Question

Find the value of \[\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right)\]
A. \[\dfrac{1}{2}\]
B. \[ - \dfrac{1}{2}\]
C. \[\dfrac{1}{4}\]
D. \[1\]

Answer 1

Hint:
We need to convert the given unknown sine angles into known sine and cosine angles so as to substitute values into the given equation and calculate the required answer.



Formula Used:
\[\sin {18^\circ } = \dfrac{{\sqrt 5 - 1}}{4},\cos {36^\circ } = \dfrac{{\sqrt 5 + 1}}{4}\]
\[\sin \left( {{{90}^\circ } - \theta } \right) = \cos \theta \]




Complete step-by-step answer:
Since we know the values of \[\sin {18^\circ }\] and \[\cos {36^\circ }\], we will try to convert the given equation\[\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right)\] into a form where we can substitute the known ones
Given,
\[\sin \left( {\dfrac{\pi }{{10}}} \right)\sin \left( {\dfrac{{3\pi }}{{10}}} \right)\]
Converting radian values into degree
\[ = \sin \left( {\dfrac{\pi }{{10}} \times \dfrac{{{{180}^ \circ }}}{\pi }} \right)\sin \left( {\dfrac{{3\pi }}{{10}} \times \dfrac{{{{180}^ \circ }}}{\pi }} \right)\]
\[ = \sin {18^\circ }\sin {54^\circ }\]
\[ = \sin {18^\circ }\sin \left( {{{90}^\circ } - {{36}^\circ }} \right)\]
Substituting using formula \[\sin \left( {{{90}^\circ } - \theta } \right) = \cos \theta \], we get
\[ = \sin {18^\circ }\cos {36^\circ }\]
Now putting the known values \[\sin {18^\circ } = \dfrac{{\sqrt 5 - 1}}{4} and \cos {36^\circ } = \dfrac{{\sqrt 5 + 1}}{4}\] , we get
\[ = \dfrac{{\sqrt 5 - 1}}{4} \cdot \dfrac{{\sqrt 5 + 1}}{4}\]
Applying the algebraic identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]
\[ = \dfrac{{5 - 1}}{{16}}\]
\[ = \dfrac{4}{{16}}\]
\[ = \dfrac{1}{4}\]
Option C is the correct answer

Additional information:
We can convert the radian to degree by multiplying \[\dfrac{{{{180}^ \circ }}}{\pi }\].
For example: \[\dfrac{\pi }{3} = \dfrac{\pi }{3} \times \dfrac{{{{180}^ \circ }}}{\pi } = {60^ \circ }\]
To convert degree to radian we multiply \[\dfrac{\pi }{{{{180}^ \circ }}}\].
For example: \[{60^ \circ } = {60^ \circ } \times \dfrac{\theta }{{{{180}^ \circ }}} = \dfrac{\pi }{3}\]



Note:
Students often confused with the formula \[\sin \left( {{{90}^\circ } + \theta } \right) = \cos \theta \] and \[\sin \left( {{{90}^\circ } + \theta } \right) = - \cos \theta \]. The correct formula is \[\sin \left( {{{90}^\circ } + \theta } \right) = \cos \theta \].