题目内容
(2013•宁波模拟)函数y=sin(x+10°)+cos(x+40°),(x∈R)的最大值是
1
1
.分析:先将函数化简,利用三角函数的性质,即可确定函数的最值.
解答:解:函数y=sin(x+10°)+cos(x+40°)
=sin(x+10°)+cos(x+10°+30°)
=sin(x+10°)+cos(x+10°)cos30°-sin(x+10°)sin30°
=
sin(x+10°)+
cos(x+10°)
=sin(x+70°)
∵y=sin(x+70°)的最大值是1
∴函数y=sin(x+10°)+cos(x+40°)(x∈R)的最大值是1
故答案为:1
=sin(x+10°)+cos(x+10°+30°)
=sin(x+10°)+cos(x+10°)cos30°-sin(x+10°)sin30°
=
| 1 |
| 2 |
| ||
| 2 |
=sin(x+70°)
∵y=sin(x+70°)的最大值是1
∴函数y=sin(x+10°)+cos(x+40°)(x∈R)的最大值是1
故答案为:1
点评:本题考查三角函数式的化简,考查三角函数的性质,属于基础题.
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