Math, asked by saleha55510, 5 months ago

prove that
2 \cos(\pi \div 13 )  \cos(9\pi \div 13)   \cos(3\pi \div 13) +  \cos(5\pi \div 13) = 0
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Answers

Answered by Anonymous
26

Answer :

To find :

  • To proof that LHS is equal to the RHS of the equation.

\sf{2\cos(\pi \div 13 )\cos(9\pi \div 13) + \cos(3\pi \div 13) + \cos(5\pi \div 13) = 0}

Knowledge required :

  • \sf{2cos(a)cos(b) = cos(a + b) + cos(a - b)}

  • \sf{cos(a) + cos(b) = 2cos\bigg(\dfrac{a + b}{2}\bigg)cos\bigg(\dfrac{a - b}{2}\bigg)}

  • The value of \sf{cos\bigg(\dfrac{\pi}{2}\bigg) = 0}

Solution :

:\implies \sf{2\cos\bigg(\dfrac{\pi}{13}\bigg)cos\bigg(\dfrac{9\pi}{13}\bigg) + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0} \\ \\

:\implies \sf{\bigg\{2\cos\bigg(\dfrac{9\pi}{13}\bigg)cos\bigg(\dfrac{\pi}{13}\bigg)\bigg\} + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0} \\ \\

By using the formula for \sf{2cos(a)cos(b)}, we get :

:\implies \sf{\bigg\{\cos\bigg(\dfrac{9\pi}{13} + \dfrac{\pi}{13}\bigg) + cos\bigg(\dfrac{9\pi}{13} - \dfrac{\pi}{13}\bigg)\bigg\} + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0} \\ \\

:\implies \sf{\bigg\{\cos\bigg(\dfrac{9\pi + \pi}{13}\bigg) + cos\bigg(\dfrac{9\pi - \pi}{13}\bigg)\bigg\} + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0} \\ \\

:\implies \sf{\bigg\{\cos\bigg(\dfrac{10\pi}{13}\bigg) + cos\bigg(\dfrac{8\pi}{13}\bigg)\bigg\} + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0} \\ \\

:\implies \sf{\bigg\{\cos\bigg(\dfrac{10\pi}{13}\bigg) + cos\bigg(\dfrac{3\pi}{13}\bigg)\bigg\} + \bigg\{cos\bigg(\dfrac{5\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg)\bigg\} = 0} \\ \\

By using the equation for \sf{cos(a) + cos(b)}, we get :

:\implies \sf{\bigg\{2\cos\bigg(\dfrac{\dfrac{10\pi}{13} + \dfrac{3\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{10\pi}{13} - \dfrac{3\pi}{13}}{2}\bigg)\bigg\} + \bigg\{2cos\bigg(\dfrac{\dfrac{8\pi}{13} + \dfrac{5\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{8\pi}{13} - \dfrac{5\pi}{13}}{2}\bigg)\bigg\} = 0} \\ \\

:\implies \sf{\bigg\{2\cos\bigg(\dfrac{\dfrac{10\pi + 3\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{10\pi - 3\pi}{13}}{2}\bigg)\bigg\} + \bigg\{2cos\bigg(\dfrac{\dfrac{8\pi + 5\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{8\pi - 5\pi}{13}}{2}\bigg)\bigg\} = 0} \\ \\

:\implies \sf{\bigg\{2\cos\bigg(\dfrac{\dfrac{13\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{7\pi}{13}}{2}\bigg)\bigg\} + \bigg\{2cos\bigg(\dfrac{\dfrac{13\pi}{13}}{2}\bigg)cos\bigg(\dfrac{\dfrac{3\pi}{13}}{2}\bigg)\bigg\} = 0} \\ \\

:\implies \sf{\bigg\{2\cos\bigg(\dfrac{pi}{2}\bigg)cos\bigg(\dfrac{7\pi}{26}\bigg)\bigg\} + \bigg\{2cos\bigg(\dfrac{\pi}{2}\bigg)cos\bigg(\dfrac{3\pi}{26}\bigg)\bigg\} = 0} \\ \\

By taking \sf{2cos\bigg(\dfrac{\pi}{2}\bigg)}, we get :

:\implies \sf{2\cos\bigg(\dfrac{\pi}{2}\bigg)\bigg\{cos\bigg(\dfrac{7\pi}{26}\bigg) + cos\bigg(\dfrac{3\pi}{26}\bigg)\bigg\} = 0} \\ \\

:\implies \sf{2(0)\bigg\{cos\bigg(\dfrac{7\pi}{26}\bigg) + cos\bigg(\dfrac{3\pi}{26}\bigg)\bigg\} = 0} \\ \\

:\implies \sf{0 = 0} \\ \\

\boxed{\therefore \sf{2\cos\bigg(\dfrac{\pi}{13}\bigg)cos\bigg(\dfrac{9\pi}{13}\bigg) + cos\bigg(\dfrac{3\pi}{13}\bigg) + cos\bigg(\dfrac{5\pi}{13}\bigg) = 0}} \\ \\

Proved !!


Cosmique: NICE
ItzArchimedes: Osm!!!
sethrollins13: Great ! ◉‿◉
NaiRa2021: thanks
Answered by riya15042006
2

Step-by-step explanation:

LHS :

2cos \frac{\pi}{13} cos  \frac{9\pi}{13}  + cos \frac{3\pi}{13}  + cos \frac{5\pi}{13}

 = 2cos \frac{\pi}{13} cos \frac{9\pi}{13}  + 2cos( \frac{ \frac{3\pi}{13} +  \frac{5\pi}{13}  }{2} )cos( \frac{ \frac{3\pi}{13} -  \frac{5\pi}{13}  }{2} )

[We used cosx + cosy = 2cos(x+y/2) cos(x-y/2) ]

 = 2cos \frac{\pi}{13} cos \frac{9\pi}{13}  + 2cos \frac{4\pi}{13} cos (\frac{ - \pi}{13} )

 = 2cos \frac{\pi}{13} cos \frac{9\pi}{13}  + 2cos \frac{4\pi}{13} cos \frac{\pi}{13}

 = 2cos \frac{\pi}{13} (cos \frac{9\pi}{13}  + cos \frac{4\pi}{13} )

 = 2cos \frac{\pi}{13} (2cos( \frac{ \frac{9\pi}{13} +  \frac{4\pi}{13}  }{2} )cos( \frac{ \frac{9\pi}{13} -  \frac{4\pi}{13}  }{2} ))

 = 2cos \frac{\pi}{13} (2cos \frac{\pi}{2} cos \frac{5\pi}{26})

 = 2cos \frac{\pi}{13}  \times 2 \times 0 \times cos \frac{5\pi}{26}

So , 0 = RHS

LHS = RHS

Hence Proved.

I hope it helps u dear friend ^_^


NaiRa2021: thank you
riya15042006: ur welcome ^_^
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