已知向量m=(cosx,-sinx),n=(cosx,sinx-23cosx),x∈R,设f(x)=m•n.(1)求函数f(x)的最小正周期.(2)若f(x)=2413,且x∈[π4,π2],求sin

题目简介

已知向量m=(cosx,-sinx),n=(cosx,sinx-23cosx),x∈R,设f(x)=m•n.(1)求函数f(x)的最小正周期.(2)若f(x)=2413,且x∈[π4,π2],求sin

题目详情

已知向量
m
=(cosx,-sinx)
n
=(cosx,sinx-2
3
cosx)
,x∈R,设f(x)=
m
n

(1)求函数f(x)的最小正周期.
(2)若f(x)=
24
13
,且x∈[
π
4
π
2
]
,求sin2x的值.
题型:解答题难度:中档来源:不详

答案

(1)∵f(x)=
m
n
=cos2x-sin2x+2
3
sinxcosx=cos2x+
3
sin2x=2sin(2x+class="stub"π
6

∴函数f(x)的最小正周期T=class="stub"2π
2

(2)∵f(x)=class="stub"24
13

∴sin(2x+class="stub"π
6
)=class="stub"12
13

又∵x∈[class="stub"π
4
,class="stub"π
2
]

∴cos(2x+class="stub"π
6
)=-
1-class="stub"12
13
=-class="stub"5
13

即sin2x=sin[(2x+class="stub"π
6
)-class="stub"π
6
]
=sin(2x+class="stub"π
6
)cosclass="stub"π
6
-cos(2x+class="stub"π
6
)sinclass="stub"π
6

=class="stub"12
13
×
3
2
-(-class="stub"5
13
)×class="stub"1
2
=
12
3
+5
26

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