
Prove the following trigonometric equation
$4\cos {12^0}\cos {48^0}\cos {72^0} = \cos {36^0}$
Answer
619.5k+ 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
Master Class 10 Computer Science: Engaging Questions & Answers for Success

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

Master Class 10 English: Engaging Questions & Answers for Success

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

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

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

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

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

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

