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

Last updated date: 20th Mar 2023
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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.

$\cos {36^0} - \cos {72^0}$
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,
$\Rightarrow \cos {36^0} - \cos {72^0} = 2\sin \left( {\dfrac{{{{36}^0} + {{72}^0}}}{2}} \right)\sin \left( {\dfrac{{{{72}^0} - {{36}^0}}}{2}} \right)$
$\Rightarrow 2\sin \left( {\dfrac{{{{36}^0} + {{72}^0}}}{2}} \right)\sin \left( {\dfrac{{{{72}^0} - {{36}^0}}}{2}} \right) = 2\sin \left( {\dfrac{{{{108}^0}}}{2}} \right)\sin \left( {\dfrac{{{{36}^0}}}{2}} \right) = 2\sin {54^0}\sin {18^0}$
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 $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
$\Rightarrow 2\sin {54^0}\sin {18^0} = 2\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right) = 2\left( {\dfrac{{5 - 1}}{{16}}} \right) = 2\left( {\dfrac{4}{{16}}} \right) = \dfrac{2}{4} = \dfrac{1}{2}$
$\Rightarrow \cos {36^0} - \cos {72^0} = \dfrac{1}{2}$