tan π z dz
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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
tanα=sinα/cosα=mcosα=sinα/m(sinα)^2+(cosα)^2=1(sinα)^2+(sinα/m)^2=1(m^2+1)/m^2*(sinα)^2=1(sinα)^2=m^
∵tanα的周期为π,这里把α看成是一个锐角,∴π-α>90°∴tan(π-α)
∵tan(x+87π)=tan(x+π+π7)=tan(x+π7)=t,∴sin(157π+x)+3cos(x−137π)sin(207π−x)−cos(x+227π)=sin(x+π7+2π)+3c
tanπ/8-ctgπ/8=(sinπ/8)/(cosπ/8)-(cosπ/8)/(sinπ/8)=[(sinπ/8)^2-(cosπ/8)^2]/(sinπ/8*cosπ/8)=(-cosπ/4)/
[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]=[(-sinα)*(-cosα)*(-tgα)]/[(-sinα)*ct
tan(a+b)=tan[(a-3/π)-(3/π-B)]=[tan(a-3/π)-tan(3/π-B)]/(1+tan(a-3/π)*tan(3/π-B))=(2-2/5)/(1+2*2/5)=8/
tanα=sinα/cosα=2s²inα/(2sinαcosα﹚=[1-cos(2α﹚]/sin﹙2α﹚∴tan(π/8)=√2-11/tan(π/12)=2+√3∴tan(π/8)+1/
tana-sina=sina/cosa-sina=sina(1/cosa-1)=sina(1-cosa)/cosa三项都大于0所以tana>sina>0tana在一个周期是增函数所以tan(sina)
根号3用两角和的正切公式展开tan(π/6-θ+π/6+θ)=a式/b式,将b式乘到等式左边再移项即得所求式子=根号3
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(
∵tan(-14π15)=a,即tan(-14π15)=tan(π-π15)=-tanπ15=-tan12°=a,∴tan12°=-a,a<0,∴sin12°=-acos12°,又sin212°+co
(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
+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)/(
∵tanπ8-cotπ8=sin(π8)cos(π8)−cos(π8)sin(π8)=−cos (π4)12sin (π4)=-2,∴答案为:-2.
三角函数与反三角函数的问题要准确结果只能用计算器了一般的数学题答案写arctan4就可以了如果非要个过程的话就要找反三角函数的函数图象了反正切函数y=arctanx的主值限在-π/2
∵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
在此我只阐述一下解题思路:你可以1,2,4项展开,同时2倍角化为单角,就可以轻松解决了!你能完成的.
tan(α-π/2)====tan(-(π/2-α))=-tan(π/2-α)=-cotα=-1/tanα
做单位圆,可以看出α∈(0,π/6)cosα>tanα>sinαtanx在(0,π/6)上是增函数,所以tan(cosα)>tan(tanα)>tan(sinα),