cos2x cos^2x*sin^2x不定积分
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 09:18:03
(1)∵函数f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ)(0<φ<π),∴f(x)=12sin2xsin∅+1+cos2x2•cos∅-12cos∅=12s
楼上都错了,图像没问题这个表达式实际是个常数,你可以运行TrigReduce[Sin[x]Sin[x+2]-Sin[x+1]^2]看看,结果为1/2(-1+Cos[2])只不过Plot的自动选择坐标系
tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su
原式=∫x*(csc^2x+1)=∫x*csc^2x+x(分开积分)前面=-x*cotx+∫cotx=-x*cotx+ln|sinx|后面=1/2x^2记得加C
sin^2(x-30)+sin^2(x+30)-sin^2(x)=(1-cos(2x-60))/2+(1-cos(2x+60))/2-sin^2x=1-1/2[cos(2x-60)+cos(2x+60
∵y=cos2xcosπ5−2sinxcosxsin6π5y=cos2xcosπ5−sin2xsin6π5=cos2xcosπ5−sin2xsinπ5=cos(2x+π5)∴2x+π5∈[2kπ-π,
你好,第一:首先将㏑(1+X)用麦克劳林公式(泰勒公式的推广)分解开就是X-(X)²/2+(X)³/3-(X)∧4+o(∧4X),第二:将㏑(1+X)中的X换为sinX就ok了,很
sin在(0,PI)是中心对称的.意思是SIN(X+PI)=-SIN(X)加绝对值就相等了.
=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答:别忘了负号再问:嗯谢谢
sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
(Ⅰ)f(x)=12sin2xsinφ+cos2xcosφ-sin(π2+φ)=12sin2xsinφ+12(1+cos2x)cosφ-12cosφ=12sin2xsinφ+12cos2xcosφ=1
sin^2x=1-cos^2x
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
sin2x+sinx=02sinxcosx+sinx=0sinx(2cosx+1)=0①党sinx=0x=kπ②党(2cosx+1)=0cosx=-1/2x=kπ/2+π/3
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
∵sin(π4-x)=513(0<x<π4),∴cos(π4-x)=1213,∴cos2xcos(π4+x)=sin(π2-2x)sin(π4-x)=2sin(π4-x)cos(π4-x)