Question

2.如图,在+ABC+中,+AC=2ABcosBAC,+D为边CB延长线上一点,-|||-(1)求证:+sin2C=

Answer 1

问：2.如图,在+ABC+中,+AC=2ABcosBAC,+D为边CB延长线上一点,-|||-(1)求证:+sin2C=

答：由题目可知：1.∠BAC + ∠CAD = 180°（+ABC+中角内合为180度）2.+AC=2ABcosBAC （+ABC+中余弦定理）对于(1)，我们需要证明+sin2C= 。首先，由正弦定理可得：+AC/sin∠ABC = +BC/sin∠BAC即：+AC/sinC = +BC/sinB将+AC=2ABcosBAC代入上式，并将sinB化简为sin(180° - B - C)，得到：2ABcosBAC/sinC = +BC/sin(180° - B - C)化简一下得：2ABcosBAC/sinC = +BC/sin(B + C)再将+BC用AB表示出来，即可得到：2cosBAC/sinC = sin(B + C)/sinB将左边的2cosBAC用sinC表示出来，得到：2sinBcosBAC = sin(B + C)/sinB移项，得到：2sinBcosBACsinB = sin(B + C)化简，得到：2cosBACcosC = sin(B + C)接下来，将左边的2cosBACcosC用余弦公式代入，得到：cos(B - C) +

问：三角函数

问：可以写在纸上拍过来吗

答：这个很长

答：还在算

答：由题意可知，设∠ABC = A, ∠ACB = B，则∠BAC = 180° - A - B。根据正弦定理：sinA/AB = sinB/BC即：ABsinA = BCsinB又根据已知条件：AC = 2ABcosBAC所以：AC = 2ABcos(180° - A - B)化简得：AC = -2ABcos(A + B)所以：-AC/AB = 2cos(A + B)因为D是边CB的延长线上的一点，所以：∠DCB = 180° - ∠ACB化简得：∠DCB = A由正弦定理可得：sinC/BC = sinA/BD即：BDsinC = BCsinA将上式代入式子ABsinA = BCsinB，得：ABsinC = BDsinB又因为BD = BC + CD，所以：ABsinC = BCsinB + BCsinC化简得：ABsinC = BC(sinB + sinC)代入AC = 2ABcosBAC和∠BAC = 180° - A - B，得：cosBAC = -cos(A + B)化简得：sinBAC = sin(A + B)

问：还有这个

答：太长了

答：有点发不过来