a.math

設p(n)為命題
p(n)=cos2Acos2^2 A....cos2^n A = sin 2^(n+1) A/ 2^n sin 2A.
n=1
L.H.Sp(1)=cos2A
R.H.Sp(1)=sin4A/(2sin2A)=2sin2Acos2A/(2sin2A)=cos2A
L.H.Sp(1)=R.H.Sp(1)
p(1)成立
假設p(k)成立
p(k)=cos2Acos2^2 A....cos2^k A = sin 2^(k+1) A/ 2^k sin 2A.

L.H.S p(k+1)=cos2Acos2^2 A....cos2^k Acos2^(k+1) A
=sin 2^(k+1) A/ 2^k sin 2A * cos2^(k+1) A
=sin 2^(k+1) A cos2^(k+1) A/ 2^k sin 2A
=[1/2 sin 2^(k+2)] / 2^k sin 2A (from sinxcosx=1/2sin2x)
=sin 2^(k+2) / 2^(k+1) sin 2A
=R.H.S p(k+1)
∴p(k+1)成立
............唔識打英文

2007-12-25 15:43:30 補充：
=sin 2^(k+2)A / 2^(k+1) sin 2A之後應該係=sin 2^[(k+1)+1]A / 2^(k+1) sin 2A∴p(k+1)成立Therefore, P(k+1) is true by assuming P(k) is true,by the principle of M.I., P(n) is true for positive integer n.