=sin²acos²b-cos²asin²b
=sin²a(1-sin²b)-(1-sin²a)sin²b
=sin²a-sin²asin²b-sin²b+sin²asin²b
=sin²a-sin²b
=右边
命题得证
证明sin(a+b)sin(a-b)=sin^2 a-sin^2 b,
=sin²a(1-sin²b)-(1-sin²a)sin²b =sin²a-sin²asin²b-sin²b+sin²asin²b =sin²a-sin²b =右边 命题得证
sin(a+b)*sin(a-b)\/sin^2 a*sin^2 b=1-tan^2 b\/tan^2 a怎么证明?
sin(a+b)*sin(a-b)\/sin^2 a*sin^2 b=(sina^2cosb^2-cosa^2sinb^2)\/sina^2*sinb^2 =1\/tanb^2-1\/tana^2=(tana^2-tanb^2)\/(tana^2tanb^2)1-tan^2 b\/tan^2 a=(tan^2a-tanb^2)\/tana^2 如果你要求的等式成立 那么(tana^2-tanb^2)\/(tana^2tanb^2)=(tan^2a-tanb^2...
sin(a+b)*sin(a-b)\/sin^2 a*sin^2 b=1-tan^2 b\/tan^2 a怎么证明?
sin(a+b)*sin(a-b)\/sin^2 a*sin^2 b=(sina^2cosb^2-cosa^2sinb^2)\/sina^2*sinb^2 =1\/tanb^2-1\/tana^2=(tana^2-tanb^2)\/(tana^2tanb^2)1-tan^2 b\/tan^2 a=(tan^2a-tanb^2)\/tana^2 如果你要求的等式成立 那么(tana^2-tanb^2)\/(tana^2tanb^2)=(tan^2a-tanb^2...
...平方正弦公式sin(a+β)sin(a-β)=sin^2a-sin^2β
sin(a+β)sin(a-β) - cos(a+β)cos(a-β) = -cos(2a) = -(1 - 2sin^2a) = 2sin^2a - 1sin(a+β)sin(a-β) + cos(a+β)cos(a-β) = cos(2β) = 1 - 2sin^2β二式相加,即可得到平方正弦公式
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+sin^2a+(1-sin^2b)=sin^2 a-sin^2 b sin²20°...
sin(a+ b) sin(a- b)=什么公式?
sin(a+b)sin(a-b)=[cos(a+b-a+b)-cos(a+b+a-b)]\/2 =[cos(2b)-cos(2a)]\/2。又由二倍角公式 [cos(2b)-cos(2a)]\/2=[1-2sin^2b-（1-2sin^2a）]\/2=[2sin^2a-2sin^2b）]\/2=sin^2 a-sin^2b。诱导公式 诱导公式是指三角函数中，利用周期性将角度比较大的三角函数，...
此式求证明sin(a+b)sin(a-b)=sin2a-sin2b 2代表的是平方
sin(a+b)=sinacosb+sinbcosa sin(a+b)=sinacosb-sinbcosa sin(a+b)sin(a-b)=(sin²acos²b-sin²bcos²a)=[sin²a(1-sin²b)-sin²b(1-sin²a)]=sin²a-sin²b
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+(...
求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B
sinαsinβ=-[cos(α+β)-cos(α-β)]\/2 积化和差后简单
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 + ...
