问题 解答题
已知f(α)=
sin(
π
2
+α)+3sin(-π-α)
2cos(
11π
2
-α)-cos(5π-α)

(Ⅰ)化简f(α);
(Ⅱ)已知tanα=3,求f(α)的值.
答案

(Ⅰ)因为f(α)=

sin(
π
2
+α)+3sin(-π-α)
2cos(
11π
2
-α)-cos(5π-α)

所以f(α)=

cosα+3sinα
-2sinα+cosα

(Ⅱ)因为tanα=3,f(α)=

cosα+3sinα
-2sinα+cosα
=
1+3tanα
-2tanα+1
=
10
-5
=-2

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