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Question
Prove the following:
cos7° cos 14° cos28° cos 56° = `sin68^circ/(16cos83^circ)`
Solution
L.H.S. = cos7° cos 14° cos28° cos 56°
= `1/(2sin7^circ)(2sin 7^circcos 7^circ)cos 14^circ cos 28^circ cos 56^circ`
= `1/(2sin7^circ)(sin 14^circ cos 14^circ cos 28^circ cos 56^circ)` ...[∵ 2 sin θ cos θ = sin 2θ]
= `1/(2(2sin 7^circ))(2 sin 14^circ cos 14^circ)cos 28^circ cos 56^circ`
= `1/(4sin7^circ)(sin 28^circ cos 28^circ cos 56^circ)`
= `1/(2(4sin7^circ))(2sin 28^circ cos 28^circ) cos 56^circ`
= `1/(8sin 7^circ)(sin 56^circ cos 56^circ)`
= `1/(2(8sin 7^circ))(2sin 56^circ cos 56^circ)`
= `1/(16 sin 7^circ)(sin112^circ)`
= `(sin(180^circ - 68^circ))/(16sin(90^circ - 83^circ))`
= `(sin 68^circ)/(16cos 83^circ)`
= R.H.S.
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