It is given that cosA is not equal to 1/2 and n is a positive integer.
(a) Prove that 2cosA - 1 = (2cos2A + 1) / (2cosA + 1)
(b) Hence, prove that (2cosA - 1) (2cos2A -1)...{2cos[2^(n-1) A] -1 } = 2cos[2^(n) A]+1 / (2cosA+1)
F.4 trigonometry---2
2010-03-09 9:40 am
回答 (1)
2010-03-09 10:15 am
✔ 最佳答案
a) (2cos2A + 1) / (2cosA + 1)= {2[2(cosA)^2 - 1] + 1} / (2cosA + 1)
= [4(cosA)^2 - 1] / (2cosA + 1)
= (2cosA - 1)(2cosA + 1) / (2cosA + 1)
= 2cosA - 1
b) (2cosA - 1) (2cos2A -1)...{2cos[2^(n-1) A] -1 }
= [(2cos2A + 1) / (2cosA + 1)] * [(2cos4A + 1) / (2cos2A + 1)] *
[(2cos8A + 1) / (2cos4A + 1)] * [(2cos16A + 1) / (2cos8A + 1)] * .... *
[(2cos(2^(n-1)) A + 1) / (2cos(2^n)A + 1)] *
[(2cos(2^n) A + 1) / (2cos(2^(n-1))A + 1)]
= 2cos[2^(n) A]+1 / (2cosA+1)
2010-03-09 02:35:36 補充：
Correction of B)
(2cosA - 1) (2cos2A -1)...{2cos[2^(n-1) A] -1 }
= [(2cos2A + 1) / (2cosA + 1)] * [(2cos4A + 1) / (2cos2A + 1)] *
[(2cos8A + 1) / (2cos4A + 1)] * [(2cos16A + 1) / (2cos8A + 1)] * .... *
2010-03-09 02:35:41 補充：
[(2cos(2^(n-2)) A + 1) / (2cos(2^(n-3)A) + 1)] *
[(2cos(2^(n-1)) A + 1) / (2cos(2^(n-2)A) + 1)] *
[(2cos(2^n) A + 1) / (2cos(2^(n-1))A + 1)]
= 2cos[2^(n) A]+1 / (2cosA+1)
收錄日期: 2021-04-13 17:08:36
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https://hk.answers.yahoo.com/question/index?qid=20100309000051KK00119