三角形ABC的三个内角满足2B=A+C则y=(cos2A+cos2C)/2+1的最小值是
问题描述:
三角形ABC的三个内角满足2B=A+C则y=(cos2A+cos2C)/2+1的最小值是
答
A+B+C=2B+B=3B=180B=60A+C=120y=(cos2A+cos2C)/2+1=2cos(2A+2C)/2cos(2A-2C)/2 /2+1=cos(A+C)cos(A-C)+1=cos120cos(A-C)+1=-cos(A-C) / 2 +1当cos(A-C)=1时则有最小值y=1-1/2=1/2