cos（a+π）
@俟芸2057：已知cos(a+?π)=3/5, π/2≤a<3π/2. 求cos2a... -
第律15510827040…… 解:因为 π/2 ≤a 所以 3π/4 ≤a +π/4 <7π/4. 又因为 cos (a +π/4) =3/5 >0, 所以 3π/2 <a +π/4 <7π/4.所以 sin (a +π/4) = -√[ 1-(3/5)^2 ] = -4/5. 所以 sin (2a +π/2) =2 sin (a +π/4) cos (a +π/4) = -24/25. 即 cos 2a = -24/25.
@俟芸2057：化简:cos(a+π)sin²(a+3π)/tan(a+π)cos³( - a - π) -
第律15510827040…… cos(a+π5261)=-cosa sin(a+3π)=sin(a+π)=-sina 所以sin²(a+3π)=sin²a tan(a+π)=tana=sina/cosa cos(-a-π)=cos(a+π)=-cosa cos³(4102-a-π)=-cos³a-cosa*1653sin²a/sina*(-cos²a)=tana 如有不懂专请追问属 望采纳
@俟芸2057：cos(a+π/4)=3/5,求 - cosa -
第律15510827040…… cosa=4/5,sina=3/5 cos(a+β)=3/5,sin(a+β)=4/5或-4/5 cosβ=cos[(a+β)-a]=cos(a+β)cosa+sin(a+β)sina 当sin(a+β)=-4/5时 cosβ=(3/5)(4/5)-(4/5)(3/5)=0,β=90度,与β是锐角矛盾(舍弃) 当sin(a+β)=4/5时 cosβ=(3/5)(4/5)+(4/5)(3/5)=24/25 sinβ=7/25 sin2β=2sinβcosβ=2(7/25)(24/25)=336/625
@俟芸2057：已知cos(a+π/4)=3/5,π/2<=a<3π/2,则cos(2a+π/4)=? -
第律15510827040…… 因为cos(a+π/4)=3/5>0,a∈ [π/2,3π/2] a+π/4∈ [3π/4,7π/4], 所以a∈ [5π/4,3π/2] cos(2a+π/2)=2[cos(a+π/4)]^2-1=-7/25 a∈ [5π/4,3π/2] 2a+π/2∈ [3π,7π/2] sin(2a+π/2)=-24/25 cos(2a+π/4)=cos(2a+π/2-π/4) =cos(2a+π/2)cosπ/4+sin(2a+π/2)sinπ/4 =(-7/25-24/25)(根号2)2 =-31倍(根号2)/50
@俟芸2057：cos(a+π/6)= - 4/5 sina=? -
第律15510827040…… sin(a+π/6)=+-3/5 sina=sin(a+π/6-π/6)=sin(a+π/6)cosπ/6-sinπ/6cos(a+π/6) sin(a+π/6)=3/5, sina=3/5x√3/2-1/2x(-4/5)=(3√3+4)/10 sin(a+π/6)=-3/5,sina=-3/5x√3/2-1/2x(-4/5)=(-3√3+4)/10
@俟芸2057：已知cos(a+π/3)=1/4,则cos a等于多少?为什么? -
第律15510827040…… 解答:cos(a+π/3)=1/4 sin(a+π/3)=±√[1-cos²(a+π/3)]=±√15/4 cosa=cos[(a+π/3)-(π/3)] =cos(a+π/3)cos(π/3)+sin(a+π/3)sin(π/3)] =(1/4)*(1/2)+(±√15/4)*(√3/2) =(1±3√5)/8
@俟芸2057：用公式证明cos(a+3π)= - cosa -
第律15510827040…… cos(a+3π)=cos(2π+π+a) 根据cos(2π+a)=cos(a) 所以cos(a+3π)=cos(2π+π+a)=cos(π+a) 根据奇变偶不变 符号看象限的原理可知cos(π+a)=-cosa 因为π+a处于第三象限cos为负值所以cosa前面有负号
@俟芸2057：cos(α - π)=cos(α+π)对吗_
第律15510827040…… 对 cos(α-π)和cos(α+π)都等于-cosα
@俟芸2057：数学题 若cosα= - 13分之12,则cos(α+π)的值为多少_
第律15510827040…… 直接利用诱导公式啊cos(π+α)=-cosα=-(-12/13)=12/13希望能帮到你,祝学习进步
@俟芸2057：已知cos(a+π)=根号10/10,a∈(0,π/2),则sin(2a - π/4)的值为 在线等,高分 -
第律15510827040…… ∵cos(a+π)=-cosa=-根号10/10 ∴cosa=√10/10 ∵a∈(0,π/2) ∴sina=√(1-cos²a)=3√10/10 ∴sin2a=2sinacosa=2*3√10/10*√10/10=3/5 cos2a=2cos²a-1=2*1/10-1=-4/5 ∴sin(2a-π/4) =sin2acosπ/4-cos2asinπ/4 =3/5*√2/2+4/5*√2/2 =7√2/10 改了一个地方,不然矛盾,不懂的地方追问
第律15510827040…… 解:因为 π/2 ≤a 所以 3π/4 ≤a +π/4 <7π/4. 又因为 cos (a +π/4) =3/5 >0, 所以 3π/2 <a +π/4 <7π/4.所以 sin (a +π/4) = -√[ 1-(3/5)^2 ] = -4/5. 所以 sin (2a +π/2) =2 sin (a +π/4) cos (a +π/4) = -24/25. 即 cos 2a = -24/25.
