Question

证明SINA+SINB+SINC

Answer 1

更新1:

不好意思，我看不懂这一步 = sin(A + B) = 2 sin[(A + B)/2] cos[(A + B)/2]

更新2:

明了!!

看以下解答时最好同时拿着「和差化积」及「积化和差」的公式表一起参看。 (1) sin C = sin(π - A - B)
　因为 三角形内角和 = A+B+C = π = sin(A + B) = 2 sin[(A + B)/2] cos[(A + B)/2] (2) sinA + sinB + sinC = (sinA + sinB) + sinC = 2 sin[(A + B)/2] cos[(A - B)/2]+ 2 sin[(A + B)/2] cos[(A + B)/2] = 2 sin[(A + B)/2] • {cos[(A + B)/2] + cos[(A - B)/2]} = 2 cos[π/2 - (A + B)/2] • {2 cos ½[(A + B)/2 + (A - B)/2] cos ½[(A + B)/2 + (A - B)/2] } = 2 cos[(π - A - B)/2]• [2 cos(A/2) cos(B/2)] = 2 cos(C/2) • 2 cos(A/2) cos(B/2) = 4 cos(A/2) cos(B/2) cos(C/2) 2007-10-27 20:11:23 补充： use sin(2x)=2sinxcosx