cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=
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cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=
cos(a-b)cos(b-c)+sin(a-b)sin(b-c)
=cos[(a-b)-(b-c)]
=cos(a+c-2b)
=cos[(a-b)-(b-c)]
=cos(a+c-2b)
cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=
cos^B-cos^C=sin^A,三角形的形状
sin²A+sin²B=cos²C
貌似不难,sin(a+b)cos(c-b)-cos(b+a)sin(b-c)sin(a-b)sin(b-c)-cos(a
在三角形中,已知,cos C/cos B=(3a-c)/b 求:sin B
sin(A+B/2)=cos(C/2)
cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b) 用三角形证明
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin
sin(B+C)=?cos(B+C)=?
若sin/a=cos/b=cos/c,则三角形是?
若sin/a=cos/b=cos/c,则△ABC是