在平面直角坐标系中,A(3,0)、B(0,3)、C(cosθ,sinθ),θ∈(π2,3π2),且|AC|=|BC|.(1)求角θ的值;(2)设α>0,0<β<π2,且α+β=23θ,求y=2-sin

题目简介

在平面直角坐标系中,A(3,0)、B(0,3)、C(cosθ,sinθ),θ∈(π2,3π2),且|AC|=|BC|.(1)求角θ的值;(2)设α>0,0<β<π2,且α+β=23θ,求y=2-sin

题目详情

在平面直角坐标系中,A(3,0)、B(0,3)、C(cosθ,sinθ),θ∈(
π
2
2
)
,且|
AC
|=|
BC
|

(1)求角θ的值;
(2)设α>0,0<β<
π
2
,且α+β=
2
3
θ
,求y=2-sin2α-cos2β的最小值.
题型:解答题难度:中档来源:不详

答案

(1)由|
AC
|=|
BC
|
得(3-cosθ)2+sin2θ=cos2θ+(3-sinθ)2,
化简得tanθ=1,
因为θ∈(class="stub"π
2
,class="stub"3π
2
)

所以θ=class="stub"5π
4

(2)α+β=class="stub"2
3
θ=class="stub"5π
6
y=2-class="stub"1-cos2α
2
-class="stub"1+cos2β
2
=1+class="stub"1
2
(cos2α-cos2β)

=1+class="stub"1
2
[cos(class="stub"5π
3
-2β)-cos2β]=1-class="stub"1
2
(
3
2
sin2β+class="stub"1
2
cos2β)=1-class="stub"1
2
sin(2β+class="stub"π
6
)

因为0<β<class="stub"π
2
class="stub"π
6
<2β+class="stub"π
6
<class="stub"7π
6
-class="stub"1
2
<sin(2β-class="stub"π
3
)≤1

所以class="stub"1
2
≤1-class="stub"1
2
sin(2β+class="stub"π
6
)<class="stub"3
4

β=class="stub"π
6
α=class="stub"2π
3
时,y取最小值,
ymin=class="stub"1
2

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