设△ABC的内角A、B、C的对边分别是a、b、c,且a=3,b=5,c=14.(Ⅰ)求cosC的值;(Ⅱ)求5-6sin(C+π3)cos2C的值.-数学

题目简介

设△ABC的内角A、B、C的对边分别是a、b、c,且a=3,b=5,c=14.(Ⅰ)求cosC的值;(Ⅱ)求5-6sin(C+π3)cos2C的值.-数学

题目详情

设△ABC的内角A、B、C的对边分别是a、b、c,且a=3,b=5,c=
14

(Ⅰ)求cosC的值;
(Ⅱ)求
5
-6sin(C+
π
3
)
cos2C
的值.
题型:解答题难度:中档来源:西城区一模

答案

(Ⅰ)由余弦定理cosC=
a2b2-c2
2ab

cosC=class="stub"9+25-14
2×3×5
=class="stub"2
3

(Ⅱ)由(Ⅰ)知cosC>0,
所以角c为锐角,所以sinC=
1-cos2C
=
5
3

5
-6sin(C+class="stub"π
3
)
cos2C
 =
5
-6(sinc×cosclass="stub"π
3
+cosc×sinclass="stub"π
3
)
2cos2C-1

5
-6(
5
3
×class="stub"1
2
+class="stub"2
3
×
3
2
)
2×class="stub"4
9
-1

=18
3

所以
5
-6sin(C+class="stub"π
3
)
cos2C
=18
3

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