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(2Cos²x%1)/2tAn(π/4%x)sin²(π/4+x)

[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²(π/4+x)] =1/2*[4cos^4(x)-4cos²x+1]/{2tan[(π/2-π/4)-x]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2tan[π/2-(π/4+x)]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2co...

tan(π/4-x)= (tanπ/4-tanx)/(1+tanπ/4tanx) = (1-tanx)(1+tanx) = (cosx-sinx)/(cosx+sinx) sin^2(π/4+x) = (√2/2sinx+√2/2cosx)^2= 1/2(sinx+cosx)^2 f(x) = [2cos^4x-2cos^2x+1/2]/[tan(π/4-x)sin^2(π/4+x)] =[2cos^4x-2cos^2x+1/2]/[ (...

你的这个式子写得太乱了,请写正规些,这样的题应该不难! 我这里给你一些提示吧: [cos²(x)]²-cos²(x)=cos²(x)[cos²(x)-1]=﹣sin²(x)cos²(x) 其它的式子看不清。

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

-√6/3 根据公式:(cosx-sinx)²=1-2sinxcosx=1-1/3=2/3 所以 cosx-sinx=+-√6/3 因为 π/4

解:lim(x->1)(1-x)tan(πx/2) =lim(y->0)[y*tan(π/2-πy/2)] (用y=1-x代换) =lim(y->0)[y*ctan(πy/2)] =lim(y->0)[y*cos(πy/2)/sin(πy/2)] =lim(y->0){[(πy/2)/sin(πy/2)]*[(2/π)*cos(πy/2)]} ={lim(y->0)[(πy/2)/sin(πy/2)]}*{lim(y->0)[(2/π)*c...

tan(x-π/4)不等于0,还有tan函数本身的定义域,所以x-π/4不等于kπ,还有x-π/4不等于kπ+π/2,你解下看看几个就可以

解:f(x)=x²sinα/3+x²cosα/2+tanα f'(x)=2xsinα/3+2xcosα/2=2x(sinα/3+cosα/2)=2x*√(1/9+1/4)*sin(α+arctan3/2)=√13/3*xsin(α+arctan3/2) f'(1)=√13/3*sin(α+arctan3/2) α范围是[0,5π/12],故α+arctan(3/2)范围是[arctan3/2,5π/...

即(1-x)sin(πx/2)/cos(πx/2) x趋于1,那么 此时sin(πx/2)趋于1,1-x趋于0 而cos(πx/2)=sin(π/2-πx/2)=sinπ/2(1-x) 由重要极限得到 (1-x)/ sinπ/2(1-x) =2/π *[π/2(1-x)]/sinπ/2(1-x) 后者趋于1,于是极限值为2/π

判断间断点的类型还是要从定义出发,肯定不会错的,求解方法是一样的

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