求证:sinA+sinB=2*sin(A+B)/2*cos(A-B)/2(即高一数学新课标必修4,140页思考题)
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求证:sinA+sinB=2*sin(A+B)/2*cos(A-B)/2(即高一数学新课标必修4,140页思考题)
求证:
sinA+sinB=2*sin(A+B)/2*cos(A-B)/2
(即高一数学新课标必修4,140页思考题)
求证:sinA+sinB=2*sin(A+B)/2*cos(A-B)/2(即高一数学新课标必修4,140页思考题)
sinA+sinB
= sin[(A+B)/2+(A-B)/2]+sin[(A+B)/2-(A-B)/2]
= sin(A+B)/2*cos(A-B)/2+cos(A+B)/2*sin(A-B)/2
+sin(A+B)/2*cos(A-B)/2-cos(A+B)/2*sin(A-B)/2
= 2*sin(A+B)/2*cos(A-B)/2
由于sin(A+B)=sinAcosB+sinBcosA;sin(A-B)=sinAcosB-sinBcosA,那么sin(a/2+b/2)=sin(a/2)cos(b/2)+cos(a/2)sin(b/2);cos(a/2-b/2)=cos(a/2)cos(b/2)+sin(a/2)sin(b/2);两式按照乘法分配律相乘,整理可得,后面的式子很繁琐,不好写,自己整整看。