Solveeit Logo
Login

Question

Question: Solve \( \cos 6^\circ \sin 24^\circ \cos 72^\circ = \\\ \left( {\text{A}} \right) - \dfrac{1}{8}...

Solve cos⁡6∘sin⁡24∘cos⁡72∘= (A)−18 (B)−14 (C)18 (D)14 \cos 6^\circ \sin 24^\circ \cos 72^\circ = \\\ \left( {\text{A}} \right) - \dfrac{1}{8} \\\ \left( {\text{B}} \right) - \dfrac{1}{4} \\\ \left( {\text{C}} \right)\dfrac{1}{8} \\\ \left( {\text{D}} \right)\dfrac{1}{4} \\\

Explanation

Solution

Since it is not possible or difficult to find the angles of the trigonometric function individually, try to bring those angles in the form of the angles which are easily available to us or which we can find out easily. You can do this by using trigonometric identities and by simple adjustments.

Complete step-by-step solution:
Now to start with this kind of question one has to always Keep in mind that the values of the trigonometric angles we do not know should always be converted into the values we know. Also bringing the trigonometric equations in the basic form makes it easy to solve the equation.
Now for the given equation we don’t know the values of cos⁡6∘,sin⁡24∘,cos⁡72∘\cos 6^\circ ,\sin 24^\circ ,\cos 72^\circ .
Hence our first aim should be to try to change the angles in such a way whose values we know or are available to us.
First we will start the solution by changing the writing the angle of cos⁡72∘\cos 72^\circ a little differently that is cos⁡(90−18)∘\cos \left( {90 - 18} \right)^\circ
So it will be written as cos⁡6∘.sin⁡24∘.cos⁡72∘=cos⁡6∘.sin⁡24∘.cos⁡(90−18)∘\cos 6^\circ .\sin 24^\circ .\cos 72^\circ = \cos 6^\circ .\sin 24^\circ .\cos \left( {90 - 18} \right)^\circ
=cos⁡6∘.sin⁡24∘.sin⁡18∘−−−= \cos 6^\circ .\sin 24^\circ .\sin 18^\circ - - -[as cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta ]
=12(2.cos⁡6.sin⁡24)sin⁡18−−−= \dfrac{1}{2}\left( {2.\cos 6.\sin 24} \right)\sin 18 - - - Multiplying and dividing by 2 so that we can use identities.=12(sin⁡(6+24)−sin⁡(6−24))sin⁡18−−−= \dfrac{1}{2}\left( {\sin \left( {6 + 24} \right) - \sin \left( {6 - 24} \right)} \right)\sin 18 - - -Using 2cos⁡a.sin⁡b=sin⁡(a+b)−sin⁡(a−b)2\cos a.\sin b = \sin \left( {a + b} \right) - \sin \left( {a - b} \right) =12(sin⁡30−sin⁡(−18))sin⁡18 = \dfrac{1}{2}\left( {\sin 30 - \sin \left( { - 18} \right)} \right)\sin 18
=12(sin⁡30+sin⁡(18))sin⁡18−−−= \dfrac{1}{2}\left( {\sin 30 + \sin \left( {18} \right)} \right)\sin 18 - - -since (sin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta )
=12(12+5−14)5−14−−−= \dfrac{1}{2}\left( {\dfrac{1}{2} + \dfrac{{\sqrt 5 - 1}}{4}} \right)\dfrac{{\sqrt 5 - 1}}{4} - - - since the value of sin⁡18\sin 18 is 5−14\dfrac{{\sqrt 5 - 1}}{4}
=12×14(2+5−1)(5−14)= \dfrac{1}{2} \times \dfrac{1}{4}\left( {2 + \sqrt 5 - 1} \right)\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)
=18(1+5)(5−14)= \dfrac{1}{8}\left( {1 + \sqrt 5 } \right)\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)
=18×4(1+5)(1−5)= \dfrac{1}{{8 \times 4}}\left( {1 + \sqrt 5 } \right)\left( {1 - \sqrt 5 } \right)
=18×4(12−(5)2)−−−= \dfrac{1}{{8 \times 4}}\left( {{1^2} - {{\left( {\sqrt 5 } \right)}^2}} \right) - - - using the formula a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
=18×4(1−5) =18×4×4 =18  = \dfrac{1}{{8 \times 4}}\left( {1 - 5} \right) \\\ = \dfrac{1}{{8 \times 4}} \times 4 \\\ = \dfrac{1}{8} \\\
Hence we got our required solution to our question.

The answer to our question is option (C)18\left( {\text{C}} \right)\dfrac{1}{8}

Note: While solving this kind of problem, one should remember that there are many ways of getting to the answer by using various trigonometric identities. This may lead to the solution getting difficult or easier depending on the path we choose. Also the identities used should be precise and one should mention about the identities used.