arctan1.02
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用微分计算近似值1)arctan1.02arctan1.02=arctan(1+0.02)~arctan1+(arctan'1)*0.02=p/4+[1/(1+1^2)]*0.02=p/4+0.01.(p是圆周率)
用微分近似值计算arctan1.02的令f(x)=arctanx则f'(x)=1/(x^2+1)而f(1)=π/4,f'(1)=1/2所以f(1.02)≈f(1)+f'(1)*(1.02-1)=π/4+1/2*0.02=π/4+0.01令F
arctan1/3+arctan1/2+arctan1=几,原题可化为:已知:tanx=1/2,tany=1/3,x,y∈(-π/2,π/2)(反正切函数定义).求x+y+π/4的值.由题意,tan(x+y)=(tanx+tany)/(1-
arctan1/2+arctan1/5+arctan1/3tana1=1/5,tana2=1/3tana3=1/2上式=a1+a2+a3tan(a1+a2)=(tana1+tana2)/(1-tana1tana2)=4/7a1+a2=arc
arctan1/3+arctan1/5+arctan1/7+arctan1/8如题令a=arctan1/3+arctan1/5b=arctan1/7+arctan1/8tan(atctanx)=x则tana=(1/3+1/5)/(1-1/3
求值:arctan1/2+arctan1/3分析:原题可化为已知:tanx=1/2,tany=1/3,x,y∈(-π/2,π/2)(反正切函数定义).求x+y的值.由题意,tan(x+y)=(tanx+tany)/(1-tanxtany)=
求证arctan1/2+arctan1/5+arctan1/8=pai/4问题等价于若tana=1/2tanb=1/5tanc=1/8则a+b+c=π/4证tan(a+b)=(tana+tanb)/(1-tanatanb)=7/9tan(a
求tan(arctan1/2+arctan1/3)设arctan1/2=α,arctan1/3=β;则有,tanα=1/2,tanβ=1/3;所以tan(arctan1/2+arctan1/3)=tan(α+β)=(tanα+tanβ)/(
利用微分求求函数值的近似值(1)e^1.01(2)arctan1.021)y=e^xdy=e^xdxx=1,dx=0.01,dy=e^1dx=0.01e所以e^1.01=e+0.01e=1.01e2)y=arctanx,dy=dx/(1+x
arctan1等于多少π/4+kπ
arctan1-arctan0=?tanπ/4=1tan0=0所以原式=π/4-0=π/4
arctan1+arctan2+arctan3=设arctan1+arctan2+arctan3=x那么tanx=tan(arctan1+arctan2+arctan3)=(tan(arctan1+arctan2)+tan(arctan3)
arctan1为什么pai/4?tanpai/4=1,arctan1是反三角函数,表示角度,也就是pai/4
tan(arctan1/5+arctan3)由tan(arctanx)=x可得原式=[tan(arctan1/5)+tan(arctan3)]/[1-tan(arctan1/5)tan(arctan3)]=8
A=arctan1/2arc就是反三角函数……这个问题还真独特arc是指圆弧的、反三角函数的……tan是tangentline的缩写意为:切线,正切故而arctan就是反正切的意思……是三角函数的反函数
arctan1/2+arctan1/3解答题求解谢谢∵tg(arctg1/2+arctg1/3)=(1/2+1/3)/[1-(1/2)×(1/3)]=1∴arctg1/2+arctg1/3=arctg1=45°
送分ing,求值:arctan1/2+arctan1/3.分析:原题可化为已知:tanx=1/2,tany=1/3,x,y∈(-π/2,π/2)(反正切函数定义).求x+y的值.由题意,tan(x+y)=(tanx+tany)/(1-tan
arctan1arctan0是等于多少?Arctan1=45(角度)或Pi/8(弧度).Arctan0=0.Arctan1=45*180i(i=0,1,-1,2,-2,...)Arctan0=180i(i=0,1,-1,2,-2,..)ps
arctan1-arctan-1等于什么?arctan1-arctan-1=π/4-(-π/4)=π/2arctan1-arctan-1=π/4-3π/4=-π/2因为arctan1=kπ+π/4因为arctan-1=kπ-π/4所以原式=
arctan1是等于arccot1吗?不是,是的有人说不是喔,他还跟我解释了全部展开是的收起