Question

已知tan(α+β)=

1

2

，tan(β-

π

4

)=

1

3

，则sin(

π

4

+α)•sin(

π

4

-α)=______．

Answer 1

∵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

．