Question

Find the value of \[\sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}\]

Answer 1

Hint: Try to break the angles inside $\sin $ in order to use its identities to solve.

Given: \[\sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}{\text{ }} \ldots \left( 1 \right)\]
Now, \[\dfrac{\pi }{{18}} = \dfrac{{{{180}^ \circ }}}{{18}} = {10^ \circ }\]
Putting the value of \[\dfrac{\pi }{{18}} = {10^0}\]in equation $\left( 1 \right)$, we get
\[
   \Rightarrow \sin \left( {{{10}^ \circ }} \right)\sin \left( {5 \times {{10}^ \circ }} \right)\sin \left( {7 \times {{10}^ \circ }} \right) \\
   \Rightarrow \sin \left( {{{10}^ \circ }} \right)\sin \left( {{{50}^ \circ }} \right)\sin \left( {{{70}^ \circ }} \right){\text{ }} \ldots \left( 2 \right) \\
\]
We know that \[{\text{2sinA}}{\text{.sinB = cos}}\left( {{\text{A - B}}} \right){\text{ - cos}}\left( {{\text{A + B}}} \right)\]
\[ \Rightarrow {\text{sinA}}{\text{.sinB}} = \dfrac{1}{2}\left[ {{\text{cos}}\left( {{\text{A - B}}} \right){\text{ - cos}}\left( {{\text{A + B}}} \right)} \right]\]
Now, comparing this identity with equation$\left( 2 \right)$, we get \[{\text{A = 7}}{{\text{0}}^0}{\text{ & B = 5}}{{\text{0}}^0}\].
Hence, substituting this in equation$\left( 2 \right)$, we get:
\[
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^ \circ }} \right)\left[ {\cos \left( {{{70}^ \circ } - {{50}^ \circ }} \right) - \cos \left( {{{70}^ \circ } + {{50}^ \circ }} \right)} \right] \\
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^ \circ }} \right)\left[ {\cos \left( {{{20}^ \circ }} \right) - \cos \left( {{{120}^ \circ }} \right)} \right] \\
\]
We know, \[\cos {120^ \circ } = - \dfrac{1}{2}\],
\[\therefore \dfrac{1}{2}\sin \left( {{{10}^ \circ }} \right)\left[ {\cos \left( {2 \times {{10}^ \circ }} \right) + \dfrac{1}{2}} \right]{\text{ }} \ldots \left( 3 \right)\]
Also, we know, \[{\text{cos2A = 1}} - {\text{2si}}{{\text{n}}^2}{\text{A}}\].
Comparing \[{\text{cos2A}}\] with \[\cos \left( {2 \times {{10}^ \circ }} \right)\] from equation $\left( 3 \right)$, we get ${\text{A}} = {10^0}$
Hence, using this identity in equation$\left( 3 \right)$, we get
\[
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^0}} \right)\left[ {1 - 2{{\sin }^2}{{10}^0} - \cos \left( {{{90}^0} + {{30}^0}} \right)} \right] \\
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^0}} \right)\left[ {1 - 2{{\sin }^2}{{10}^0} + \dfrac{1}{2}} \right]{\text{ }}\left\{ {\because \cos {{120}^0} = - \dfrac{1}{2}} \right\} \\
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^0}} \right)\left[ {\dfrac{3}{2} - 2{{\sin }^2}{{10}^0}} \right] \\
   \Rightarrow \dfrac{1}{2}\sin \left( {{{10}^0}} \right)\left[ {\dfrac{{3 - 4{{\sin }^2}{{10}^0}}}{2}} \right] \\
   \Rightarrow \dfrac{{3\sin {{10}^0} - 4{{\sin }^3}{{10}^0}}}{2}{\text{ }} \ldots \left( 4 \right) \\
\]
Now, we know that \[{\text{3sin}}\theta - {\text{4si}}{{\text{n}}^3}\theta = \sin 3\theta \].
Using this identity in equation$\left( 4 \right)$, we get:
\[
   \Rightarrow \dfrac{{3\sin {{10}^0} - {\text{4si}}{{\text{n}}^3}{{10}^0}}}{2} \\
   \Rightarrow \dfrac{{\sin 3 \times {{10}^0}}}{2} \\
   \Rightarrow \dfrac{{\sin {{30}^0}}}{2} \\
   \Rightarrow \dfrac{1}{4}{\text{ }}\left\{ {\because \sin {{30}^0} = \dfrac{1}{2}} \right\} \\
\]

Note- Whenever you see complicated trigonometric terms together, always try to break them by using trigonometric relations and formulas and try to reduce the power and find the relations between them.