已知sin(nπ+π/2+x)=-1/2,n∈Z

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/09 19:51:34

已知sin(nπ+π/2+x)=-1/2,n∈Z
(1)求cosx的值
求sin2x-3的值

sin(nπ+π/2 + x) = - sin(π/2 + x) = - cosx
根据题意则有－cosx = - 1/2
所以 cosx = 1/2
由cosx = 1/2 知 x是第一或第四象限的角
当 x是第一象限的角时 sinx = √（1 －cos^2 x) = √[1 - (1/2)^2] =√3/2
sin2x - 3 = 2sinx*cosx - 3
= 2*(√3/2)*(1/2)-3
= √3/2 - 3
= (√3 -6)/2
当 x是第四象限的角时 sinx = -√( 1 - cos^2 x) = -√[ 1 - (1/2)^2] = -√3/2
sin2x - 3 = 2sinx*cosx - 3
= 2*(-√3/2)*(1/2) - 3
= -√3/2 - 3
=-(√3+6)/2
再问： ��ô֪��sin(n��+��/2 + x) = - sin(��/2 + x) = - cosx Ӧ�÷��nΪ��ż��ʱ��
再答： ��ǶԵģ�ȷʵӦ��ۡ� ��nΪ��ʱ�� sin(n��+��/2 + x) = - sin(��/2 + x) =cosx = -1/2 �� x�Ƕ��޵Ľ� ��x�Ƕ��޵Ľ�ʱ��sinx = ��(1- cos^2 x ) =��[1-(-1/2)^2 ]=��3/2 sin2x - 3 = 2sinx*cosx - 3 = 2 * (��3/2)*(-1/2) - 3 = -(��3 + 6)/2 ��x�ǵ��޵Ľ�ʱ,sinx = -��(1-cos^2 x) = -��[1 - (-1/2)^2] =-��3/2 sin2x - 3 = 2sinx*cosx - 3 = 2*(-��3/2)*(-1/2) - 3 = (��3 - 6)/2 ��nΪż��ʱ�� sin(n��+��/2 + x) = sin(��/2 + x) =-cosx = - 1/2 ��ԣ�cosx = 1/2 ��ˣ�x��һ��޵Ľǣ� ��xΪ��һ��޵Ľ�ʱ��sinx = ��(1 - cos^2 x) = �� [ 1 - (1/2)^2 ]= ��3/2 ��ԣ� sin2x - 3 = 2sinx*cosx - 3 = 2*(��3/2)*(1/2) - 3 = (��3 - 6)/2 ��xΪ��޵Ľ�ʱ��sinx = -��(1 - cos^2 x ) = -��[1 - (1/2)^2] = -��3/2 ��ԣ�sin2x - 3 = 2sinx * cosx - 3 = 2*(-��3/2)*(1/2) - 3 = - ( ��3 + 6)/2

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