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
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![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](/uploads/image/z/8768955-3-5.jpg?t=%E8%AF%81%E6%98%8EcosA%EF%BC%8BcosB%EF%BC%8BcosC%3D1%2B4sin%28A%2F2%29sin%28B%2F2%29sin%28C%2F2%29%E8%AF%81%E6%98%8E%EF%BC%9AcosA%EF%BC%8BcosB%EF%BC%8BcosC%3D1%2B4sin%28A%2F2%29sin%28B%2F2%29sin%28C%2F2%29%E5%B0%BD%E9%87%8F%E8%AF%A6%E7%BB%86%E4%B8%80%E7%82%B9%E9%80%89%E5%81%9AcosA%2BcosB%2BcosC%3D1%2B4sin%28A%2F2%29sin%28B%2F2%29sin%28C%2F2%29+cos%28180-B-C%29%2BcosB%2BcosC%3D1%2B2sin%28A%2F2%29%5B2sin%28B%2F2%29sin%28C%2F2%29%5D+cos%28180-B-C%29%2BcosB%2BcosC)
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
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)
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
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)
汗 愁啊 不懂