y3df+the+tan汉化版

晏非卿3385cos(9π/4)+tan( - 11π/3)+sin1125° 怎么算? -
惠虾录13269248764 ______[答案] cos(9π/4)+tan(-11π/3)+sin1125° =cos(2π+π/4)+tan(-12π/3+π/3)+sin(1080°+45°) =cosπ/4+tanπ/3+sin45° =二分之根号2+根号3+二分之根号2 =根号2+根号3

晏非卿3385化简tan∝+cos∝÷(1+sin∝) -
惠虾录13269248764 ______[答案] tanx+cosx/(1+sinx) =(sinx)/cosx+cosx(1-sinx)/(cosx)^2 =(sinxcosx+cosx-sinxcosx)/(cosx)^2 =1/cosx

晏非卿3385已知(1 - tanθ)/(2+tanθ)=1,求证tan2θ= - 4tan(θ+π/4) 想得到就不会来问了!乱死了! -
惠虾录13269248764 ______[答案] (1-tanθ)/(2+tanθ)=1 所以1-tanθ=2+tanθ tanθ=-1/2 因此tan2θ=2tanθ/(1-tan^θ)=-1/(1-1/4)=-4/3 tan(θ+∏/4)=(tanθ+tan∏/4)/(1-tanθtan∏/4)=(-1/2+1)/(1+1/2)=1/3 所以tan2θ=-4tan(θ+∏/4)

晏非卿33853tan°cos30°+tan45°+sin30° - sin60°cos60°怎么解? -
惠虾录13269248764 ______[答案] 解 3tan30°cos30°+tan45°+sin30°-sin60°cos60° =3*√3/3*√3/2+1+1/2-√3/2*1/2 =3/2+1+1/2-√3/4 =3-√3/4

晏非卿3385已知tan(4分之派+a)=2分之1,怎样求tana的值?和求1+cos2a分之sin2a - cos2a的值怎么求? -
惠虾录13269248764 ______[答案] 就是用两角和化式tan(α+β)=tanα+tanβ/1-tanαtanβ,直接用换元,tan(4分之派)=1,设tana=x,tan(4分之派+a)= (1+x) / (1-1*x)=1/2x= -1/3

晏非卿3385已知tan(α+π\4)=2,求cos2α+3sin^2α的值 -
惠虾录13269248764 ______[答案] tan(α+π\4)=(tanα+tanπ\4)/(1-tanαtanπ\4)=2 求得tanα=1/3 cos2α+3sin^2α=1-2sin^2α+3sin^2α=1+sin^2α=11/10

晏非卿3385任意角三角函数 化简 tanα(cosα - sinα)+﹙sinα+tanα﹚/(cotα+cosα) -
惠虾录13269248764 ______[答案] tanα ,oct α拆成Sin a/Cosa形式,通分、利用和差公式慢慢算.你不做永远也不会.

晏非卿3385已知角α的终边过点P(3,4),则sinα+cosα+tanα=( ) -
惠虾录13269248764 ______[答案] r=5, sinα=y/r=4/5 cosα=x/r=3/5 tanα=y/x =4/3 sinα+cosα+tanα=4/5+3/5+4/3=41/15

晏非卿3385证明tan(阿尔法+四分之派)+tan(阿尔法+四分之三派)=2tan2阿尔法 -
惠虾录13269248764 ______[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))...

晏非卿3385tan29°+tan31°+√3tan29°tan31° -
惠虾录13269248764 ______[答案] tan(a+b)=(tana+tanb)/(1-tanatanb) tana+tanb=(1-tanatanb)tan(a+b) tan29°+tan31°+√3tan29°tan31° =(1-tan29°tan31°)*tan(29+31°)+√3tan29°tan31° =√3