解三角方程

圖片參考：http://hk.geocities.com/abc2004y2k/MATHS/maths8.jpg

(a)(i)
cos4θ=0
4θ=90
θ=22.5
通解是θ=360n+22.5 或 θ=360n-22.5
(ii)
cos4θ
=cos^2(2θ)-sin^2(2θ)
=cos^2(2θ)-(1-cos^2(2θ))
=2cos^2(2θ)-1
=2[(cos^2θ-sin^2θ)]^2-1
=2[2cos^2θ-1]^2-1
=2(4cos^4θ-4cos^2θ+1]-1
=8cos^4θ-8cos^2θ+1
(iii)
cos4(3π/8)=0
8cos^4(3π/8)-8cos^2(3π/8)+1=0
cos^2(3π/8)=1/16[8-√64-32]
cos^2(3π/8)=1/16[8-4√2]
cos^2(3π/8)=1/2+√2/4=(2-√2)/4
cos(3π/8)=1/2+√2/4=√(2-√2)/2
(b)(i)
let u=arctanx v=arctany
tan(u+v)
=(tanu+tanv)/(1-tanutanv)
=(x+y)/(1-xy)
arctanx+arctany=arctan[(x+y)/(1-xy)]
(ii)
arctanx+arctany=arctan[(x+y)/(1-xy)]
arctanx+arctany+arctanz
=arctan[(x+y)/(1-xy)]+arctanz
=arctan{[[(x+y)/(1-xy)]+z]/[1-[(x+y)/(1-xy)z]}
=arctan{[[(x+y)/(1-xy)]+z]/[1-[(x+y)/(1-xy)z]}
arctanx+arctany+arctanz=0
arctan{[[(x+y)/(1-xy)]+z]/[1-[(x+y)/(1-xy)z]}=0
{[[(x+y)/(1-xy)]+z]/[1-[(x+y)/(1-xy)z]}=0
[(x+y)/(1-xy)]+z=0
x+y=-z(1-xy)
x+y+z=xyz

2007-02-10 14:56:20 補充：
(a) 漏了cos4θ=04θ=270θ=67.5通解是θ=360n+67.5 或 θ=360n-67.5