已知f(α)=sin(π2+α)+3sin(-π-α)2cos(11π2-α)-cos(5π-α).(Ⅰ)化简f(α);(Ⅱ)已知tanα=3,求f(α)的值.-数学

题目简介

已知f(α)=sin(π2+α)+3sin(-π-α)2cos(11π2-α)-cos(5π-α).(Ⅰ)化简f(α);(Ⅱ)已知tanα=3,求f(α)的值.-数学

题目详情

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.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)因为f(α)=
sin(class="stub"π
2
+α)+3sin(-π-α)
2cos(class="stub"11π
2
-α)-cos(5π-α)

所以f(α)=class="stub"cosα+3sinα
-2sinα+cosα

(Ⅱ)因为tanα=3,f(α)=class="stub"cosα+3sinα
-2sinα+cosα
=class="stub"1+3tanα
-2tanα+1
=class="stub"10
-5
=-2

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