1a)By using the idrntity 2sinAcosB=sin(A+B)+sin(A-B),show that 2sin(x)cos(p-x)=sin(2x-p)+sin p
b)hence find the general solution of the equation sin x+2sin xcos(p-X)-sinp=0terms of p.
f.4 a maths
2007-03-27 6:55 am
回答 (2)
2007-03-27 7:08 am
✔ 最佳答案
a.2sin(x)cos(p-x)
=sin(x+p-x)+sin(x-p+x)
=sin(2x-p)+sinp
b.
sin(x)+2sin(x)cos(p-X)-sinp=0
sin(2x-p)+sinp-sinp=0
sin(2x-p)=0
2x-p=180°n
2x=180°n+p
x=90°n+p/2
2007-03-26 23:24:36 補充：
sinθ=aθ=180°n (-1)^nαNow, α=0Therefore, θ=180°n
2007-03-27 7:13 am
(a)2sin(x)cos(p-x)
=sin(x+p-x)+sin(x-p+x)
=sin p+sin (2x-p)
=sin(2x-p)+sin p
(b)sin x+2sin xcos(p-X)-sinp=0 (in terms of p)
sin x+2sin xcos(p-X)=sinp
By (a),
2sin(x)cos(p-x)=sin(2x-p)+sin p
sin(2x-p)+sin p+sinx=sinp
sin(2x-p)=sin(-x)
Bty genernal solution,
2x-p=-(-1)^nx+n(pi) (n is any integer)
=(-1)^nx+n(pi)
2007-03-26 23:17:26 補充：
x=[ (-1)^n x+n(pi) +p]/2
=sin(x+p-x)+sin(x-p+x)
=sin p+sin (2x-p)
=sin(2x-p)+sin p
(b)sin x+2sin xcos(p-X)-sinp=0 (in terms of p)
sin x+2sin xcos(p-X)=sinp
By (a),
2sin(x)cos(p-x)=sin(2x-p)+sin p
sin(2x-p)+sin p+sinx=sinp
sin(2x-p)=sin(-x)
Bty genernal solution,
2x-p=-(-1)^nx+n(pi) (n is any integer)
=(-1)^nx+n(pi)
2007-03-26 23:17:26 補充：
x=[ (-1)^n x+n(pi) +p]/2
收錄日期: 2021-04-12 23:36:06
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