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sin 5A+ 2sin 8A+sin11Asin 8A+ 2sin 11A+sin 14Asin 8Asin11A10. Show that |
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Answer» sin 5A + 2 sin 8A + sin 11A] / [sin 8A + 2 sin 11A + sin 14A] = [ (sin 5A + sin 11 A) + 2 sin 8A] / [(sin 8A + sin 14A) + 2 sin 11A] (Rearranging) = [ 2 sin (11A + 5A)/2 * sin (11A - 5A)/2 + 2 sin 8A] / [ 2 sin (14 A + 8A)/2 * sin (14A - 8A)/2 + 2 sin 11A] { Using sin C + sin D = 2 sin (C + D)/2 * sin (C - D)/2 } = [ 2 sin 16A/2 * sin 6A/2 + 2 sin 8A] / [ 2 sin 22A/2 * sin 6A/2 + 2 sin 11A] =2[ sin 8A * sin 3A + sin 8A] /2[ sin 11A * sin 3A + sin 11A] ( Taking 2 common in numerator and denominator) = sin 8A[ sin 3A + 1]/ sin 11A[ sin 3A + 1]( Taking sin 8A common in numerator sin 11A common in denominarator) = sin 8A / sin11A |
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