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

(Ⅰ)化简f(α);
(Ⅱ)若α是第三象限角,且cos(α-
2
)=
2
2
3
,求f(α)的值.
答案

(Ⅰ)f(α)=

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

=

sinα(-cosα)(-sinα)(-sinα)
(-cosα)sinαsinαcosα
=tanα

(Ⅱ)cos(α-

2
)=-sinα=
2
2
3

∴sinα=-

2
2
3

∵α是第三象限角

∴cosα=-

1-
8
9
=-
1
3

∴f(α)=tanα=

sinα
cosα
=2
2

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