Question

Find the value of \[\sin 12^\circ \sin 48^\circ \sin 54^\circ = \]
A.\[\dfrac{1}{{16}}\]
B.\[\dfrac{1}{{32}}\]
C.\[\dfrac{1}{8}\]
D.\[\dfrac{1}{4}\]

Answer 1

Hint: Here, we will multiply and divide the given expression by 2 so as to make it in the form of a trigonometric identity. Then using the trigonometric formulas, we will simplify the expression. Again, we will rewrite the equation in the form of a trigonometric identity by multiplying and dividing it by 2. Then using the trigonometric formulas, we will simplify the expression to get the required value.

Formula Used: We will use the following formulas:
1.\[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]
2.\[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\]

Complete step-by-step answer:
We have to find the value of: \[\sin 12^\circ \sin 48^\circ \sin 54^\circ \]
Now, this can also be written as:
\[\left( {\sin 48^\circ \sin 12^\circ } \right)\sin 54^\circ \]
Now, multiplying and dividing by 2,
\[ \Rightarrow \left( {\sin 48^\circ \sin 12^\circ } \right)\sin 54^\circ = \dfrac{1}{2}\left( {2\sin 48^\circ \sin 12^\circ } \right)\sin 54^\circ \]
Now, using the formula: \[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\], we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{2}\left[ {\cos \left( {48^\circ - 12^\circ } \right) - \cos \left( {48^\circ + 12^\circ } \right)} \right]\sin 54^\circ \]
Subtracting the angles in the bracket, we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{2}\left[ {\cos 36^\circ - \cos 60^\circ } \right]\sin 54^\circ \]
According to the trigonometric tables, we know that: \[\cos 60^\circ = \dfrac{1}{2}\].
Substituting \[\cos 60^\circ = \dfrac{1}{2}\] in the above equation, we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{2}\left[ {\cos 36^\circ - \dfrac{1}{2}} \right]\sin 54^\circ \]
Multiplying the terms, we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{2}\left[ {\sin 54^\circ \cos 36^\circ - \dfrac{1}{2}\sin 54^\circ } \right]\]
Again, multiplying and dividing the RHS by 2, we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {2\sin 54^\circ \cos 36^\circ - \sin 54^\circ } \right]\]
Now, using the formula, \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\], we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {\sin \left( {54^\circ + 36^\circ } \right) + \sin \left( {54^\circ - 36^\circ } \right) - \sin 54^\circ } \right]\]
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {\sin 90^\circ + \sin 18^\circ - \sin 54^\circ } \right]\]
We know that the trigonometric tables, \[\sin 90^\circ = 1\], \[\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}\] and \[\sin 54^\circ = \dfrac{{\sqrt 5 + 1}}{4}\].
Hence, substituting these values in the above equation, we get,
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {1 + \dfrac{{\sqrt 5 - 1}}{4} - \dfrac{{\sqrt 5 + 1}}{4}} \right]\]
Subtracting the terms, we get
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {1 + \dfrac{{\sqrt 5 - 1 - \sqrt 5 - 1}}{4}} \right]\]
Solving further, we get,
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{4}\left[ {1 - \dfrac{1}{2}} \right] = \dfrac{1}{4}\left[ {\dfrac{1}{2}} \right]\]
\[ \Rightarrow \sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{8}\]
Therefore, the required value of \[\sin 12^\circ \sin 48^\circ \sin 54^\circ = \dfrac{1}{8}\]
Hence, option C is the correct answer.

Note: While solving this question, we have taken the first two angles in brackets because they make a sum equal to \[60^\circ \] which is an angle whose value is given in the trigonometric table. Hence, when we will use the required trigonometric formula in the bracket formed, and then we will find one angle by its value. Also, it is really important to take sine with a larger angle as first term and the sine with smaller angle as second term while solving the product. This is to avoid the use of quadrants in this question to remove the negative sign further.