tan(π 4-t)=
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1.左=tan(x/2+π/4)+tan(x/2-π/4)=tan[(x/2+π/4)+(x/2-π/4)][1-tan(x/2-π/4)tan(x/2+π/4)]=tanx[1-(-1)]=2tan
这需要用到万能公式tanα=tan[2*(α/2)]=2tan(α/2)/[1-tan(α/2)^2]=[2tan(α/2)]/[1-(tanα/2)^2]tan(a+π/4)=(tana+1)/(1
tan(π/12)+1/tan(π/12)=sin(π/12)/cos(π/12)+cos(π/12)/sin(π/12)=[sin(π/12)的平方+cos(π/12)的平方]/sin(π/12)c
利用tan(A+B)展开即可.
∵tanα的周期为π,这里把α看成是一个锐角,∴π-α>90°∴tan(π-α)
(1)∵tanα=2,∴tan(α+π4)=tanα+11−tanα=2+11−2=-3.(2)∵tanα=2,∴6sinα+cosα3sinα−2cosα=6tanα+13tanα−2=12+16−
tan(X/2+π/4)+tan(x/2-π/4)=(tanx/2+1)/(1-tanx/2)+(tanx/2-1)/(1+tanx/2)=[(tanx/2+1)^2-(tanx/2-1)^2]/[(
tan(α+π/4)=-3/5(tanα+1)/(1-tanα)=-3/5tanα=-4tan(α-π/4)=(tanα-1)/(1+tanα)=5/3tan(α-π/4)=-cot(α-π/4+π/
[1-tan(π/4-A)]/[1+tan(π/4+A)]=[tan(π/4)-tan(π/4-A)]/[1+tan(π/4)*tan(π/4+A)]=tan(π/4-π/4/4+A)=tan(A)此
假设地理纬度为φ,指时针的高度为H,要刻划的时间与正午的差值为T;时间线与指时针的夹角为A,距离为D只是一个三角函数的公式,是用来具体运算,没有太大意义
设tan(x+π/4)=t则t属于(-∞,+∞)当t=2值域是(-∞,-2]并[2,+∞)因为y=t+1/t在(-∞,-1)并(1,+∞)上是单调递增的而tan(-π/4+kπ)=-1tan(π/4+
分子把平方展开之后整个式子化为4tan(x/2)/[1-(tan(x/2))^2]=2{tan(x/2)+tan(x/2)/[1-(tan(x/2))×(tan(x/2))]}=2tanx再问:。。=
tan(α+π/4)+tan(α+3π/4)=(tanα+tanπ/4)/(1-tanαtanπ/4)+(tanα+tan3π/4)/(1-tanα+tan3π/4)=(tanα+1)/(1-tanα
tan(a+π/4)=(tana+tan(π/4))/[1-tana*tan(π/4)]=(3+1)/(1-3*1)=-2tan(a-π/4)=(tana-tan(π/4))/[1+tana*tan(
(cosa-sina)^2=(cosa)^2+(sina)^2-2sinacosa=1-sin(2a)sin2a=2tana/(1+tana)^2=6/4^2=3/8(cos-sina)^2=5/8π
原式=lim{x->0}{tan(sinx)-tan(tanx)[1+cos(tanx)-1]}/(tanx-sinx)=lim{x->0}{tan(sinx-tanx)[1+tan(sinx)tan
+π/4=a+b-(a-π/4)tan(b+π/4)=tan{a+b-(a-π/4)}={tan(a+b)-tan(a-π/4)}/{1+tan(a+b)tan(a-π/4)}=(3/4-1/2)/(
tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)ta
∵tan(a+π4)=tana+11−tana=13∴tana=-12因此,(sina−cosa)2cos2a=sin2a−2sinacosa+cos2acos2a−sin2a分子分母都除以cos2a
tan(π/4+a)=(1+tana)/(1-tana)=-1/2tana=-3sina=3根号10/10[sin2a-2(cosa)^2]/1+tana=[sin2a-2+2(sina)^2]/1+