cos4x等于
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因为1-cos4x/2sin^2x+xtan^2x=1-(1-2sin^2x)/2sin^2x+sin^2x/cos^2x=2sin^2x/2sin^2x+sin^2x/cos^2x=2/1+1/co
y=7-4sinxcosx+4cos2x-4cos4x=7-2sin2x+4cos2x(1-cos2x)=7-2sin2x+4cos2xsin2x=7-2sin2x+sin22x=(1-sin2x)2
cos4x-cos2x=02(cos2x)^2-cos2x-1=0(2cos2x+1)(cos2x-1)=0cos2x=-1/2或cos2x=12x=2kπ-π/3或2x=2kπ+π/3或2x=2kπ
解析:已知sin(x-0.75π)*cos(x-0.25π)=-0.25则sin(x-0.25π-0.5π)*cos(x-0.25π)=-0.25即-sin[0.5π-(x-0.25π)]*cos(x
教授,你不行,我才高一(不要怀疑我的数学,不比大一的差),全部搞定!ABCDB,否是是是否是,CCB错的应该极少,你再看看吧
y=sin2x+cos4x=sin2x+1-2(sin2x)^2=-2(sin2x-1/2)^2+3/2所以最小正周期应为T=2π/2=π选A
sin2x-cos4x=2*sinx*cosx-1+8(sinx*cosx)^2由题:2sinx*cosx=m^2-1带进去
cot2x=cos2x/sin2x=(2cos2xcos2x)/2sin2xcos2x=(1+cos4x)/sin4x
以角度30为例:sin120+cos120=3^0.5/2-0.51-2sin60cos60=1-2*(3^0.5/2)*0.5=1-3^0.5/2不等于sin120+cos120题目不正确,无解若题
=cos2;2xcos2;x-1cos2;2xcos2;x-cos2;2xcos2;x=cos2;2xcos2;x-1=(1cos4x)/2(1cos2x)/2-1=(1/2)(cos4xcos2x
先积化和差1/2(cos3x+cosx)=1/2(cos7x+cosx)即cos3x-cos7x=0再和差化积2sin2xsin5x=0即x=kπ/2或x=kπ/5
点击图片就可以看清楚,加油!
(Ⅰ)由题意知,f(x)=cos4x-2sinxcosx-sin4x=(cos2x+sin2x)(cos2x−sin2x)−sin2x=cos2x−sin2x=2cos(2x+π4)∴f(x)的最小正
(sin4x)/(1+cos4x)*(cos2x)/(1+cos2x)*(cosx)/(1+cosx)=(2sin2xcos2x)/(1+2cos²2x-1)*(cos2x)/(1+cos2
(1)证明:tan^2(x)+1/[tan^2(x)]=[sin^2(x)/cos^2(x)]+[cos^2(x)/sin^2(x)]={[sin^4(x)+cos^4(x)]/[sin^2(x)co
1+cos2x=2cos²分母为cos²2x*cos²x*(1+cosx)分子为4sinx*cos²2x*cos²x化简为4sinx/(1+cosx)
解题思路:利用三角函数的公式及性质求解。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inclu
tan^2x+cot^2x=tan^2x+1/tan^2x=[(sinx)^4+(cosx)^4]/(sinxcosx)^2={[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2}
∵y=sin4x+cos4x=(1−cos2x2)2+(1+cos2x2)2=2+2cos22x4=14cos4x+34,∴ymin=-14+34=12,ymax=14+34=1,故答案为:[12,1