Question

Prove that 2cospi/13cos9pi/13+cos3pi/13+cos5pi/13=0

Answer 1

Answer:

$\displaystyle{2\cos\dfrac{\pi}{13}\cos\dfrac{9\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}=0 \ [\text{Proved}]}$

Step-by-step explanation

$\display \text{Given :}$

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

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

$\display \text{Using formula : }$

$\display \text{2 cos A cos B = cos ( A + B ) + cos ( A - B )}$

$\displaystyle{\implies\left(\cos\dfrac{10\pi}{13}+\cos\dfrac{8\pi}{13}\right)+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}}\\\\\\\displaystyle{\implies\cos\dfrac{10\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{8\pi}{13}+\cos\dfrac{5\pi}{13}}$

$\display \text{Now using product formula :}$

$\displaystyle{\cos \ \text{C}+\cos \ \text{D}=2\cos\left(\dfrac{\text{C + D}}{2}\right).\cos\left(\dfrac{\text{C - D}}{2}\right)}$

$\displaystyle{\implies\cos\dfrac{10\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{8\pi}{13}+\cos\dfrac{5\pi}{13}}\\\\\\\displaystyle{\implies\cos\left(\dfrac{13\pi}{26}\right).\cos\left(\dfrac{7\pi}{26}\right)+\cos\left(\dfrac{13\pi}{26}\right).\cos\left(\dfrac{3\pi}{26}\right)}\\\\\\\displaystyle{\implies\cos\left(\dfrac{\pi}{2}\right).\cos\left(\dfrac{7\pi}{26}\right)+\cos\left(\dfrac{\pi}{2}\right).\cos\left(\dfrac{3\pi}{26}\right)}\\\\\\\display \text{We know \cos\dfrac{\pi}{2}=0 }$

$\displaystyle{\implies0\times\cos\left(\dfrac{7\pi}{26}\right)+0\times\cos\left(\dfrac{3\pi}{26}\right)}\\\\\\\displaystyle{\implies0}$

$\display \text{L.H.S. = R.H.S}$ .

$\display \text{Hence proved.}$