Question

cos方a+cos^2c cos方b=1-sina sin c求角b

在△ABC中,边abc的对角分别为ABC,且sin^2A+sin^2C--sinA*sinC=sin^2B.
（1）求角B的值（2）求2cos^2A+cos(A--C)的范围

Answer 1

(1)正弦定理：a/sinA=b/sinB=c/sinC=2R
∴由sin²A+sin²C-sinAsinC=sin²B得：a²+c²-ac=b²
∴cosB=(a²+c²-b²)/2ac=ac/2ac=1/2 【余弦定理】
而B∈(0,π),∴B=π/3
(2) C=π-B-A=2π/3-A
∴2cos²A+cos(A-C)=1+cos2A+cos(A-2π/3+A)
=cos2A+cos(2A-2π/3)+1
=cos2A+cos2A*(-1/2)+sin2A*(√3/2)+1
=(1/2)*cos2A+(√3/2)*sin2A+1
=sin(2A+π/6)+1
而∵0