已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb=
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/25 10:45:33
x){}K54m
Mu] y1$1/QH$$铨CΆ[jD N-K$n
CM*]$UpU@c5 l"[#}S.PRhēOOvtOMˀ v~qAbȏ
已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb=
已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb=
已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb=
cos(a+b)=cosacosb-sinasinb=1/5(1)
cos(a-b)=cosacosb+sinasinb=3/5(2)
(1)+(2) cosacosb=2/5
(2)-(1) sinasinb=1/5
下式除以上式得
tanatanb=1/2
已知cos(a+b)cos(a-b)=1/3,则cos^2a-sin^2b的值是?
Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2
已知锐角a、b满足cos a=4/5,tan(a-b)=-1/3,求cos b
已知cos(a+b)=-1/3,cos2a=-5/13,a,b均为锐角,求cos(a-b)
已知cos(a+b)=4/5,cos(a-b)=-4/5,3π/2
.已知Cos(a+b)=4/5,cos(a-b)= -4/5,3/2π
已知cos(a+b)cos(a-b)=1/3 ,求(cosa)^2-(cosb)^2
sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是
已知cos(a+b)cos(a-b)=1/4,则cos^2a+cos^2b=是(cosa)^2
已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb=
已知cos(a+b)=1/5、cos(a-b)=3/5,求tana .tanb的值?
已知cos(a+b)=1/5,cos(a-b)=3/5,求tanatanb的值
已知cos(a+b)=1/3,cos(a-b)=1/5,求tanatanb的值
已知cos(a+b)=1/3,cos(a-b)=1/5,求tana*tanb的值
已知cos(a+B)=1/3,cos(a-B)=1/5,求tana x tanB
已知cos(a +b) =1/3 ,cos (a-b) =1/5 ,求cosacosb 的值
已知cos(a+b)=2/3,cos(a-b)=1/5,则tanatanb=?
已知:5cos(a-b/2)+7cos(b/2)=0 求:tan(a/2)*tan((a-b)/2)已知 5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].由5cos(a-b/2)+7cos(b/2)=0得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=05{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=010[cos^2(a/2)cos(b