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

Last updated date: 20th Mar 2023

•

Total views: 306k

•

Views today: 7.84k

Answer

Verified

306k+ views

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.

“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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE