Question

# Prove that $2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} = 0$

Verified
129.6k+ views
Hint: Here we use the property of trigonometry $2\cos A\cos B = \cos (A + B) + \cos (A - B)$ to solve first part of the equation and then we group together two pairs of values in cos on which we apply the formula $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$.

First we solve $2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}}$by using the identity $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
Where $A = \dfrac{\pi }{{13}},B = \dfrac{{9\pi }}{{13}}$
So, by substituting the values we can write.
$\Rightarrow 2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} = \cos (\dfrac{\pi }{{13}} + \dfrac{{9\pi }}{{13}}) + \cos (\dfrac{\pi }{{13}} - \dfrac{{9\pi }}{{13}})$
Take LCM of the angles within the bracket.
$\Rightarrow 2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} = \cos (\dfrac{{\pi + 9\pi }}{{13}}) + \cos (\dfrac{{\pi - 9\pi }}{{13}}) \\ \Rightarrow 2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} = \cos (\dfrac{{10\pi }}{{13}}) + \cos (\dfrac{{ - 8\pi }}{{13}}) \\$
Since we know cosine is an even function, which means that $\cos ( - x) = \cos (x)$
Here value of $x = - \dfrac{{8\pi }}{{13}}$
So, substitute the value of $\cos ( - \dfrac{{8\pi }}{{13}}) = \cos (\dfrac{{8\pi }}{{13}})$in the equation.
$\Rightarrow 2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} = \cos (\dfrac{{10\pi }}{{13}}) + \cos (\dfrac{{8\pi }}{{13}})$ … (i)
Now we substitute the value from equation (i) on the left hand side of the equation.
$\Rightarrow 2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} = \cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}$ … (ii)
Now we use the formula $\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})$in two separate pairs.
We make pairs such that when we add the numerator it gets cancelled by the denominator and gives us an angle whose cosine is known to us.
We make a pair of $\left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right)$and other pair of $\left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right)$
First we solve $\left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right)$, where $A = \dfrac{{10\pi }}{{13}},B = \dfrac{{3\pi }}{{13}}$
$\Rightarrow \left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{10\pi }}{{13}} + \dfrac{{3\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{10\pi }}{{13}} - \dfrac{{3\pi }}{{13}}}}{2}} \right)$
Taking LCM in the numerator of the angles inside the bracket.
$\Rightarrow \left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{10\pi + 3\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{10\pi - 3\pi }}{{13}}}}{2}} \right) \\ \Rightarrow \left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{13\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{7\pi }}{{13}}}}{2}} \right) \\$
Cancelling the same term from numerator and denominator in the angle.
$\Rightarrow \left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{{7\pi }}{{26}}} \right)$
Since we know $\cos \dfrac{\pi }{2} = 0$
$\Rightarrow \left( {\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}}} \right) = 2 \times 0 \times \cos \left( {\dfrac{{7\pi }}{{26}}} \right) = 0$ … (iii)
Now we solve$\left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right)$, where $A = \dfrac{{8\pi }}{{13}},B = \dfrac{{5\pi }}{{13}}$
$\Rightarrow \left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{8\pi }}{{13}} + \dfrac{{5\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{8\pi }}{{13}} - \dfrac{{5\pi }}{{13}}}}{2}} \right)$
Taking LCM in the numerator of the angles inside the bracket.
$\Rightarrow \left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{8\pi + 5\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{8\pi - 5\pi }}{{13}}}}{2}} \right)$
$\Rightarrow \left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{{\dfrac{{13\pi }}{{13}}}}{2}} \right)\cos \left( {\dfrac{{\dfrac{{3\pi }}{{13}}}}{2}} \right)$
Cancelling the same term from numerator and denominator in the angle.
$\Rightarrow \left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right) = 2\cos \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{{3\pi }}{{26}}} \right)$
Since we know $\cos \dfrac{\pi }{2} = 0$
$\Rightarrow \left( {\cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}} \right) = 2 \times 0 \times \cos \left( {\dfrac{{3\pi }}{{26}}} \right) = 0$ … (iv)
Now we substitute the values from equation (iii) and (iv) in equation (ii)
$\Rightarrow \cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} = 0 + 0 = 0$
which is equal to the right hand side of the equation.
Hence Proved.

Note: Students many times make mistake of writing the value of $\cos ( - \dfrac{{8\pi }}{{13}}) = - \cos (\dfrac{{8\pi }}{{13}})$ because they think that negative sign comes out of the angle which is wrong, we always classify if the function is an odd or even function and then find if sign vanishes or comes out. In this case cosine is an even function, so negative sign vanishes.