Question

prove that cos A+cos (4 pi/3-A)+cos(4 pi/3+A)=0

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \cos A + \cos \bigg( \frac{4\pi}{3} - A \bigg) +\cos \bigg( \frac{4\pi}{3} + A \bigg) = 0}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \cos C + \cos D =2 \cos \bigg( \frac{C + D}{2} \bigg) \cos \bigg( \frac{C - D}{2} \bigg) }$

EVALUATION

LHS

$\displaystyle \sf{ = \cos A + \cos \bigg( \frac{4\pi}{3} - A \bigg) +\cos \bigg( \frac{4\pi}{3} + A \bigg) }$

$\displaystyle \sf{ = \cos A + \cos \bigg( {240}^{ \circ} - A \bigg) +\cos \bigg( {240}^{ \circ} + A \bigg) }$

$\displaystyle \sf{ = \cos A +2 \cos \bigg( \frac{ {240}^{ \circ} + A + {240}^{ \circ} - A }{2} \bigg) \cos \bigg( \frac{ {240}^{ \circ} + A - {240}^{ \circ} + A }{2} \bigg)}$

$\displaystyle \sf{ = \cos A +2 \cos \bigg( \frac{ {480}^{ \circ} }{2} \bigg) \cos \bigg( \frac{ 2 A }{2} \bigg)}$

$\displaystyle \sf{ = \cos A +2 \cos {240}^{ \circ} \cos A}$

$\displaystyle \sf{ = \cos A - 2 \cos {60}^{ \circ} \cos A}$

$\displaystyle \sf{ = \cos A - 2 \times \frac{1}{2} \times \cos A}$

$\displaystyle \sf{ = \cos A - \cos A}$

$\displaystyle \sf{ = 0}$

= RHS

Hence proved

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