题目内容

(1)已知cos(x+
π
6
)=
1
4
,求cos(
6
-x)+cos2(
π
3
-x)
的值;
(2)计算:sin
π
6
+cos2
π
4
cosπ+3tan2
π
6
+cos
π
3
-sin
π
2
(1)∵cos(x+
π
6
)=
1
4
,∴cos(
6
-x)=cos[π-(x+
π
6
)]=-cos(x+
π
6
)=-
1
4

cos(
π
3
-x)=cos[
π
2
-(x+
π
6
)]=sin(x+
π
6
)

cos2(
π
3
-x)=sin2(x+
π
6
)=1-cos2(x+
π
6
)=1-
1
16
=
15
16

cos(
6
-x)+cos2(
π
3
-x)
=-
1
4
+
15
16
=
11
16

(2)sin
π
6
+cos2
π
4
cosπ+3tan2
π
6
+cos
π
3
-sin
π
2
=
1
2
+
1
2
×(-1)+3×
1
3
+
1
2
-1
=
1
2
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