搜狗做题网1 cos^2x*sin^2x的不定积分
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lim(x→∞)[sin(2/x)+cos(1/x)]^x的极限.需要详细步骤.lim(x→∞)[sin(2/x)+cos(1/x)]^x=lim(x→∞)[1+（sin(2/x)+cos(1/x)-
由于x属于[(-π/2),(π/2)]则:定义域关于原点对称则:f(-x)=[1+sin(-x)-cos(-x)]/[1+cos(-x+sin(-x)]=[1-sinx-cosx]/[1+cosx-s
f(x)=1/2sinx+1/2cosx(二倍角的正弦、余弦公式)=根号2/2（sinxcos45°+cosxsin45°）=根号2/2sin（x+45°）（1）f(a)=根号2/2sin(a+45°
整理方程,得y=1+2sinxcosx+2(cosx)^2利用降幂公式和二倍角公式,得y=sin2x+cos2x+2再利用辅助角公式,得y=根号2*sin(2x+π/4)+2所以当2x+π/4属于[2
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sin^2x+cos^2x=1
=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答：别忘了负号再问：嗯谢谢
题目应该是当x逼近到0得时候,limx^2*cos(1/x)=0lim(sin(x^2*cos(1/x)))/x=lim(x^2*cos(1/x))/x=lim(x*cos(1/x))=0再问：你用罗
1-sin²x=cosx²(1+sinx)(1-sinx)=cos²x所以(1+sinx)/cosx=cosx/(1-sinx)=1/2所以cosx/(sinx-1)=-
左边=(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(
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
3/2cosx-3/2(sinx)^2
因为sin[2*(x/2)]=2sin(x/2)cos(x/2)所以x-sin(x/2)cos(x/2)=x-1/2sinx导数为1-1/2cosx
∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x
2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)
证明：∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
2（√3/2sinx+1/2cosx)=2sin(x+π/6）√2*√2（√2/2sinx-√2/2cosx)=2sin(x-π/4）（3）解;2√2(1/2cosx-√3/2sinx)=2√2cos
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