)(1+2sinxcosx)=(sin²x+cos²x-2sinxcosx)(sin²x+cos²x+2sinxcosx)=(sinx-
(1)∵函数f(x)=cos2x-sin2x+23sinxcosx=cos2x+3sin2x=2sin(2x+π6),∵x∈[0,π2],∴(2x+π6)∈[π6,7π6],∴sin(2x+π6)∈[
(1)∵f(x)=23sinxcosx-3sin2x-cos2x+3=3sin2x-3•1−cos2x2-1+cos2x2+3=3sin2x-cos2x+1=2sin(2x+π6)+1,∵x∈[0,π
sinx-cosx=sinxcosx两边平方(sinx)^2-2sinxcosx+(cosx)^2=(sinxcosx)^21-2sinxcosx=(sinxcosx)^2(sinxcosx)^2+2
①由已知f(x)=2cos2x+23sinxcosx=cos2x+1+3sin2x=2sin(2x+π6)+1∴T=2π2=π &
(Ⅰ)f(x)=23sinxcosx-2sin2x+1=3sin2x+cos2x=2sin(2x+π6),∴T=2π2=π.(Ⅱ)∵x∈[0,π2],∴π6≤2x+π6≤7π6,∴当2x+π6=π2时
(1)函数f(x)=2cos2x+23sinxcosx-1(x∈R)=cos2x+3sin2x=2(12cos2x+32sin2x)=2sin(π6+2x),∴周期T=2πω=2π2=π.(2)由&n
楼上结果有误,应为sinx=5/13cosx=12/13其实这道题的结果可以猜出来.169=13*1313*13=12*12+5*55*12=60