sir i had a ques: show that- cos6.cos42.cos66.cos78= 1/16

Asked by  | 30th May, 2012, 04:33: PM

Expert Answer:

cos6.cos42.cos66.cos78
= (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 + sin54 sin36 / cos18]
= (1/4) [- 1/4 + (cos18 - cos90) / 2cos18]
= (1/4) [- 1/4 + 1/2]
= 1/16

Answered by  | 31st May, 2012, 10:18: AM

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