cos6.cos42.cos66.cos78

Hi Shubh,
The easiest way would be to group cos6 with cos66
and cos42 with cos78.
So cos6cos66 = (1/2)(cos72 + cos60)
where cos72 = (√5 - 1)/4, cos60 = 1/2
Similarly cos42cos78 = (1/2)(cos36+cos120)
where cos36 = (√5+1)/4, cos120 = -1/2
Substitute, and get the answer... you should get: 3/8 as the answer....

Best Regards,
Ashwin (IIT Madras).

Approved

Dear shubh mishra,
cos6.cos42.cos66.cos78

= (cos 6) (cos 42)(cos 66) (cos 78)
= (1/4) * [2cos6 cos66] * [2cos42 cos78]
= (1/4) * (cos72 + cos60) * (cos120 + cos36)
= (1/4) (cos72 + 1/2) * (- 1/2 + cos36)
= (1/4) [- 1/4 + (1/2) (cos36 - cos72) + cos36cos72]
= (1/4) [- 1/4 + sin54sin18 + sin54 sin18]
= (1/4) [- 1/4 + 2sin54 sin18 cos18 / cos18]
= (1/4) [- 1/4 + 2sin54 sin36 / cos18]
= (1/4) [- 1/4 + (cos18 - cos90) / 2cos18]
= (1/4) [- 1/4 + 1/2]
= 1/16

This is the answer !!!
All the very best & good luck to you dear..
Hope this helped you immensely...

Regards,
AskIITians Expert,
Godfrey Classic Prince
IIT-M
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