赞 llyyjj2613 幼苗
共回答了22个问题采纳率：100% 举报
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
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
1年前 追问
1
ô֪sin(n+/2 + x) = - sin(/2 + x) = - cosx Ӧ÷nΪżʱ
ǶԵģȷʵӦۡ nΪʱ sin(n+/2 + x) = - sin(/2 + x) =cosx = -1/2 xǶĽ xǶĽʱsinx = (1- cos^2 x ) =[1-(-1/2)^2 ]=3/2 sin2x - 3 = 2sinx*cosx - 3 = 2 * (3/2)*(-1/2) - 3 = -(3 + 6)/2 xǵĽʱ,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 nΪżʱ sin(n+/2 + x) = sin(/2 + x) =-cosx = - 1/2 ԣcosx = 1/2 ˣxһĽǣ xΪһĽʱsinx = (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 xΪĽʱsinx = -(1 - cos^2 x ) = -[1 - (1/2)^2] = -3/2 ԣsin2x - 3 = 2sinx * cosx - 3 = 2*(-3/2)*(1/2) - 3 = - ( 3 + 6)/2
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