# Prove the following trigonometric equation

$4\cos {12^0}\cos {48^0}\cos {72^0} = \cos {36^0}$

Last updated date: 16th Mar 2023

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Answer

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Hint- Use different trigonometric identities of combination of angles in order to solve the question, also keep in mind the RHS part for manipulation.

Complete step-by-step solution -

Given that: to prove $4\cos {12^0}\cos {48^0}\cos {72^0} = \cos {36^0}$

Since we know that

$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$

Taking the LHS part and proceeding further

$

LHS = 4\cos {12^0}\cos {48^0}\cos {72^0} \\

= 2 \times \left( {2\cos {{12}^0}\cos {{48}^0}} \right) \times \cos {72^0} \\

$

With the help of above formula simplifying the middle term

$

= 2 \times \left[ {\cos \left( {{{48}^0} + {{12}^0}} \right) + \cos \left( {{{48}^0} - {{12}^0}} \right)} \right] \times \cos {72^0} \\

= 2 \times \left[ {\cos \left( {{{60}^0}} \right) + \cos \left( {{{36}^0}} \right)} \right] \times \cos {72^0} \\

= 2\cos {60^0}\cos {72^0} + 2\cos {36^0} \times \cos {72^0} \\

$

Now substituting the value of know trigonometric quantity in the above equation

$

= 2 \times \dfrac{1}{2} \times \cos {72^0} + 2\cos {36^0}\cos {72^0}{\text{ }}\left[ {\because \cos {{60}^0} = \dfrac{1}{2}} \right] \\

= \cos {72^0} + 2\cos {36^0}\cos {72^0} \\

$

Again using the formula in second part

$

= \cos {72^0} + \left[ {\cos \left( {{{72}^0} + {{36}^0}} \right) + \cos \left( {{{72}^0} - {{36}^0}} \right)} \right] \\

= \cos {72^0} + \left[ {\cos {{108}^0} + \cos {{36}^0}} \right] \\

$

Also we know that

$

\cos \left( {180 - \theta } \right) = - \cos \theta \\

\Rightarrow \cos {108^0} = \cos \left( {{{180}^0} - {{72}^0}} \right) = - \cos {72^0} \\

$

So, substituting the value in above equation we have

$

= \cos {72^0} + \cos {108^0} + \cos {36^0} \\

= \cos {72^0} - \cos {72^0} + \cos {36^0}{\text{ }}\left[ {\because \cos {{108}^0} = - \cos {{72}^0}({\text{proved above}})} \right] \\

= \cos {36^0} \\

$

Which is equal to the RHS.

Hence, the given trigonometric equation is proved.

Note- In order to solve types of complex problems including some random angle values always try to use the trigonometric identities in order to solve the problem. Never try to find the value of such trigonometric terms. Whenever while solving if some known values of trigonometric terms appear, put into the values of such terms. Also keep in mind the part to be proved for an easy solution.

Complete step-by-step solution -

Given that: to prove $4\cos {12^0}\cos {48^0}\cos {72^0} = \cos {36^0}$

Since we know that

$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$

Taking the LHS part and proceeding further

$

LHS = 4\cos {12^0}\cos {48^0}\cos {72^0} \\

= 2 \times \left( {2\cos {{12}^0}\cos {{48}^0}} \right) \times \cos {72^0} \\

$

With the help of above formula simplifying the middle term

$

= 2 \times \left[ {\cos \left( {{{48}^0} + {{12}^0}} \right) + \cos \left( {{{48}^0} - {{12}^0}} \right)} \right] \times \cos {72^0} \\

= 2 \times \left[ {\cos \left( {{{60}^0}} \right) + \cos \left( {{{36}^0}} \right)} \right] \times \cos {72^0} \\

= 2\cos {60^0}\cos {72^0} + 2\cos {36^0} \times \cos {72^0} \\

$

Now substituting the value of know trigonometric quantity in the above equation

$

= 2 \times \dfrac{1}{2} \times \cos {72^0} + 2\cos {36^0}\cos {72^0}{\text{ }}\left[ {\because \cos {{60}^0} = \dfrac{1}{2}} \right] \\

= \cos {72^0} + 2\cos {36^0}\cos {72^0} \\

$

Again using the formula in second part

$

= \cos {72^0} + \left[ {\cos \left( {{{72}^0} + {{36}^0}} \right) + \cos \left( {{{72}^0} - {{36}^0}} \right)} \right] \\

= \cos {72^0} + \left[ {\cos {{108}^0} + \cos {{36}^0}} \right] \\

$

Also we know that

$

\cos \left( {180 - \theta } \right) = - \cos \theta \\

\Rightarrow \cos {108^0} = \cos \left( {{{180}^0} - {{72}^0}} \right) = - \cos {72^0} \\

$

So, substituting the value in above equation we have

$

= \cos {72^0} + \cos {108^0} + \cos {36^0} \\

= \cos {72^0} - \cos {72^0} + \cos {36^0}{\text{ }}\left[ {\because \cos {{108}^0} = - \cos {{72}^0}({\text{proved above}})} \right] \\

= \cos {36^0} \\

$

Which is equal to the RHS.

Hence, the given trigonometric equation is proved.

Note- In order to solve types of complex problems including some random angle values always try to use the trigonometric identities in order to solve the problem. Never try to find the value of such trigonometric terms. Whenever while solving if some known values of trigonometric terms appear, put into the values of such terms. Also keep in mind the part to be proved for an easy solution.

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