已知:0<α<π2<β<π,cos(β-π4)=13,sin(α+β)=45.(1)求sin2β的值;(2)求cos(α+π4)的值.-数学

题目简介

已知:0<α<π2<β<π,cos(β-π4)=13,sin(α+β)=45.(1)求sin2β的值;(2)求cos(α+π4)的值.-数学

题目详情

已知:0<α<
π
2
<β<π,cos(β-
π
4
)=
1
3
,sin(α+β)=
4
5

(1)求sin2β的值;
(2)求cos(α+
π
4
)的值.
题型:解答题难度:中档来源:南京二模

答案

(1)法一:∵cos(β-class="stub"π
4
)=cosclass="stub"π
4
cosβ+sinclass="stub"π
4
sinβ
=
2
2
cosβ+
2
2
sinβ=class="stub"1
3

∴cosβ+sinβ=
2
3

∴1+sin2β=class="stub"2
9
,∴sin2β=-class="stub"7
9

法二:sin2β=cos(class="stub"π
2
-2β)
=2cos2(β-class="stub"π
4
)-1=-class="stub"7
9

(2)∵0<α<class="stub"π
2
<β<π,∴class="stub"π
4
<β-class="stub"π
4
class="stub"3π
4
class="stub"π
2
<α+β<class="stub"3π
2

∴sin(β-class="stub"π
4
)>0,cos(α+β)<0.
∵cos(β-class="stub"π
4
)=class="stub"1
3
,sin(α+β)=class="stub"4
5

∴sin(β-class="stub"π
4
)=
2
2
3
,cos(α+β)=-class="stub"3
5

∴cos(α+class="stub"π
4
)=cos[(α+β)-(β-class="stub"π
4
)]
=cos(α+β)cos(β-class="stub"π
4
)+sin(α+β)sin(β-class="stub"π
4

=-class="stub"3
5
×class="stub"1
3
+class="stub"4
5
×
2
2
3
=
8
2
-3
15

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