Question

已知cos(α+β)cosβ+sin(α+β)sinβ=1/3,且α∈(3π/2,2π),求cos（2π+π/4）的值

Answer 1

cos(α+β)cosβ+sin(α+β)sinβ=1/3，可用复角公式整合
cos[(α+β)-β]=1/3
cosα=1/3
α∈(3π/2,2π)，机、即α在第四个象限
sinα=-√(3²-1)/3
sinα=-(2√2)/3
cos(2α+π/4),打错了个π为α吧
=cos2αcos(π/4)-sin2αsin(π/4)
=(2cos²α-1)*(√2/2)-2sinαcosα*(√2/2)
=[2(1/3)²-1]*(√2/2)+2[(2√2)/3]*(1/3)*(√2/2)
=(-7√2)/18+4/9
=(8-7√2)/18

Answer 2

cos(a+b)cosb+sin(a+b)sinb=1/3
--->cos[(a+b)-b]=cosa=1/3.
3pi/2<a<2pi--->sina=-2√2/3
3pi/2<a<2pi--->3pi<2a<4pi--->sin2a=2sinacosa=-2√2/9
cos2a=2(cosa)^2-1=-7/9.
cos(2a+pi/4)=cos2acos(pi/4)-sin2asin(pi/4)
=-7/9*√2/2-(-2√2/9）*√2/2
=(-7√2+8)/18.

Answer 3

cos(a+b)cosb+sin(a+b)sinb=1/3
--->cos[(a+b)-b]=cosa=1/3.
3pi/2<a<2pi--->sina=-2√2/3
3pi/2<a<2pi--->3pi<2a<4pi--->sin2a=2sinacosa=-2√2/9
cos2a=2(cosa)^2-1=-7/9.
cos(2a+pi/4)=cos2acos(pi/4)-sin2asin(pi/4)
=-7/9*√2/2-(-2√2/9）*√2/2
=(-7√2+8)/18.

Answer 4

题目有问题