(1)求sin50°(1+3tan10°)的值.(2)若α,β∈(0,π2),cos(α-β2)=32,sin(α2-β)=-12,求cos(α+β)的值.-数学

题目简介

(1)求sin50°(1+3tan10°)的值.(2)若α,β∈(0,π2),cos(α-β2)=32,sin(α2-β)=-12,求cos(α+β)的值.-数学

题目详情

(1)求sin50°(1+
3
tan10°)
的值.
(2)若α,β∈(0,
π
2
)
cos(α-
β
2
)=
3
2
sin(
α
2
-β)=-
1
2
,求cos(α+β)的值.
题型:解答题难度:中档来源:不详

答案

(1)原式
=sin500
cos100+
3
sin100
cos100

=sin500
2sin(100+300)
cos100
=sin500
2cos500
cos100
=
sin1000
cos100
=
sin800
cos100
=1

(2)∵-class="stub"π
4
<-class="stub"β
2
<0∴-class="stub"π
4
<α-class="stub"β
2
<class="stub"π
2

cos(α-class="stub"β
2
)=
3
2
∴α-class="stub"β
2
=±class="stub"π
6

0<class="stub"α
2
<class="stub"π
4
,-class="stub"π
2
<-β<0∴-class="stub"π
2
<class="stub"α
2
-β<class="stub"π
4

sin(class="stub"α
2
-β)=-class="stub"1
2
∴class="stub"α
2
-β=-class="stub"π
6

∴①-②得 class="stub"α+β
2
=class="stub"π
3
或0

α+β=class="stub"2π
3

cos(α+β)=cosclass="stub"2π
3
=-class="stub"1
2

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