Question

高中三角函数题

在三角形ABC中，若a+c=2b，试求cosA+cosC-cosAcosC+1/3sinAsinC的值。

Answer 1

∵2b=a+c
∴由正弦定理得：2sinB=sinA+sinC
利用和化积公式：sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
∴2sin(B/2)cos(B/2) = sin[(A+C)/2]cos[(A-C)/2]
∵ 在三角形ABC中，∠A+∠B+∠C=π
∴B/2 = (A+C)/2 为锐角
∴2sin(B/2) = cos[(A-C)/2]
2sin[(π/2) - (A+C)/2] = cos[(A-C)/2]
2cos[(A+C)/2] = cos[(A-C)/2]
则：原式=2cos[(A+C)/2]cos[(A-C)/2] - (1/2)[cos(A+C) + cos(A-C)] + (1/3)*(-1/2)[cos(A+C) + cos(A-C)]
=2cos[(A+C)/2]cos[(A-C)/2] - (2/3)cos(A+C) - (1/3)cos(A-C)
=4[cos (A+C)/2]^2 - (2/3){2[cos (A+C)/2]^2 - 1]} - (1/3){2[cos (A-C)/2]^2 - 1}
=1