(1/x²-cos²x/sin²x)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 08:18:51
求证:(sin x+cos x−1)(sin x−cos x+1)sin&nbs

证明:左边=[sin x−(1−cos x)](sin x+1−cos x)sin 2x=(2sin x2cosx2−2sin2 

证明 (tan xsin x)/(tan x-sin x)-(1+cos x)/sin x

证明因为:tanx=sinx/cosx所以cosx=sinx/tanx(tanxsinx)/(tanx-sinx)分子分母同时除以tanx=sinx/(1-sinx/tanx)=sinx/(1-cos

Matlab编程问题 cos(x*y)*cos(x*(1-y))-0.5x*sin(x*y)*sin(x*(1-y))=

symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)

证明:(1 + tan x) / (sin x + cos x) = 1 / cos x

(1+tanx)/(sinx+cosx)=(1+sinx/cosx)/(sinx+cosx)=(sinx+cosx)/cosx/(sinx+cosx)=1/cosx

Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y)

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

lim(sin(x^2*cos(1/x)))/x怎么做?

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

证明(1-2sin x cos x )/(cos^2x-sin^2x)=(1-tan x)/(1+tan x)

左边=(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x

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

2cos x (sin x -cos x)+1

2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)

已知f(x)=sin(x/2) + cos(x/2) +[cos(x/2)]^2-1/2

你确定第一个符号是加号不是乘号?

化简f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1

(1)f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1=2cos(x/2)·sin(x/2)+2cos^2(x/2)-1=sinx+cosx(倍角公式)=√2sin(x+π/4

sin(1/x)-cos(1/x)/x

该函数在x=0处的左右极限都没有比如x=1/(2npi+pi/2)时,f(x)=1x=1/(2npi-pi/2)时,f(x)=-1取n->无穷大所以在x=0处没有右极限,左极限同理

求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2

证明:∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co

已知函数f(x)=2Cos x(Sin x-Cos x)+1

f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π

(x*sin x*cos x)的导数

(x*sinx*cosx)'=(1/2xsin2x)'=1/2(sin2x+xcos2x*2)=1/2sin2x+xcos2x

∫(1-sin/x+cos)dx不定积分

可用凑微分法如图积分.经济数学团队帮你解答,请及时采纳.

化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x

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

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

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