已知向量a=(sinx,-1),b=(cosx,32).(1)当a∥b时,求cos2x-3sin2x的值.(2)求f(x)=(a+b)•b的最小正周期和单调递增区间.-数学

题目简介

已知向量a=(sinx,-1),b=(cosx,32).(1)当a∥b时,求cos2x-3sin2x的值.(2)求f(x)=(a+b)•b的最小正周期和单调递增区间.-数学

题目详情

已知向量
a
=(sinx,-1),
b
=(cosx,
3
2
)

(1)当
a
b
时,求cos2x-3sin2x
的值.
(2)求f(x)=(
a
+
b
)•
b
的最小正周期和单调递增区间.
题型:解答题难度:中档来源:不详

答案

(1)∵
a
b
a
=(sinx,-1),
b
=(cosx,class="stub"3
2
)

class="stub"3
2
sinx+cosx=0
…(2分)
tanx=-class="stub"2
3
…(3分)
cos2x-3sin2x=
cos2x-6sinxcosx
sin2x+cos2x
=class="stub"1-6tanx
1+tan2x

=
1-6×(-class="stub"2
3
)
1+(-class="stub"2
3
)
2
=class="stub"1+4
1+class="stub"4
9
=class="stub"5
class="stub"13
9
=class="stub"45
13
(5分)
(2)∵
a
=(sinx,-1),
b
=(cosx,class="stub"3
2
)

a
+
b
=(sinx+cosx,class="stub"1
2
)
…(6分)
f(x)=(
a
+
b
)•
b
=(sinx+cosx)cosx+class="stub"3
4
=class="stub"1
2
(sin2x+cos2x)+class="stub"5
4
=
2
2
sin(2x+class="stub"π
4
)+class="stub"5
4
…(8分)
∴最小正周期为π…(9分)
2kπ-class="stub"π
2
≤2x+class="stub"π
2
≤2kπ+class="stub"π
2
,得kπ-class="stub"3
8
π≤x≤kπ+class="stub"π
8

故f(x)的单调递增区间为[kπ-class="stub"3
8
π,kπ+class="stub"π
8
]k∈Z
…(10分)

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