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Prove the following: cos7° cos14° cos28° cos56° = sin⁡68∘16⁢cos⁡83∘ - Mathematics and Statistics

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प्रश्न

Prove the following:

cos7° cos14° cos28° cos56° = `sin68^circ/(16cos83^circ)`

बेरीज
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उत्तर

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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Trigonometric Functions of Triple Angle
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - 2 - Exercise 3.3 [पृष्ठ ४८]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 3 Trigonometry - 2
Exercise 3.3 | Q 3. (xiii) | पृष्ठ ४८

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