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Prove that cos10°+cos110°+cos130°=0 |
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Answer» Use the formulacosC+cosD:2cos(C+D/2)cos(C-D/2)Use this in the question:2cos(10+110/2)cos(10-110/2)+cos130=02cos60°cos(-50°)+cos130°=02×1/2.cos50+cos130=0cos50+cos130=02cos(50+130/2)cos(50-130/2)=02cos90.cos40=02×0.cos40=00=0H.P Thku cos 10 + cos 110 + cos 130 = (cos 10 + cos 110) + cos 130=\xa0{tex}2cos \\frac{(110+10)}{2}\\times cos\\frac{(110-10)}{2}{/tex}\xa0+ cos 130 {using cos C + cos D}= {tex}2cos \\frac {(C+D)}{2}{/tex}{tex}\\times{/tex}{tex}cos\\frac {(C-D)}{2}{/tex}={tex}2cos\\frac{120}{2}{/tex}{tex}\\times{/tex}{tex}cos\\frac {100}{2}{/tex}+ cos (180 - 50) (As 130 = 180 - 50)= 2 cos 60 {tex}\\times{/tex} cos 50 - cos 50 (Using cos (180 - A) = - cos A)= 2 {tex}\\times{/tex}{tex}\\frac{1}{2}{/tex}{tex}\\times{/tex} cos 50 - cos 50 ( As cos 60 = {tex}\\frac{1}{2}{/tex})= cos 50o - cos 50o = 0
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