12求证sin3A=3sinA-4Sin^3A?求证cos3a=4cos^3a-3cosa?

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12求证sin3A=3sinA-4Sin^3A?求证cos3a=4cos^3a-3cosa?
12
求证sin3A=3sinA-4Sin^3A?
求证cos3a=4cos^3a-3cosa?

12求证sin3A=3sinA-4Sin^3A?求证cos3a=4cos^3a-3cosa?
sin3A=sin(2A+A)=sin2AcosA+cos2AsinA
=(2sinAcosA)cosA+(2cos^2A-1)sinA
=4sinAcos^2A-sinA
=4sinA(1-sin^2A)-sinA
=3sinA-4sin^3A.
cos3a=cos(2a+a)=cos2acosa-sin2asina
=(2cos^2a-1)cosa-2sin^2acosa
=2cos^3a-cosa-2(1-cos^2a)cosa
=4cos^3a-3cosa.
求证完毕!

sin3A=sin(2A+A)
=sin2AcosA+cos2AsinA
=2sinAcosAcosA+(1-2(sinA)^2)sinA
=sinA(2(cosA)^2+1-2(sinA)^2)
=sinA(2(1-(sinA)^2)+1-2(sinA)^2)
=sinA(3-4(sinA)^2)
=3sinA-4(sinA)^3
cos3...

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sin3A=sin(2A+A)
=sin2AcosA+cos2AsinA
=2sinAcosAcosA+(1-2(sinA)^2)sinA
=sinA(2(cosA)^2+1-2(sinA)^2)
=sinA(2(1-(sinA)^2)+1-2(sinA)^2)
=sinA(3-4(sinA)^2)
=3sinA-4(sinA)^3
cos3A=cos(2A+A)
=cos2AcosA-sin2AsinA
=(2(cosA)^2-1)cosA-2sinAcosAsinA
=cosA(2(cosA)^2-1-2(sinA)^2)
=cosA(2(cosA)^2-1-2(1-(cosA)^2)
=cosA(4(cosA)^2-3)
=4(cosA)^3-3cosA

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12求证sin3A=3sinA-4Sin^3A?求证cos3a=4cos^3a-3cosa? 求证:(1)sin3a=3sina-4sin^3a(2)cos3a=4cos^3a-3cosa 求证sin3a*cos^3a+cos3a*sin^3a=3/4sin4a 求证：sin3a*cos^3a+cos3a*sin^3a=3/4sin4a 求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2] 证明sin3A=3sinA-4sin3次方A 已知sina+sinβ=√3（cosβ-cosa）,a,β∈（0,π/2）.求证：sin3a+sin3β=0 求证:(sinA+sin2A+sin3A)/(cosA+cos2A+cos3A)=tan2A 提问一道高中知识三角函数题3sinA-4sinA*sinA*sinA=sin3A需要推导过程求证:sin3a*sin^3a+cos3a*cos^3a=cos^3 2a 求证cos3A*cos^3 A+sin3A*sin^3 A=cos^3 2A这题好绕啊~ 求证:sin3a·sin^3 a+cos3a·cos^3 a=cos^2 a 三倍角公式推导,举个例子：sin3a=sin(2a+a)=sin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina=3sina-4sin³asin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina就这两步之间怎么推？求证：(sin3A+4cos2A+sinA-4)/(3cosA+cos3A-4sin2A)=tanA应该是用到三倍角公式的,可是我就是做不了求证：(sin3A+4cos2A+sinA-4)/(3cosA+cos3A-4sin2A)=tanA应该是用到三倍角公式的,可是我就是做不了 sin3a/sina-cos3a/cosa=? sina=sin3a,求sin2A 二、化简：1、（sina+cosa）² 2、cos⒋a-sin⒋a(4是四次方)三、证明：1、 sin3a=3sina-4sin⒊a2、 cos3a=4cos⒊a-cosa3、已知sin(a+b)=1/2,sin(a-b)=1/3求证（1）sinacosb=5cosasinb,（2）tana=5tanb四、已知sin(a-b)cosa-c