三国志战略版+s5开荒转型
@卜眉5458：三国志战略版倚天剑特技如何获取 -
徐食19826922031…… 首先介绍倚天剑的【奸雄】特技:曹操装备时,战斗首回合我军群体(2人)获得洞察状态.S5开荒阵容基本是核弹张和三势陆逊,而本套阵容的开荒思路是在于不需要转型,不需要三势陆的20统御,没有核弹张阵容开荒遇到7,8级地时的乏力...
@卜眉5458：计算cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π
徐食19826922031…… (cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π) = (cos5分之π+cos5分之4π + +cos5分之2π+cos5分之3π) = 2cos[(5分之π+5分之4π)/2]*cos[(5分之π - 5分之4π)] + 2cos[(5分之2π+5分之3π)/2]*cos[(5分之2π - 5分之3π)] = 2cos(π/2)*cos[(5分之π - 5分之4π)] + 2cos(π/2)*cos[(5分之2π - 5分之3π)] = 0; 用到cos(a) + cos(b) = 2cos[(a+b)/2]*cos[(a-b)/2]
@卜眉5458：求值cosπ/9+cos3π/9+cos5π/9+cos7π/9 -
徐食19826922031…… cosπ/9+cos3π/9+cos5π/9+cos7π/9=cos3π/9+cos(π-π/9)+cos(π-4π/9)+cos(π-2π/9)=cosπ/3-(cos8π/9+cos...
@卜眉5458：cos5/12πcosπ/12+cosπ/12sinπ/6=? -
徐食19826922031…… cos5/12πcosπ/12+cosπ/12sinπ/6=cosπ/12(cos5/12π+sinπ/6)=cosπ/12(cos(π/2-π/12)+sinπ/6)=cosπ/12(sin(π/12)+2*sin(π/12)*cos(π/12))=sin(π/12)*cos(π/12)*(1+2*cos(π/12))=1/2*sin(π/6)*(1+2*cos(π/12))=1/4*(1+2*cos(π/12))
@卜眉5458：1/2 - sin平方19π/12=多少啊 -
徐食19826922031…… 1/2-sin²(19π/12)=(1/2)[1-2sin²(19π/12)]=(1/2)cos(19π/6)=(1/2)cos(3π+π/6)=-(1/2)cos(π/6)=-(1/2)(√3/2)=-√3/4.
@卜眉5458：(sina5π/12+cos5π/12)(sin5π/12 - cos5π/12) -
徐食19826922031…… (sin5π/12+cos5π/12)(sin5π/12-cos5π/12)=sin²5π/12-cos²5π/12 =(1-cos5π/6)/2-(1+cos5π/6)/2 =(1+√3/2)/2-(1-√ 3/2)/2 =√3/2
@卜眉5458：《三国志战略版》零氪怎么搭配? -
徐食19826922031…… 零氪开荒推荐:孙坚、孙策、关平.获取难度:这三个武将组合获得要求低,对零氪玩家友好.强度方面:不仅输出强力,还带有治疗、控制这样的额外功能,非常实用. 《三国志战略版》零氪开荒阵容组合搭配攻略: 零氪开荒推荐:孙坚、...
@卜眉5458：√1 - sin10° -
徐食19826922031…… √(1-sin10°) =√(sin5°-cos5°)^2 =cos5°-sin5°
@卜眉5458：cos5π/13+cos3π/13+cos10π/13+cos8π/13=0 pai=π -
徐食19826922031…… cos x + cos(pai - x) = cos x - cos x = 0 所以cos5π/13 +cos8π/13 = 0 cos3π/13+cos10π/13 = 0
@卜眉5458：设{an}为等差数列,若a1+a5+a9=π,则tan(a2+a8)的值为 - ----- -
徐食19826922031…… 在等差数列{an}中,当m+n=p+q(m,n,p,q∈N+)时,am+an=ap+aq. 因为{an}为等差数列,且a1+a5+a9=π,所以有a5= π 3 ,所以a2+a8=2a5=2π 3 ,所以tan(a2+a8)=tan2π 3 =- 3 故答案为:- 3 .
