Prove the following trigonometric equation
$4\cos {12^0}\cos {48^0}\cos {72^0} = \cos {36^0}$
Answer
639.9k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the name of Japan Parliament?

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

