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Question
Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.
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Solution
LHS = cos 20° cos 40° cos 60° cos 80°
`= cos 20° cos 40° (1/2) cos 80° [∵ cos 60° = 1/2]`
`= 1/2 (cos 20° cos 40° cos 80°)`
`= 1/2 ((2 sin 20^circ)/(2 sin 20^circ))` (cos 20° cos 40° cos 80°)
[multiply and divide by 2 sin 20°]
`= 1/2 (((2 sin 20^circ cos 20^circ)cos 40^circ cos 80^circ)/(2 sin 20^circ))`
`= 1/2 (sin (2 xx 20^circ) cos 40^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 (sin 40^circ cos 40^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx (2 sin 40^circ cos 40^circ)/(2 sin 20^circ) xx cos 80^circ`
[multiply and divide by 2]
`= 1/2 1/2 xx ((sin 2 xx 40^circ)cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx (sin 80^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx 1/2 ((2 sin 80^circ cos 80^circ))/(2 sin 20^circ)`
`= 1/2 1/8 xx (sin 160^circ)/(sin 20^circ)`
`= 1/8 xx (sin (180^circ - 20^circ))/(sin 20^circ)`
`= 1/2 1/8 xx (sin 20^circ)/(sin 20^circ)`
`= 1/2 1/8 xx 1`
`= 1/2 (1/8) = 1/16`
Hence Proved.
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