问题 解答题
已知f(x)=
sin2x-2sin2x
1-tanx

(Ⅰ)求函数f(x)的定义域和最小正周期;
(Ⅱ)当cos(
π
4
+x)=
3
5
时,求f(x)的值.
答案

(Ⅰ)由1-tanx≠0得x≠kπ+

π
4
(k∈Z).又x≠kπ+
π
2
(k∈Z)

∴函数的定义域为{x|x∈R,且x≠kπ+

π
4
,x≠kπ+
π
2
(k∈Z)}.

f(x)=

sin2x-2sin2x
1-tanx
=
cosx•2sinx(cosx-sinx)
cosx-sinx
=sin2x,

∴f(x)的最小正周期为π(7分)

(Ⅱ)∵cos(

π
4
+x)=
3
5

f(x)=sin2x=-cos(2x+

π
2
)=-2cos2(x+
π
4
)+1=-2×
9
25
+1=
7
25

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