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Prove the following : sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33° - Mathematics and Statistics

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प्रश्न

Prove the following :

sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°

योग

उत्तर

L.H.S. = sin 18°·cos 39° + sin 6°·cos 15°

= `1/2(2 cos 39^circ*sin18^circ +  2cos15^circ*sin6^circ)`

= `1/2[{sin(39^circ + 18^circ) - sin(39^circ - 18^circ)} + {sin(15^circ + 6^circ) - sin(15^circ - 6^circ)}]`

= `1/2[sin57^circ - sin21^circ +  sin21^circ -  sin9^circ]`

= `1/2(sin57^circ -  sin9^circ)`

= `1/2 xx 2cos((57^circ + 9^circ)/2)*sin((57^circ - 9^circ)/2)`

= cos 33°·sin 24°

= sin 24°·cos 33°

= R.H.S.

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अध्याय 3: Trigonometry - 2 - Exercise 3.4 [पृष्ठ ५१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 3 Trigonometry - 2
Exercise 3.4 | Q 2. (iv) | पृष्ठ ५१
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