@俟芸2057：化简:cos(a+π)sin²(a+3π)/tan(a+π)cos³( - a - π) -
第律15510827040…… cos(a+π5261)=-cosa sin(a+3π)=sin(a+π)=-sina 所以sin²(a+3π)=sin²a tan(a+π)=tana=sina/cosa cos(-a-π)=cos(a+π)=-cosa cos³(4102-a-π)=-cos³a-cosa*1653sin²a/sina*(-cos²a)=tana 如有不懂专请追问属 望采纳
@俟芸2057：cos(a+π/4)=3/5,求 - cosa -
第律15510827040…… cosa=4/5,sina=3/5 cos(a+β)=3/5,sin(a+β)=4/5或-4/5 cosβ=cos[(a+β)-a]=cos(a+β)cosa+sin(a+β)sina 当sin(a+β)=-4/5时 cosβ=(3/5)(4/5)-(4/5)(3/5)=0,β=90度,与β是锐角矛盾(舍弃) 当sin(a+β)=4/5时 cosβ=(3/5)(4/5)+(4/5)(3/5)=24/25 sinβ=7/25 sin2β=2sinβcosβ=2(7/25)(24/25)=336/625
@俟芸2057：已知cos(a+π/4)=3/5,π/2<=a<3π/2,则cos(2a+π/4)=? -
第律15510827040…… 因为cos(a+π/4)=3/5>0,a∈ [π/2,3π/2] a+π/4∈ [3π/4,7π/4], 所以a∈ [5π/4,3π/2] cos(2a+π/2)=2[cos(a+π/4)]^2-1=-7/25 a∈ [5π/4,3π/2] 2a+π/2∈ [3π,7π/2] sin(2a+π/2)=-24/25 cos(2a+π/4)=cos(2a+π/2-π/4) =cos(2a+π/2)cosπ/4+sin(2a+π/2)sinπ/4 =(-7/25-24/25)(根号2)2 =-31倍(根号2)/50
@俟芸2057：cos(a+π/6)= - 4/5 sina=? -
第律15510827040…… sin(a+π/6)=+-3/5 sina=sin(a+π/6-π/6)=sin(a+π/6)cosπ/6-sinπ/6cos(a+π/6) sin(a+π/6)=3/5, sina=3/5x√3/2-1/2x(-4/5)=(3√3+4)/10 sin(a+π/6)=-3/5,sina=-3/5x√3/2-1/2x(-4/5)=(-3√3+4)/10
@俟芸2057：已知cos(a+π/3)=1/4,则cos a等于多少?为什么? -
第律15510827040…… 解答:cos(a+π/3)=1/4 sin(a+π/3)=±√[1-cos²(a+π/3)]=±√15/4 cosa=cos[(a+π/3)-(π/3)] =cos(a+π/3)cos(π/3)+sin(a+π/3)sin(π/3)] =(1/4)*(1/2)+(±√15/4)*(√3/2) =(1±3√5)/8
@俟芸2057：用公式证明cos(a+3π)= - cosa -
第律15510827040…… cos(a+3π)=cos(2π+π+a) 根据cos(2π+a)=cos(a) 所以cos(a+3π)=cos(2π+π+a)=cos(π+a) 根据奇变偶不变 符号看象限的原理可知cos(π+a)=-cosa 因为π+a处于第三象限cos为负值所以cosa前面有负号
@俟芸2057：cos(α - π)=cos(α+π)对吗_
第律15510827040…… 对 cos(α-π)和cos(α+π)都等于-cosα
@俟芸2057：数学题 若cosα= - 13分之12,则cos(α+π)的值为多少_
第律15510827040…… 直接利用诱导公式啊cos(π+α)=-cosα=-(-12/13)=12/13希望能帮到你,祝学习进步
@俟芸2057：已知cos(a+π)=根号10/10,a∈(0,π/2),则sin(2a - π/4)的值为 在线等,高分 -
第律15510827040…… ∵cos(a+π)=-cosa=-根号10/10 ∴cosa=√10/10 ∵a∈(0,π/2) ∴sina=√(1-cos²a)=3√10/10 ∴sin2a=2sinacosa=2*3√10/10*√10/10=3/5 cos2a=2cos²a-1=2*1/10-1=-4/5 ∴sin(2a-π/4) =sin2acosπ/4-cos2asinπ/4 =3/5*√2/2+4/5*√2/2 =7√2/10 改了一个地方,不然矛盾,不懂的地方追问
