问题 解答题
已知 f(α)=
sin(2π-α)cos(π+α)cos(
π
2
+α)cos(
11π
2
-α)
2sin(3π+α)sin(-π-α)sin(
2
+α)

①化简f(α).
②若sinα是方程10x2+x-3=0的根,且α在第三象限,求f(α)的值.
③若a=-
25
4
π
,求f(α)的值.
答案

f(α)=

sin(2π-α)cos(π+α)cos(
π
2
+α)cos(
11π
2
-α)
2sin(3π+α)sin(-π-α)sin(
2
+α)

=

sin(2π-α)cos(π+α)cos(
π
2
+α)cos[6π-(
π
2
+α)]
2sin[2π+(π+α)]sin[-(π+α)]sin[4π+(
π
2
+α)]

=

(-sinα)(-cosα)(-sinα)(-sinα)
2(-sinα)sinαcosα

=-

1
2
sinα;…(4分)

②由方程10x2+x-3=0,解得:x1=

1
2
x2=-
3
5

又α在第三象限,∴sinα=-

3
5

f(α)=-

1
2
sinα=-
1
2
×(-
3
5
)=
3
10
;…(8分)

(3)当a=-

25
4
π时,f(α)=-
1
2
sin(-
25
4
π)=-
1
2
×sin(-6π-
π
4
)=-
1
2
×sin(-
π
4
)=
2
4
.…(12分)

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