问题 填空题
已知tan(α+β)=
1
2
,tan(β-
π
4
)=
1
3
,则sin(
π
4
+α)•sin(
π
4
-α)
=______.
答案

tan(α+β)=

1
2
,tan(β-
π
4
)=
1
3
,∴tan(α+
π
4
)
=tan[(α+β)-(β-
π
4
)]
=
tan(α+β)-tan(β-
π
4
)
1+tan(α+β)tan(β-
π
4
)
=
1
2
-
1
3
1+
1
2
×
1
3
=
1
7

tanα=tan(α+

π
4
-
π
4
)=
tan(α+
π
4
)-tan
π
4
1+tan(α+
π
4
)tan
π
4
=
1
7
-1
1+
1
7
=-
3
4

sin(

π
4
+α)•sin(
π
4
-α)=
2
2
(cosα+sinα)•
2
2
(cosα-sinα)
=
1
2
(cos2α-sin2α)
=
1
2
×
cos2α-sin2α
cos2α+sin2α
=
1
2
×
1-tan2α
1+tan2α
=
1
2
×
1-(-
3
4
)2
1+(-
3
4
)2
=
7
50

故答案为

7
50

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