Question

三角函数值域问题

y=cos2x+3sinx最值
要详细过程

Answer 1

y=cos2x+3sinx=(1-2sin^2x)+3sinx=-2(sin^2x-3/2sinx)+1
=-2(sinx-3/4)^2+9/8+1=-2(sinx-3/4)^2+17/8
-7/4<=sinx-3/4<=1/4
所以 0<=(sinx-3/4)^2<=49/16
max=17/8
min=-2*49/16+17/8=-32/8=-4

Answer 2

因为：y=cos2x+3sinx
所以：y=1-2sin²x+3sinx
将sinx设为 t
所以：y=-2t²+3t+1
根据方程画出图像
因为：-1＜sinx＜1
所以：-1＜t＜1
看图就可得到最值

Answer 3

y=cos2x+3sinx=(1-2sin²x)+3sinx=-2(sin²x-3/2sinx)+1
=-2(sinx-3/4)²+9/8+1=-2(sinx-3/4)²+17/8
显然sinx=¾时y取最大值17/8
sinx=-1时y取最小值=-2×49/16+17/8=-32/8=-4 。

Answer 4

y=cos2x+3sinx=(1-2sin^2x)+3sinx=-2(sin^2x-3/2sinx)+1
=-2(sinx-3/4)^2+9/8+1=-2(sinx-3/4)^2+17/8
-7/4<=sinx-3/4<=1/4 0<=(sinx-3/4)^2<=49/16
max=17/8
min=-2*49/16+17/8=-32/8=-4