已知a、β是锐角,a+β≠π/2,且满足tanβ=(sin2a)/(3-cos2a).证明:tan(a+β)=2tana
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/24 02:02:24
已知a、β是锐角,a+β≠π/2,且满足tanβ=(sin2a)/(3-cos2a).证明:tan(a+β)=2tana
![已知a、β是锐角,a+β≠π/2,且满足tanβ=(sin2a)/(3-cos2a).证明:tan(a+β)=2tana](/uploads/image/z/7530388-52-8.jpg?t=%E5%B7%B2%E7%9F%A5a%E3%80%81%CE%B2%E6%98%AF%E9%94%90%E8%A7%92%2Ca%2B%CE%B2%E2%89%A0%CF%80%2F2%2C%E4%B8%94%E6%BB%A1%E8%B6%B3tan%CE%B2%3D%28sin2a%29%2F%283-cos2a%29.%E8%AF%81%E6%98%8E%EF%BC%9Atan%28a%2B%CE%B2%29%3D2tana)
tan(a+β)=(tana+tanβ)/(1-tana *tanβ),将tanβ=(sin2a)/(3-cos2a)带入原式得,[tana+sin2a/(3-cos2a)]/{1-tana*[(sin2a)/(3-cos2a)]},分母下边的化简得1-(sina/cosa)*(2sina*cosa/3-cos2a)=2/(3-cos2a)
分母上边的化简得(3tana-tana*cos2a+sin2a)/(3-cos2a)=4tana/(3-cos2a)
所以得tan(a+β)=[4tana/(3-cos2a)]/2/(3-cos2a)=2tana
分母上边的化简得(3tana-tana*cos2a+sin2a)/(3-cos2a)=4tana/(3-cos2a)
所以得tan(a+β)=[4tana/(3-cos2a)]/2/(3-cos2a)=2tana
已知a、β是锐角,a+β≠π/2,且满足tanβ=(sin2a)/(3-cos2a).证明:tan(a+β)=2tana
已知a为锐角且tan(∏/4+a)=2求sin2a+cos2a的值
已知tan(π/4+a)=2,求tana的值,求sin2a+sina平方+cos2a的值
已知a在(90,180),且满足2sin²a=sin2a+cos2a 求tana
已知tan(a+π|4)=3+2根号2,求1-cos2a|sin2a
已知A、B是锐角,求证(tan(π+A)+tan(-B))/(1/tan(3π-A)+tan(π/2-B))=tanA*
已知∩a为锐角,且tana=3分之根号3,则根号下sin2a-2sinacosa+cos2a=
已知锐角A.B满足tan(A+B)=2tanA,则tanA的最大值为?
已知tan a/2=1/2 求1+sin2a/1+sin2a+cos2a
.已知tan a/2=1/2,求1+sin2a/1+sin2a+cos2a的值
已知tana=2,tanβ=3,a、β均为锐角,求证 a+β=135°
已知tan(a+β)=2tan(α-β),求sin2β/sin2a.