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Question
AnswersFind the value of \[\sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}\]
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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.
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.