徐食19826922031…… 首先介绍倚天剑的【奸雄】特技:曹操装备时,战斗首回合我军群体(2人)获得洞察状态.S5开荒阵容基本是核弹张和三势陆逊,而本套阵容的开荒思路是在于不需要转型,不需要三势陆的20统御,没有核弹张阵容开荒遇到7,8级地时的乏力...
@卜眉5458：计算cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π
徐食19826922031…… (cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π) = (cos5分之π+cos5分之4π + +cos5分之2π+cos5分之3π) = 2cos[(5分之π+5分之4π)/2]*cos[(5分之π - 5分之4π)] + 2cos[(5分之2π+5分之3π)/2]*cos[(5分之2π - 5分之3π)] = 2cos(π/2)*cos[(5分之π - 5分之4π)] + 2cos(π/2)*cos[(5分之2π - 5分之3π)] = 0; 用到cos(a) + cos(b) = 2cos[(a+b)/2]*cos[(a-b)/2]
@卜眉5458：求值cosπ/9+cos3π/9+cos5π/9+cos7π/9 -
徐食19826922031…… cosπ/9+cos3π/9+cos5π/9+cos7π/9=cos3π/9+cos(π-π/9)+cos(π-4π/9)+cos(π-2π/9)=cosπ/3-(cos8π/9+cos...
@卜眉5458：cos5/12πcosπ/12+cosπ/12sinπ/6=? -
徐食19826922031…… cos5/12πcosπ/12+cosπ/12sinπ/6=cosπ/12(cos5/12π+sinπ/6)=cosπ/12(cos(π/2-π/12)+sinπ/6)=cosπ/12(sin(π/12)+2*sin(π/12)*cos(π/12))=sin(π/12)*cos(π/12)*(1+2*cos(π/12))=1/2*sin(π/6)*(1+2*cos(π/12))=1/4*(1+2*cos(π/12))
@卜眉5458：1/2 - sin平方19π/12=多少啊 -
徐食19826922031…… 1/2-sin²(19π/12)=(1/2)[1-2sin²(19π/12)]=(1/2)cos(19π/6)=(1/2)cos(3π+π/6)=-(1/2)cos(π/6)=-(1/2)(√3/2)=-√3/4.
@卜眉5458：(sina5π/12+cos5π/12)(sin5π/12 - cos5π/12) -
徐食19826922031…… (sin5π/12+cos5π/12)(sin5π/12-cos5π/12)=sin²5π/12-cos²5π/12 =(1-cos5π/6)/2-(1+cos5π/6)/2 =(1+√3/2)/2-(1-√ 3/2)/2 =√3/2
@卜眉5458：《三国志战略版》零氪怎么搭配? -
徐食19826922031…… 零氪开荒推荐:孙坚、孙策、关平.获取难度:这三个武将组合获得要求低,对零氪玩家友好.强度方面:不仅输出强力,还带有治疗、控制这样的额外功能,非常实用. 《三国志战略版》零氪开荒阵容组合搭配攻略: 零氪开荒推荐:孙坚、...
@卜眉5458：√1 - sin10° -
徐食19826922031…… √(1-sin10°) =√(sin5°-cos5°)^2 =cos5°-sin5°
@卜眉5458：cos5π/13+cos3π/13+cos10π/13+cos8π/13=0 pai=π -
徐食19826922031…… cos x + cos(pai - x) = cos x - cos x = 0 所以cos5π/13 +cos8π/13 = 0 cos3π/13+cos10π/13 = 0
@卜眉5458：设{an}为等差数列,若a1+a5+a9=π,则tan(a2+a8)的值为 - ----- -
徐食19826922031…… 在等差数列{an}中,当m+n=p+q(m,n,p,q∈N+)时,am+an=ap+aq. 因为{an}为等差数列,且a1+a5+a9=π,所以有a5= π 3 ,所以a2+a8=2a5=2π 3 ,所以tan(a2+a8)=tan2π 3 =- 3 故答案为:- 3 .
