Question

Find the value of $\cos {36^0} - \cos {72^0}$?

Answer 1

Hint – In this question we have to find the value of the given expression. The basic trigonometric identities like $\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right)$, along with the algebraic identities will help will in simplifying the given expression.

“Complete step-by-step answer:”
Given equation is
$\cos {36^0} - \cos {72^0}$
As we know $\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right)$ so, use this property in above equation we have,
$\Rightarrow \cos {36^0} - \cos {72^0} = 2\sin \left( {\dfrac{{{{36}^0} + {{72}^0}}}{2}} \right)\sin \left( {\dfrac{{{{72}^0} - {{36}^0}}}{2}} \right)$
Now simplify the above equation we have,
$\Rightarrow 2\sin \left( {\dfrac{{{{36}^0} + {{72}^0}}}{2}} \right)\sin \left( {\dfrac{{{{72}^0} - {{36}^0}}}{2}} \right) = 2\sin \left( {\dfrac{{{{108}^0}}}{2}} \right)\sin \left( {\dfrac{{{{36}^0}}}{2}} \right) = 2\sin {54^0}\sin {18^0}$
Now we know that $\sin {54^0} = \dfrac{{\sqrt 5 + 1}}{4},{\text{ }}\sin {18^0} = \dfrac{{\sqrt 5 - 1}}{4}$ so, substitute this value in above equation we have,
$\Rightarrow 2\sin {54^0}\sin {18^0} = 2\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)$
Now simplify this using property $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
$\Rightarrow 2\sin {54^0}\sin {18^0} = 2\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right) = 2\left( {\dfrac{{5 - 1}}{{16}}} \right) = 2\left( {\dfrac{4}{{16}}} \right) = \dfrac{2}{4} = \dfrac{1}{2}$
$\Rightarrow \cos {36^0} - \cos {72^0} = \dfrac{1}{2}$

Note – Whenever we face such types of questions the key concept is simply to have the gist of basic trigonometric identities, application of these identities combined with algebraic identities helps simplify such trigonometry problems and will get you on the right track to reach the answer.