已知在△中,abc成等差数列,B=π/4,求cos A-cos C

abc成等差数列,B=π/4,A+C=3π/4sinA,sinB,sinC成等差数列,sinB=√2/4,sinA+sinC=2sinB=√2(sinA+sinC)^2=2设cosA-cosC=tt^2=(cosA-cosC)^2(cosA-cosC)^2+(sinA+sinC)^2=t^2+2sinA * sinA+cosA * cosA=1;sinC * sinC...t^2=(cosA-cosC)^2 这一步往下的部分好乱，都看不懂。(cosA-cosC)^2+(sinA+sinC)^2=t^2+2;(cosA-cosC)^2+(sinA+sinC)^2=[(cosA)^2+(cosC )^2- 2cosA * cosC ]+[ (sinA) ^2+(sinC)^2+2sinA * sinC ]=sinA * sinA+cosA * cosA+sinC * sinC+cosC * cosC+2sinA * sinC - 2cosA * cosC=2+2sinA * sinC - 2cosA * cosC=2-2(cosA * cosC-sinA * sinC)=2-2cos(A+C)=2-2cos(π-B)=2-2cos3π/4=2+√2(cosA-cosC)^2+(sinA+sinC)^2=t^2+2=2+√2;t=±2^(1/4)=cosA-cosC是，太乱了。整理如上。