tanatanb=1为什么a b=派
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tan(A+B)=sin(A+B)/cos(A+B)=(sinAcosB+sinBcosA)/(cosAcosB-sinAsinB)分子,分母同时除以cosAcosB得:=(sinA/cosA+sin
1.倍角公式:cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)3+cos4a-4cos2a=3+(2cos^2(2a)-1)-4(1-2sin^2(a))
tanA+tanB=sinA/cosA+sinB/cosB=(sinAcosB+cosAsinB)/(cosAcosB)=sin(A+B)/(cosAcosB)=[sin(A+B)/cos(A+B)]
(1)tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A+B)(1-tanAtanB)=tanA+tanB=根号3tanAtanB-根号3tan(A+B)=-根号3,tan(
tanC=tan(派-A-B)=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-1所以C=135度
1)由cos2θ=1-2[(sinθ)^2]可得(sinθ)^2=(1-cos2θ)/2即sinθ=根号下(1-cos2θ)/2将θ换成θ/2可得:sin(θ/2)=根号下(1-cosθ)/2同理,由
1-tanAtanB0,故A、B都为锐角,此时tanAtanB>0,从而cosAcosB>0,两边同乘cosAcosB得,cosAcosB-sinAsinB
cos(a+b)=cosacosb-sinasinb=1/2cos(b-a)=cosacosb+sinasinb=1/2两式相加得cosacosb=1/2两式相减得sinasinb=0tgatgb=(
两角和的正切公式的变形
用sin(A+B)除以cos(A+B),再把两角和的正余弦公式代入就可以
tan(A+B)=sin(A+B)/cos(A+B)=(sinAcosB+sinBcosA)/(cosAcosB-sinAsinB)分子,分母同时除以cosAcosB得:=(sinA/cosA+sin
cos(a+b)=1/3cosacosb-sinasinb=1/3cos(a-b)=1/4cosacosb+sinasinb=1/4相加:cosacosb=7/24相减:sinasinb=-1/24相
tanA+tanB+tanAtanB=1(tanA+tanB)/(1-tanAtanB)=1tan(A+B)=1A+B+C=πTAN(A+B)=-tanC=1C=3π/4
不相等,正确的式子应该是tan(A+B)=tanA+tanB+tanAtanBtan(A+B)推倒的方式如下:∵tan(A+B)=(tanA+tanB)/(1-tanAtanB)tanA+tanB=(
tanA+tanB=1+tanAtanB→(tanA+tanB)/(1-tanAtanB)=(1+tanAtanB)/(1-tanAtanB)=(cosAcosB+sinAsinB)/(cosAcos
sin(A+B)/cos(A-B)=(sinAcosB+cosAsinB)/(cosAcosB+sinAsinB)=[(sinAcosB+cosAsinB)/cosAcosB]/[(cosAcosB+
这个本来就是公式推公式sin(A-B)=sinAcosB-cosAsinBcos(A-B)=cosAcosB+sinAsinBtan(A-B)=(sinAcosB-cosAsinB)/(cosAcos
先给你做第一题吧tanAtanB=tanAtanC+tanBtanCsinA/cosA*sinB/cosB=sinA/cosA*sinC/cosC+sinB/cosB*sinC/cosCsinAsin
在△ABC中,tanAtanB=1,即sinAsinB=cosAcosB,∴cosAcosB-sinAsinB=0,∴cos(A+B)=0,∴A+B=π2,即C=π2∴sin(C−π6)=cosπ6=
1、tanAtanB=tanA+tanB+1tanAtanB-1=tanA+tanB则:tan(A+B)=[tanA+tanB]/[1-tanAtanB]=-1因为A、B为锐角,则:A+B=3π/4,