(1)已知tanα=13,求12sinαcosα+cos2α的值;(2)化简:tan(π-α)cos(2π-α)sin(-α+32π)cos(-α-π)sin(-π-α).-高二数学

题目简介

(1)已知tanα=13,求12sinαcosα+cos2α的值;(2)化简:tan(π-α)cos(2π-α)sin(-α+32π)cos(-α-π)sin(-π-α).-高二数学

题目详情

(1)已知tanα=
1
3
,求
1
2sinαcosα+cos2α
的值;
(2)化简:
tan(π-α)cos(2π-α)sin(-α+
3
2
π)
cos(-α-π)sin(-π-α)
题型:解答题难度:中档来源:不详

答案

(1)∵tanα=class="stub"1
3

∴原式=
sin2α+cos2α
2sinαcosα+cos2α-sin2α
=
tan2α+1
2tanα+1-tan2α
=
class="stub"1
9
+1
2×class="stub"1
3
+1-class="stub"1
9
=class="stub"5
7

(2)
tan(π-α)cos(2π-α)sin(-α+class="stub"3
2
π)
cos(-α-π)sin(-π-α)
=
-tanαcos(-α)sin(α+class="stub"1
2
π)
-cosαsinα
=
class="stub"sinα
cosα
cosα
-sinα
=-1.

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