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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: sin 20° sin 40° sin 80° = 38

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प्रश्न

Prove the following:

sin 20° sin 40° sin 80° = `sqrt(3)/8`

बेरीज
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उत्तर

L.H.S. = sin 20° sin 40° sin 80° 

= sin 20°. sin 40°. sin 80°

= `1/2(2.sin40^circ.sin20^circ).sin80^circ`

= `1/2[cos(40^circ - 20^circ) - cos(40^circ + 20^circ)].sin80^circ`

= `1/2(cos20^circ - cos60^circ)sin80^circ`

= `1/2cos20^circ.sin80^circ - 1/2 cos60^circ.sin80^circ`

= `1/(2 xx 2)(2sin80^circ.cos20^circ) - 1/2(1/2).sin80^circ`

= `1/4.[sin(80^circ + 20^circ) + sin(80^circ - 20^circ)] - 1/4.sin80^circ`

= `1/4.(sin100^circ + sin60^circ) - 1/4.sin80^circ`

= `1/4sin 100^circ + 1/4sin 60^circ - 1/4sin 80^circ`

= `1/4.sin(180^circ - 80^circ) + 1/4 xx sqrt(3)/2 - 1/4.sin80^circ`

= `1/4sin80^circ + sqrt(3)/8 - 1/4sin80^circ` ...[∵ sin (180° – θ) = sin θ]

= `sqrt(3)/8`

= R.H.S.

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पाठ 3: Trigonometry - 2 - Miscellaneous Exercise 3 [पृष्ठ ५८]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
पाठ 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II. (22) | पृष्ठ ५८

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