the+bad+tan2汉化
和媚哪3714tan2θ=﹣2√2,θ∈﹙π/4,π/2﹚,求[2cos^2(θ/2) - sinθ - 1]/[√3sin﹙π/3+θ﹚sin﹙π/3 - θ﹚] -
和伦素15936699223 ______[答案] ∵tan2θ=2tanθ/(1-tan²θ)=﹣2√2 ∴tanθ/(1-tan²θ)=√2 ∴√2tan²θ+tanθ-√2=0 ∴(tanθ-√2)(√2tanθ-1)=0 ∴tanθ=√2或tanθ=√2/2 ∵θ∈﹙π/4,π/2﹚∴tanθ>1 ∴tanθ=√2 ∵sec²θ=1+tan²θ=3 ∴cos²θ=1/3 sin²θ=2/3 ∵θ∈﹙π/4,π/2﹚ ∴cosθ=√...
和媚哪3714(1+tan1°)(1+tan2°)(1+tan3°)···(1+tan45°)=? -
和伦素15936699223 ______[答案] tan 45=tan [x+(45-x)]=[tan x+tan (45-x)]/[1-tan x*tan (45-x)]=1所以tan x+tan (45-x)+tan x*tan (45-x)=1(1+tan x)*[1+tan (45-x)]=1+tan x*tan (45-x)+tan x+tan (45-x)=2则:原式=[(1+tan 1)*(1+tan 44)]*[(1+...
和媚哪3714已知tan 2分之a=2,tan 2分之b=3分之1,则tan(a+b)= -
和伦素15936699223 ______[答案] 解tana=(2tana/2)/(1-tan²a/2)=4/(1-4)=-4/3tanb=(2tanb/2)/(1-tan²b/2)=2/3*(9/8)=3/4∴tan(a+b)=(tana+tanb)/(1-tanatanb)=(-4/3+3/4)/(1+1)=(-16/12+9/12)*(1/2)=-7/12*1/2=-7/24
和媚哪3714求证明一道数学题证明:3+tan1tan2+tan2tan3=tan3/tan1 注:都是角度制的,不是弧度制的. -
和伦素15936699223 ______[答案] 3+tan1tan2+tan2tan3=1+tan1tan2+1+tan2tan3+1= (tan2-tan1)/tan(2-1)+ (tan3-tan2)/tan(3-2)+1=((tan2-tan1)+ (tan3-tan2))/tan1+1=(tan3-tan1)/tan1+1=tan3 /tan1-tan1 /tan1+1=tan3/tan1
和媚哪3714The+menu+bad+only+10+dishes+and+the+service+was+not+gootd+at+all! -
和伦素15936699223 ______[答案] 句子本身有语法错误,猜大意应该是:菜单很差,只有十道菜,服务也非常不好.
和媚哪3714证明tan(阿尔法+四分之派)+tan(阿尔法+四分之三派)=2tan2阿尔法 -
和伦素15936699223 ______[答案] tan(a+π/4)+tan(a+3π/4)=tan(a+π/4)+tan(π/2+a+π/4)=tan(a+π/4)-cot(a+π/4)=sin(a+π/4)/cos(a+π/4)-cos(a+π/4)/sin(a+π/4)=(sin²(a+π/4)-cos²(a+π/4))/(sin(a+π/4)cos(a+π/4))...
和媚哪3714求arc.tan2+arc.tan3的值. -
和伦素15936699223 ______[答案] tan(a)=2 tan(b)=3 求a+b tan(a+b)=(2+3)/(1-2x3)=-1 a+b=arctan(-1)=派-arctan1
和媚哪3714sinα+2cosα=根号10/2 则tan2α=? -
和伦素15936699223 ______[答案] sinα+2cosα=√10/2 两边平方可得: ∴sin²α+4cos²α+4sinαcosα=5/2 ∴【sin²α+4sinαcosα+4cos²α】/(sin²α+cos²α)=5/2 左边分子分母同除以cos²α可得: 【tan²α+4tanα+4】/(tan²α+1)=5/2 化简可得: 3tan²α-8tanα-3=0 解得: tanα=-1/3或...
和媚哪3714已知 - π/2和伦素15936699223 ______[答案] 解由tan2a=-4/3 知tan2a=2tana/[1-tan^2a]=-4/3 即3tana=2tan^2a-2 即2tan^2a-3tana-2=0 即(2tana+1)(tana-2)=0 即tana=-1/2或tana=2 由-π/2
和媚哪3714三角函数的计算tan(2arctan2+arctan1)=7请问这是怎么算的? -
和伦素15936699223 ______[答案] tan(2arctan2+arctan1)={tan(2arctan2)+tan(arctan1)}/1-tan(2arctan2)tan(arctan1) tan(2arctan2)=2tan(arctan2)/1-tan(arctan2)^2=2*2/1-2^2=-4/3 tan(2arctan2+arctan1)=(-4/3+1)/1-(-4/3)=-1/7 你确定是+7吗?