证明cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/14 02:57:49

证明cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)
证明：cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)尽量详细一点选做cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2) cos(180-B-C)+cosB+cosC=1+2sin(A/2)[2sin(B/2)sin(C/2)] cos(180-B-C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)] -cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)] -cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[cos(B/2-C/2)-cos(B/2+C/2)] -cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)cos(B/2-C/2)-2[cos(B/2+C/2)]^2 cosB+cosC=2cos(B/2+C/2)cos(B/2-C/2)

应该是在三角形中吧 ,要不然没法算
cosA+cosB+cosC
=2cos[(A+B)/2]cos[(A-B)/2]+cosC
=2cos[(π-C)/2]cos[(A-B)/2]+cosC
=2sin(c/2)cos[(A-B)/2]+1-2[sin(C/2)]^2
=1+2sin(c/2){cos[(A-B)/2]-[sin(C/2)]}
=1+2sin(c/2){cos[(A-B)/2]-[sin(π -A-B)/2]}
=1+2sin(c/2){cos[(A-B)/2]-[cos[(A+B)/2]}
=1+2sin(c/2)[-2sin(A/2)sin(-B/2)]
=1+4sin(A/2)sin(B/2)sin(C/2)

三角形ABC中证明 COSA+COSB+COSC=1+4SIN(A/2)*SIN(B/2)*SIN(C/2) 在三角形ABC中,内角A,B,c的对边a,b,c.已知（2c-a)/b=(cosA-2cosC）/cosB.1、求sin 2sinC-sinA/sinB=cosA-2cosC/cosB 如何整理求得sin（A+B）=2sin（B+C）已知tanC=(sinA+sinB)/(cosA+cosB),sin(B-A)=cosC.（1）求A,C；（2）S△AB 在三角形ABC中,tanC=(sinA+sinB)/(cosA+cosB),sin(B-A)=cosC【1】求A,C 为什么cosa-cosb=-2sin[(a-b)/2]*sin[(a+b)/2], 若sin(A+B)sin(B-A)=m,则(cosA)^2-(cosB)^2=? 1.若sin(A+B)sin(B-A)=M,则(cosA)^2-(cosB)^2等于? 已知sina+cosb=1/2,cosa+sinb=3/4,求sin(a+b) 【证明】Sin A+sin B=2Sin 22 sina+cosb=1/2,cosa+sinb=1/3 求sin(a+b)的值 sina(cosa)^3=1/4sin 2a+1/8sin 4a证明过程