已知0〈a〈π,tana=-2求2sina2-sinacosa cosa2
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 12:54:05
![已知0〈a〈π,tana=-2求2sina2-sinacosa cosa2](/uploads/image/f/4203944-8-4.jpg?t=%E5%B7%B2%E7%9F%A50%E3%80%88a%E3%80%88%CF%80%2Ctana%3D-2%E6%B1%822sina2-sinacosa+cosa2)
(5x+3)(x-2)=0∴tana=-0.6或tana=2我没看错题目的表达式的话……化简得,原式=[cosa*(-cosa)*(tana)^2]/[sina*(-sina)*(-cota)]=[-
tana/2+1/(tana/2)=5/2解上述方程,得tana/2=1/2或tana/2=2(舍去).已知0<a<π/2,可知,sina/2=1/√5,cosa/2=2/√5.sina=2sina/
解题思路:考查了同角三角函数的基本关系式,及其应用。考查了根式的运算解题过程:
sin²2a+sin2acosa-cos2a=1(2sinacosa)^2+2sina(cosa)^2-2(cosa)^2+1=12(cosa)^2[2(sina)^2+sina-1]=02
tanA=tan[(B+A)-B]=[tan(B+A)-tanB]/[1+tan2BtanB]=tanB/[1+2(tanB)^2]=1/[1/tanB+2tanB]≤1/(2√2)=√2/4
tan(a-π/4)=(tana-tanπ/4)/(1+tana*tanπ/4)=(2-1)/(1+2*1)=2/3
tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+si
将tan(a+b)化简,易知tana*tanb=1/2
tan(a+π/4)=(tana+tanπ/4)/1-tana*tanπ/4=(2+1)/1-2*1=-3
A∈(-π/2,0)说明角A是第四象限角tan是负的,sin(A+π/2)=1/3可知cosA=1/3则sinA=-根号(1-cosA的平方)=2倍根号2/3,从而tanA=sinA/cosA=-2倍
因为tana/2=1/2,所以tana=(2*tana/2)/[1-(tana/2)^2]=(2*1/2)/(1-1/4)=4/3,因为0
1.∵tan(a/2)=2∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3∴tan(a+π/4)=[tana+tan(π/4)]/[1-tan
1(√3)(tanAtanB+a)+2tanA+3tanB=0①(tanA+tanB)/(1-tanAtanB)=tan(A+B)=1/(√3)3(tanA+tanB)+(√3)(tanAtanB-1
sina+cosa=tana-1/tanasina+cosa=sina/cosa-cosa/sinasina+cosa=(sin²a-cos²a)/sinacosa1=(sina-
∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6
别做了没用、以后生活也用不着
2tana×sina=32sin^2a-3cosa=02cos^2a+3cosa-2=0(2cosa-1)(cosa+2)=0cosa=1/2,a=-π/3,cosa=-2(舍)cos(a-π/6)=
解;:(tana/2)/(1-tana/2的平方)=1/4∴2tana/2/(1-tan²a/2)=1/2∴tana=1/2∵a,b∈(0,π/4)∴2sina=cosa∴sina=√5/5
tan(π/4-a)=1-tana/1+tana=根号5;cot(π/4+a)=cot(π/2-(π/4-a))=tan(π/4-a)=根号5;故选B
sin(2π+a)cos(-π+a)/cos(-a)tana=sin(a)cos(π-a)/cos(a)(sina/cosa)=-sinacosa/sina=-cosa=-1/4