问题
选择题
函数y=cos2x+sin(
|
答案
∵y=f(x)=cos2x+sin(
-x)π 2
=cos2x+cosx
=2cos2x+cosx-1
=2(cosx+
)2-1 4
,9 8
∵f(-x)=f(x),
∴y=f(x)为偶函数,又-1≤cosx≤1,
∴当cosx=-
时,f(x)min=-1 4
;9 8
当cosx=1时,f(x)max=2.
故选C.
函数y=cos2x+sin(
|
∵y=f(x)=cos2x+sin(
-x)π 2
=cos2x+cosx
=2cos2x+cosx-1
=2(cosx+
)2-1 4
,9 8
∵f(-x)=f(x),
∴y=f(x)为偶函数,又-1≤cosx≤1,
∴当cosx=-
时,f(x)min=-1 4
;9 8
当cosx=1时,f(x)max=2.
故选C.