求值cos^2α+cos^2(α+120)+cos^2(α+240) 2 ;y=(sinα+1)(cosα+1)求最大最小值

题目

求值cos^2α+cos^2(α+120)+cos^2(α+240) 2 ;y=(sinα+1)(cosα+1)求最大最小值
2;是第二题的意思

答案

1.原式=cos²a+cos²(a+120°)+cos²(a-120°)
=cos²a+1/2[1+cos(2a+240°)]+1/2[1+cos(2a-240°)]
=cos²a+1+1/2[cos(2a+240°)+cos(2a-240°)]
=cos²a+1+cos2acos240°
=3/2+1/2cos2a-1/2cos2a
=3/2.
2.y=sinacosa+sina+cosa+1
令sina+cosa=x,则sinacosa=1/2[(sina+cosa)²-1]=1/2(x²-1).
y=1/2(x²-1)+x+1
=1/2x²+x+1/2
=1/2(x+1)².
当x=-1时取最小值0；当x=根号2时取最大值3/2+根号2.