问题
解答题
已知 f(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=-
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答案
①f(α)=sin(2π-α)cos(π+α)cos(
+α)cos(π 2
-α)11π 2 2sin(3π+α)sin(-π-α)sin(
+α)9π 2
=sin(2π-α)cos(π+α)cos(
+α)cos[6π-(π 2
+α)]π 2 2sin[2π+(π+α)]sin[-(π+α)]sin[4π+(
+α)]π 2
=(-sinα)(-cosα)(-sinα)(-sinα) 2(-sinα)sinαcosα
=-
sinα;…(4分)1 2
②由方程10x2+x-3=0,解得:x1=
,x2=-1 2
,3 5
又α在第三象限,∴sinα=-
,3 5
则f(α)=-
sinα=-1 2
×(-1 2
)=3 5
;…(8分)3 10
(3)当a=-
π时,f(α)=-25 4
sin(-1 2
π)=-25 4
×sin(-6π-1 2
)=-π 4
×sin(-1 2
)=π 4
.…(12分)2 4