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find the value of cosAcos2Acos3A....cos999A if A=2pi upon 1999

find the value of cosAcos2Acos3A....cos999A if                              A=2pi upon 1999

Grade:12

1 Answers

Vikas TU
14149 Points
7 years ago
cosAcos2Acos3A....cos999A be the eqn. let (1)
now multiply by sinAsin2Asin3A......sin999A in both numerator and denominator,
we get,
=> (sinAcosA*sin2Acos2A*sin3Acos3A*............sin999Acos999A)/sinAsin2Asin3A......sin999A
Multiply by 2^999 in both numerator and denominator.
=> (2sinAcosA*2sin2Acos2A*2sin3Acos3A*......2sin999Acos999A)/2^999*sinAsin2Asin3A......sin999A
=> (sin2A*sin4A*sin6A......sin1998A)/(2^999)*(sinA*sin2A*sin3A.........sin999A)
solve it further and put A= 2pi/1999.

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