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化简y=sin2x/1%Cosx

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

已经是最简了。

>> clear >> x=-10:0.01:10; >> y=sin(2*x).*cos(x); >> plot(x,y,'color','r','linewidth',3); >> grid on

求y=cosx/cos2x的二阶导数 ,并求二阶导数的零点 解:dy/dx=(-cos2xsinx+2cosxsin2x)/cos²2x 令d²y/dx²=[cos²2x(2sin2xsinx-cos2xcosx-2sinxsin2x+4cosxcos2x)+4cos2xsin2x(-cos2xsinx+2cosxsin2x)]/cos⁴(2x)=0 约分...

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

y=二分之根三sin2x+cosx^2 =二分之根三sin2x+(cos2x+1)/2=二分之根三sin2x+cos2x/2 +1/2=sin(2x+30度)+1/2所以T=π

cosx+cos2x+...+cosnx =1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] 先乘以2后除以2 =[cos(n+1)x/2][cos((n-1)x/2)+cos(((n-3)x/2)+...+cos((n-(2n-1))x/2) 和差化积 =[cos(n+1)x/2/sin(x/2)]*[sin(x/2)*cos((n-1)x/2)+sin(x/2)*cos...

y=sinx*cosx-(√3/2)*cos2x+1 =(1/2)*sin2x-(√3/2)*cos2x+1 =sin(2x-π/3)+1 所以最大值是1+1=2,最小值是-1+1=0 最小正周期T=2π/2=π 如果不懂,请Hi我,祝学习愉快!

y=(sinx四次方+cosx四次方+sinx平方cosx平方)/(2-sin2x) ={(sinx平方+cosx平方)²-sinx平方cosx平方}/(2-sin2x) =(1-sinx平方cosx平方)/(2-sin2x) ={(1+sinxcosx)(1-sinxcosx)}/2(1-sinxcosx) =(1+sinxcosx)/2=1/4sin2x+1/...

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