Math, asked by FazeelKarkhi, 9 months ago

Prove this trigonometry question.

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Answers

Answered by Rohit18Bhadauria

To Prove:

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

Solution:

We know that,

$\pink{\underline{\boxed{\bf{cosC+cosD=2cos\dfrac{C+D}{2}cos\dfrac{C-D}{2}}}}}$

$\purple{\underline{\boxed{\bf{cos(-x)=cosx}}}}$

$\orange{\underline{\boxed{\bf{cos\dfrac{\pi}{2}=0}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Here,

$\rm{L.H.S.=2cos\dfrac{\pi}{13}cos\dfrac{9\pi}{13}+cos\dfrac{3\pi}{13}+cos\dfrac{5\pi}{13}}$

$\sf{L.H.S.=2cos\dfrac{\pi}{13}cos\dfrac{9\pi}{13}+2cos\dfrac{\frac{3\pi+5\pi}{13}}{2}cos\dfrac{\frac{3\pi-5\pi}{13}}{2}}$

$\sf{L.H.S.=2cos\dfrac{\pi}{13}cos\dfrac{9\pi}{13}+2cos\dfrac{8\pi}{2\times13}cos\dfrac{-2\pi}{2\times13}}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}cos\dfrac{9\pi}{13}+2cos\dfrac{4\pi}{13}cos\dfrac{-\pi}{13}}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}cos\dfrac{9\pi}{13}+2cos\dfrac{4\pi}{13}cos\dfrac{\pi}{13}}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(cos\dfrac{9\pi}{13}+2cos\dfrac{4\pi}{13}\bigg)}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(2cos\dfrac{\frac{9\pi+4\pi}{13}}{2}cos\dfrac{\frac{9\pi-4\pi}{13}}{2}\bigg)}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(2cos\dfrac{13\pi}{2\times13}cos\dfrac{5\pi}{2\times13}\bigg)}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(2cos\dfrac{\pi}{2}cos\dfrac{5\pi}{26}\bigg)}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(2(0)cos\dfrac{5\pi}{26}\bigg)}$

$\rm{L.H.S.=2cos\dfrac{\pi}{13}\bigg(0\bigg)}$

$\rm{L.H.S.=0}$

$\rm\green{L.H.S.=R.H.S.}$

$\bigstar\underline{\underline{\rm\red{Hence\ Proved}}}$

Answered by MuskanJoshi14

Step-by-step explanation:

To Prove:

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

Solution: