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Question
Show that `cos pi/15 cos (2pi)/15 cos (3pi)/15 cos (4pi)/15 cos (5pi)/15 cos (6pi)/15 cos (7pi)/15 = 1/128`
Solution
`(pi/15 = 12^circ)`
L.H.S = cos 12° cos 24° cos 36° cos 48° cos 60° cos 72° cos 84° .....(1)
Consider (we know that)
cos A cos(60° + A) cos(60° – A)
= `cos "A" [cos^2 60^circ - sin^2"A"]`
= `cos "A"[1/4 - (1 - cos^2"A")]`
cos A cos(60° + A) cos(60° – A) = `1/4 cos 3"A"`
= `cos "A"[cos^2"A" - 3/4]`
= `(4cos^3 "A" - 3 cos "A")/4`
cos 12° cos 72° cos 48° = `1/4 cos 3(12^circ)`
= `1/4 cos 36^circ`
= `1/4[(sqrt(5) + 1)/4]`
Similarly cos 24° cos 84° cos 36° = `1/4 cos3 (12^circ)`
= `1/4 cos 72^circ`
= `1/4 cos(90^circ - 18^circ)`
= `1/4 sin 18^circ`
= `1/4[(sqrt(5) - 1)/4]`
(1) ⇒ L.H.S = `1/4[(sqrt(5) + 1)/4] * 1/4[(sqrt(5) - 1)/4] * 1/2`
= `1/4((sqrt(5) + 1)/4 * (sqrt(5) - 1)/4) * 1/2`
= `(5 - 1)/(128 xx 4)`
= `1/128`
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