这道题怎么做,sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b)
(cosa)^2=1,把所有的正弦化为余弦,再乘进去,可以消掉两项;最后得原式为:(cosb)^2-(cosa)^2;最后,由余弦和角公式得:(cosb)^2=1\/2【(cos2b)1】,所以原式可化为:1\/2【(cos2b)-(cos2a)】,把中间两项一倒,可证最后答案。
这道题怎么做,sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b) 给...
cos2a-cos2b=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)-(cos(a+b)cos(a-b)+sin(a+b)sin(a-b))=-2sin(a+b)sin(a-b)所以 sin(a+b)sin(a-b)=-1\/2(cos2a-cos2b)
sin(a+b)sin(a-b)怎么化成 负二分之一(cos2a-cos2b)
cos2a-cos2b=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)-(cos(a+b)cos(a-b)+sin(a+b)sin(a-b))=-2sin(a+b)sin(a-b)所以 sin(a+b)sin(a-b)=-1\/2(cos2a-cos2b)
已知sin(a+b)*sin(a-b)=m,求cos^2a-cos^2b的值
根据积化和差公式:sinαsinβ=-1\/2[cos(α+β)-cos(α-β)]sin(a+b)sin(a-b)=-1\/2[cos(a+b+a-b)-cos(a+b-a-b)]=-1\/2(cos2a-cos2b)=-1\/2[(2cos^2a-1)-(2cos^2b-1)]=-1\/2(2cos^2a-2cos^2b)=-1\/2[2(cos^2a-cos^2b)]=-(cos^2a-cos^2b)所以cos^2a-...
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
sin(a+b)sin(a-b)=
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,...
...cos2A)=sinBsin(A+B)化简成sin(A+B)sin(A-B)=sinBsin(A+B)_百度知...
1\/2(cos2B-cos2A)=1\/2*-2sin[(2B+2A)\/2]sin[(2B-2A)\/2] =1\/2*-2sin(A+B)sin(B-A) =sin(A+B)sin(A-B)
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)...
已知sin(A+B)·sin(B-A)=m
sin(A+B)·sin(B-A)直接积化和差sin(A+B)·sin(B-A)=-1\/2(cos2B-cos2A)=m cos2A-cos2B=-2m
...sin^A+sin^B=sin(A+B)sin(A-B)是怎么解得,sin^2A—sin^2B 是A...
- cosAsinB)= sin^2 Acos^2 B - cos^2 Asin^2 B - -...不相等啊,1,得不到啊,也许是我错了。从右向左推,结果:(sinA+sinB)(sinA-sinB),0,三角函数公式的问题 sin^A+sin^B=sin(A+B)sin(A-B)是怎么解得,sin^2A—sin^2B 是A B的平方,对不起打错了 就是这个,平方,