F.4 TRIGONOMETRY M2

2011-03-19 2:09 am

give that 2sinAcos3A=sin4A=sin2A and 2sinAcos5A= sin6A-sin4A

hence prove cosA+cos3A+cos5A+cos7A+cos9A =sin10A/2sinA where sina is not equal to 0

回答 (1)

用戶2053

2011-03-19 2:32 am

✔ 最佳答案

2sinAcos3A = sin4A - sin2A
and
2sinAcos5A = sin6A - sin4ASimilarly ,
2sinAcosA = sin2A - sin0A
2sinAcos7A = sin8A - sin6A
2sinAcos9A = sin10A - sin8A
Summing up :
2sinAcosA + 2sinAcos3A + 2sinAcos5A + 2sinAcos7A + 2sinAcos9A
=
(sin2A - sin0A) + (sin4A - sin2A) + (sin6A - sin4A) + (sin8A - sin6A)
+ (sin10A - sin8A)
2sinA (cosA + cos3A + cos5A + cos7A + cos9A) = - sin0A + sin10A
cosA + cos3A + cos5A + cos7A + cos9A = sin10A / (2sinA)where sinA is not equal to 0.

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