已知f(α)=sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)2sin(3π+α)sin(-π-α)sin(9π2+α).①化简f(α).②若sinα是方程10x2+x-3=

题目简介

已知f(α)=sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)2sin(3π+α)sin(-π-α)sin(9π2+α).①化简f(α).②若sinα是方程10x2+x-3=

题目详情

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.
题型:解答题难度:中档来源:不详

答案

f(α)=
sin(2π-α)cos(π+α)cos(class="stub"π
2
+α)cos(class="stub"11π
2
-α)
2sin(3π+α)sin(-π-α)sin(class="stub"9π
2
+α)

=
sin(2π-α)cos(π+α)cos(class="stub"π
2
+α)cos[6π-(class="stub"π
2
+α)]
2sin[2π+(π+α)]sin[-(π+α)]sin[4π+(class="stub"π
2
+α)]

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

=-class="stub"1
2
sinα;…(4分)
②由方程10x2+x-3=0,解得:x1=class="stub"1
2
x2=-class="stub"3
5

又α在第三象限,∴sinα=-class="stub"3
5

f(α)=-class="stub"1
2
sinα=-class="stub"1
2
×(-class="stub"3
5
)=class="stub"3
10
;…(8分)
(3)当a=-class="stub"25
4
π
时,f(α)=-class="stub"1
2
sin(-class="stub"25
4
π)=-class="stub"1
2
×sin(-6π-class="stub"π
4
)=-class="stub"1
2
×sin(-class="stub"π
4
)=
2
4
.…(12分)

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