Question

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or modicators and maths Aryabhatta
$\pink{2cos \frac{\pi}{13} \cos \frac{9\pi}{13}+ \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0}$
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Answer 1

Given:-

$\pink{2cos \frac{\pi}{13} \cos \frac{9\pi}{13}+ \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0}$

To Assume:-

• Prove it

Solution:-

$\pink{2cos \frac{\pi}{13} \cos \frac{9\pi}{13}+ \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0}$

Taking L.H.S

$\blue{2cos \frac{\pi}{13} \cos \frac{9\pi}{13}+ \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0}$

$2 \cos \times cos \: y \: = cos(x + y) + cos(x - y) \\ putting \: \: \: x = \frac{9\pi}{13} \: and \: \frac{\pi}{13} \\ 2cos \: \: \frac{9\pi}{13} cos \: \frac{\pi}{13} = \cos( \frac{9\pi}{13} + \frac{\pi}{13}) + cos( \frac{9\pi}{13} + \frac{\pi}{13} ) = cos( \frac{10\pi}{13} ) + cos( \frac{8\pi}{13} ) \\ = (cos \frac{10\pi}{13} + cos \frac{8\pi}{13} ) + cos \frac{3\pi}{13} + cos \frac{5\pi}{13}$

$cos( \frac{10\pi}{13} + cos \: \frac{3\pi}{13} + (cos \frac{8\pi}{13} + cos \: \frac{5\pi}{13} )$

$Using \: cos \: x + cos \: y = 2 \: cos \: \: \frac{x + y}{2} \: \: cos \: \: \frac{x - y}{2} \\ (2cos \: \: (\frac{ \frac{10\pi}{13} + \frac{3\pi}{13} }{10} ) .cos \: ( \frac{ \frac{10\pi}{13} - \frac{3\pi}{13}}{2} )) + (2 \: cos \: ( \frac{ \frac{8\pi}{13} + \frac{5\pi}{13} }{2} ).cos ( \frac{ \frac{8\pi}{13} - \frac{5\pi}{13} }{2} )) \\ (2 \: cos \: ( \frac{ \frac{13\pi}{13} }{2 } ).cos \: \: ( \frac{ \frac{7\pi}{13} }{2} )) + (2cos \: \frac{ \frac{13\pi}{13} }{2} .cos \frac{ \frac{3\pi}{13} }{2} )$

$(2cos \frac{\pi}{2} .cos \frac{7\pi}{26} ) + (2cos \frac{\pi}{2} .cos \frac{3\pi}{26} ) \\ 2cos \frac{\pi}{2} (cos \frac{7\pi}{26} + cos \frac{3\pi}{26} ) \\ 2 \times 0(cos \frac{7\pi}{26} + cos \frac{3\pi}{26} ) \: \: \: \: \: (cos \frac{\pi}{2} = 0) \\ = 0$

R.H.S

$\small\color{black}\boxed{\colorbox{white}{∴Hence,L.H.S =R.H.S}}$

$\small \color{hotpink}\boxed{\colorbox{white}{Hence Proved}}$

Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:2cos \dfrac{\pi}{13} \cos \dfrac{9\pi}{13}+ \cos \dfrac{3\pi}{13} + \cos \dfrac{5\pi}{13} = 0$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:2cos \dfrac{\pi}{13} \cos \dfrac{9\pi}{13}+ \cos \dfrac{3\pi}{13} + \cos \dfrac{5\pi}{13}$

We know,

$\red{ \boxed{ \sf{ \:2cosxcosy = cos(x + y) + cos(x - y)}}}$

So, using this identity, we get

$\rm \: = \: cos\bigg(\dfrac{9\pi}{13} + \dfrac{\pi}{13} \bigg) + cos\bigg(\dfrac{9\pi}{13} - \dfrac{\pi}{13} \bigg) + cos\dfrac{3\pi}{13} + cos\dfrac{5\pi}{13}$

$\rm \: = \: cos\bigg(\dfrac{10\pi}{13} \bigg) + cos\bigg(\dfrac{8\pi}{13} \bigg) + cos\dfrac{3\pi}{13} + cos\dfrac{5\pi}{13}$

can be rewritten as

$\rm \: = \: cos\bigg(\pi - \dfrac{3\pi}{13} \bigg) + cos\bigg(\pi - \dfrac{5\pi}{13} \bigg) + cos\dfrac{3\pi}{13} + cos\dfrac{5\pi}{13}$

We know,

$\red{ \boxed{ \sf{ \:cos(\pi - x) = - \: cosx}}}$

So, using this, we get

$\rm \: = - \cancel{\: cos\bigg(\dfrac{3\pi}{13} \bigg)} -\cancel{ cos\bigg(\dfrac{5\pi}{13} \bigg)} +\cancel{ cos\dfrac{3\pi}{13}} + \cancel{cos\dfrac{5\pi}{13}}$

$\rm \: = \: 0$

Hence,

$\red{ \boxed{ \sf{ \:2cos \frac{\pi}{13} \cos \frac{9\pi}{13}+ \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:2sinxcosy = sin(x + y) + sin(x - y)}}}$

$\red{ \boxed{ \sf{ \:2sinxsiny = cos(x - y) - cos(x + y)}}}$

$\red{ \boxed{ \sf{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg]}}}$