=(sinacosb+cosasinb)(sinacosb-cosasinb)
=(sinacosb)^2-(cosasinb)^2
=(cosb)^2[1-(cosa)^2]-(cosa)^2[1-(cosb)^2]
=(cosb)^2-(cosb)^2(cosa)^2-(cosa)^2+(cosa)^2(cosb)^2
=(cosb)^2-(cosa)^2
=(cosb)^2-[1-(sina)^2]
=(sina)^2+(cosb)^2-1
=(sina)^2-(sinb)^2
题目有问题，请楼主认真核对!
=(-1/2)(1-2sin²a+2cos²b-1)
=sin²a-cos²b
求证sin(a+b)sin(a-b)=sina^2-cosb^2
证明：sin(a+b)sin(a-b)=(sinacosb+cosasinb)(sinacosb-cosasinb)=(sinacosb)^2-(cosasinb)^2 =(cosb)^2[1-(cosa)^2]-(cosa)^2[1-(cosb)^2]=(cosb)^2-(cosb)^2(cosa)^2-(cosa)^2+(cosa)^2(cosb)^2 =(cosb)^2-(cosa)^2 =(cosb)^2-[1-(sina)^2]=(sina)^2+(...
证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,
左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b =sin²a(1-sin²b)-(1-sin²a)sin²b =sin²a-sin²asin²b-sin²b+sin²asin²b =sin²a-sin²b =右边 命题得证 ...
证sin(a+b)sin(a-b)=cos^2b-cos^2a
sin(a+b)sin(a-b)=（sinacosb+cosasinb)(sinacosb-cosasinb)=sin^2acos^2b-cos^2asin^2b =(1-cos^2a)cos^2b-cos^2a(1-cos^2b)=cos^2b-cos^2acos^2b-cos^2a+cos^2acos^2b =cos^2b-cos^2a
若sin(A+B)sin(B-A)=m,则(cosA)^2-(cosB)^2=?
解答：-m=sin(a+b)sin(a-b)=(sinacosb+cosasinb)(sinacosb-cosasinb)=(sina)^2(cosb)^2-(cosa)^2(sinb)^2 =(sina)^2(cosb)^2-[1-(sina)^2](sinb)^2 =(sina)^2[(cosb)^2+(sinb)^2]-(sinb)^2 =(sina)^2-(sinb)^2 =1-(cosa)^2-1+(cosb)^2=-[(cosa)^2-(cos...
已知sin(a+b)sin(a-b)=2m(m不等于0),则(cosa)^2-(cosb)^2等于
解：sin（a+b）sin（a-b）=-1\/2（cos2a-cos2b）又 cos2a＝2(cosa)^2－1 cos2b＝2(cosb)^2－1 ∴sin（a+b）sin（a-b）=-1\/2(2(cosa)^2－1-2(cosb)^2+1)=-1\/2(2(cosa)^2-2(cosb)^2)=(cosa)^2-(cosb)^2 ∵sin（a+b）sin（a-b）=2m ∴（cosa）^2-（cosb）...
sin(α+β)+ sin(α-β)等于什么?
sin(A+B)= sinAcosB+cosAsinB……(1)sin(A-B)= sinAcosB-cosAsinB……(2)(1)+(2)可得：2sinAcosB = sin(A+B)+ sin(A-B)A=(x-a)\/2 B=(a+x)\/2 A+B= x A-B= a sinx+sina=2sin[(x-a)\/2]cos[(a+x)\/2]
和差化积公式推导过程是什么?
sin(a-b)=sina*cosb-cosa*sinb 我们把两式相加就得到sin(a+b)+sin(a-b)=2sina*cosb 所以，sina*cosb=(sin(a+b)+sin(a-b))\/2 同理，若把两式相减，就得到cosa*sinb=(sin(a+b)-sin(a-b))\/2 同样的，我们还知道cos(a+b)=cosa*cosb-sina*sinb cos(a-b)=cosa*cosb+sina*...
证明sin(a+b)sin(a-b)=sin²a-sin² b,并用该式计算sin²20°+s...
sin(a+b)sin(a-b)=(sinacosb+sinbcosa)(sinacosb-sinbcosa) =(sinacosb)^2+sinasinbcosacosb-sinasinbcosacosb-(sinbcosa)^2 =(sinacosb)^2-(1-sin^2a)(1-cos^2b) =(sinacosb)^2-[1-sin^2a-cos^2b+(sinacosb)^2] =-1...
三角函数公式的问题sin^A+sin^B=sin(A+B)sin(A-B)是怎么解得,sin^2A...
sin(A+B)sin(A-B)=-1\/2*[cos(A+B+A-B)-cos(A+B-A+B)]=-1\/2[cos2A-cos2B]=-1\/2[1-2sin^2A-(1-2sin^2B)=-1\/2[-2sin^2A+2sin^2B]=sin^2A-sin^2B,2,让我想想,2,sin^A什么意思？,1,到底是哪个,1,把后面用积化和差公式变形再做,1,sin(A+B) = sinAcosB + ...
高一数学题 求证sin(a+b)cos(a-b)-cos(a+b)sin(a-b)=sin2b
令A=a+b B=a-b 则sin（a+b）cos（a-b）-cos（a+b）sin（a-b）=sinAcosB-cosAsinB =sin(A-B)=sin[(a+b)-(a-b)]=sin2a 不要被题目给你的形式所迷惑
