Prove that cos10°+cos110°+cos130°=0
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Posted by Aparna Sinha 5 years, 10 months ago
- 3 answers
Sia ? 5 years, 10 months ago
cos 10 + cos 110 + cos 130 = (cos 10 + cos 110) + cos 130
= {tex}2cos \frac{(110+10)}{2}\times cos\frac{(110-10)}{2}{/tex} + 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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Iram Hasan Hasan 5 years, 10 months ago
1Thank You