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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。

这道题用到积化和差公式 公式如下:sinαcosβ=1/2[sin(α+β)+sin(α-β)] 将原式代入可得cosxsin2x=1/2(sin3x+sinx) 所以根据函数式可得它是奇函数

>> 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...

变形=2sinxcosx/[cosx+(1-cos2x)/2]dx =-2cosx[cosx+(1-cos2x)/2]d(cosx) 令t=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)]/(...

y=sin2x-2cosx(x∈R)=1-cos2x-2cosx=-(cosx+1)2+2,故当 cosx=-1时,y有最小值等于-2,故答案为:-2.

f(x) = sin²x+sinxcosx = 1/2(1-cos2x)+1/2sin2x = 1/2sin2x-1/2cos2x+1/2 = √2/2sin(2x-π/4) + 1/2 = (√2+1)/2 sin(2x-π/4) = 1 2x-π/4 = 2kπ+π/2 2x = 2kπ+3π/4 x = kπ+3π/8 x(n+1)- x(n) = π 图像与y=(√2+1)/2的两个相邻交点的距离为π

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