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Prove the following: sin47° + sin61° − sin11° − sin25° = cos7°

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Question

Prove the following:

sin47° + sin61° − sin11° − sin25° = cos7°

Sum
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Solution

L.H.S. = sin47° + sin61° − sin11° − sin25°

= (sin47° − sin25°) + (sin61° − sin11°)

= `2cos((47^circ + 25^circ)/2)*sin ((47^circ - 25^circ)/2) + 2cos((61^circ + 11^circ)/2)*sin((61^circ - 11^circ)/2)`

= 2cos36° · sin11° + 2cos36° · sin25°

= 2cos36° (sin25° + sin11°)

= `2cos36^circ xx 2sin((25^circ + 11^circ)/2)*cos((25^circ - 11^circ)/2)`

= 4cos36° · sin18° · cos7°

= `4 xx (sqrt(5) + 1)/4 xx (sqrt(5) - 1)/4 xx cos7^circ   ...[because cos36^circ = (sqrt(5) + 1)/4, sin 18^circ = (sqrt(5) - 1)/4]`

= `(4(5 - 1))/16 cos7^circ`

= cos7°

= R.H.S.

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Chapter 3: Trigonometry - 2 - Miscellaneous Exercise 3 [Page 58]

APPEARS IN

Balbharati Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
Chapter 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II. (29) | Page 58

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