2305
Shaswata Roy
·2013-02-28 06:05:29
1sin(Ï€/7) - 1sin(3Ï€/7)
Converting the numerator into product form we get,
2cos(2Ï€/7)sin(Ï€/7)sin(3Ï€/7)sin(Ï€/7)
= 2sin(3Ï€/14)sin(3Ï€/7) (since cos(2Ï€/7)= sin(3Ï€/14))
= 2sin(3Ï€/14)2sin(3Ï€/14)cos(3Ï€/14)
= 1cos(3Ï€/14)
= 1sin(2Ï€/7)
Hence,cosec (Ï€/7) = cosec(2Ï€/7) + cosec(3Ï€/7).
31
Gourav Bose
·2013-10-14 20:21:54
1
sin(Ï€/7) - 1
sin(3Ï€/7)
Converting the numerator into product form we get,
2cos(2Ï€/7)sin(Ï€/7)
sin(3Ï€/7)sin(Ï€/7)
= 2sin(3Ï€/14)
sin(3Ï€/7) (since cos(2Ï€/7)= sin(3Ï€/14))
= 2sin(3Ï€/14)
2sin(3Ï€/14)cos(3Ï€/14)
= 1
cos(3Ï€/14)
= 1
sin(2Ï€/7)
Hence,cosec (Ï€/7) = cosec(2Ï€/7) + cosec(3Ï€/7).