题目内容
函数y=sin2x-cosx+1最小值为
0
0
.分析:化正弦为余弦,然后利用配方法求最值.
解答:解:y=sin2x-cosx+1
=1-cos2x-cosx+1
=-cos2x-cosx+2
=-(cosx+
)2+
.
当cosx=1时,ymin=-(1+
)2+
=0.
故答案为:0.
=1-cos2x-cosx+1
=-cos2x-cosx+2
=-(cosx+
| 1 |
| 2 |
| 9 |
| 4 |
当cosx=1时,ymin=-(1+
| 1 |
| 2 |
| 9 |
| 4 |
故答案为:0.
点评:本题考查了复合函数的单调性,训练了利用配方法求二次函数的最值,是中档题.
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