问题 填空题
已知函数f(x)=(1+cos2x)sin2x,x∈R.若f(α)=
1
4
,则f(α+
π
8
)
=______.
答案

f(x)=(1+cos2x)sin2x=(1+cos2x)

1-cos2x
2
=
1-cos22x
2
=
1
4
-
1
4
cos4x

因为f(α)=

1
4
,即
1
4
-
1
4
cos4α=
1
4
,解得α=
4
+
π
8
,k∈Z,

所以f(α+

π
8
)=
1
4
-
1
4
cos4(α+
π
8
)=
1
4
-
1
4
cos4(
4
+
π
4
)=
1
4
-
1
4
cos(kπ+π),

当k为偶数时,f(α+

π
8
)=
1
2
,当k为奇数时,f(α+
π
8
)=0,

所以f(α+

π
8
)=
1
2
或0,

故答案为:

1
2
或0.

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