Find the value of $ m $ if $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} = \dfrac{1}{m} $ find $ m $ .
Answer
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Hint: Let us say there is an angle A. Then sine of angle A will be cosine of $ \left( {{{90}^ \circ } - A} \right) $ . 90 degrees can also be written as $ \dfrac{\pi }{2} $ radians. Use this info and below mentioned formulas to find the value of LHS of the given trigonometric equation. And then equate the obtained value with RHS to find the value of m.
Formulas used:
1) $ \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) $
2) $ \sin \left( {\pi - \theta } \right) = \sin \theta $
3) $ \cos \left( { - \theta } \right) = \cos \theta $
4) $ 2\sin \theta \cos \theta = \sin \left( {2\theta } \right) $
Complete step-by-step answer:
The given trigonometric equation is $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} = \dfrac{1}{m} $
Considering the LHS:
$ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $
$ \sin \dfrac{{7\pi }}{{14}} $ is equal to $ \sin \dfrac{\pi }{2} $ and the value of $ \sin \dfrac{\pi }{2} $ is 1.
On substituting the above value in the given equation, we get
$
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}} \times \left( 1 \right) \times \sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} \\
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} \\
$
And we already know that $ \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) $
Therefore, we can write sine functions in terms of cosine functions.
$
\sin \dfrac{{3\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{{14}}} \right) = \cos \dfrac{{2\pi }}{7} \\
\sin \dfrac{{5\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{5\pi }}{{14}}} \right) = \cos \dfrac{\pi }{7} \\
\sin \dfrac{{9\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{9\pi }}{{14}}} \right) = \cos \dfrac{{ - \pi }}{7} = \cos \dfrac{\pi }{7} \;
\sin \dfrac{{11\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{11\pi }}{{14}}} \right) = \cos \dfrac{{ - 2\pi }}{7} = \cos \dfrac{{2\pi }}{7} \\
\sin \dfrac{{13\pi }}{{14}} = \sin \left( {\pi - \dfrac{{13\pi }}{{14}}} \right) = \sin \dfrac{\pi }{{14}} \;
$
On substituting the above obtained values in $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $ , we get
$
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} = \sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\sin \dfrac{\pi }{{14}} \\
\Rightarrow {\left( {\sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} \;
$
To get a form $ 2\sin \theta \cos \theta = \sin \left( {2\theta } \right) $ , we have to multiply and divide with $ \cos \theta $
So, first multiply and divide with $ 2\cos \dfrac{\pi }{{14}} $
$
\Rightarrow {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\cos \dfrac{\pi }{{14}}\sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} \\
2\sin \theta \cos \theta = \sin 2\theta\Rightarrow 2\cos \dfrac{\pi }{{14}}\sin \dfrac{\pi }{{14}} = \sin \left( {2 \times \dfrac{\pi }{{14}}} \right) = \sin \dfrac{\pi }{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} = {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
$
Multiply and divide with 2.
$
\Rightarrow {\left( {\dfrac{1}{{\left( {2 \times 2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7} = \sin \left( {2 \times \dfrac{\pi }{7}} \right) = \sin \dfrac{{2\pi }}{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {4\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
$
Multiply and divide with 2 again.
$
\Rightarrow {\left( {\dfrac{1}{{\left( {4 \times 2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
2\sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7} = \sin \left( {2 \times \dfrac{{2\pi }}{7}} \right) = \sin \dfrac{{4\pi }}{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {8\cos \dfrac{\pi }{{14}}} \right)}}\sin \dfrac{{4\pi }}{7}} \right)^2} \;
$
$ \sin \dfrac{{4\pi }}{7} $ can be written as $ \cos \left( {\dfrac{\pi }{2} - \dfrac{{4\pi }}{7}} \right) = \cos \dfrac{{ - \pi }}{{14}} = \cos \dfrac{\pi }{{14}} $
Therefore, substituting the obtained value of $ \sin \dfrac{{4\pi }}{7} $ , we get
$
\left( {\dfrac{1}{{8\cos \left( {\dfrac{\pi }{{14}}} \right)}} \times \cos \dfrac{\pi }{{14}}} \right) \\
\Rightarrow {\left( {\dfrac{1}{8}} \right)^2} = \dfrac{1}{{64}} \;
$
Therefore, the value of $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $ is $ \dfrac{1}{{64}} $
But the given value is $ \dfrac{1}{m} $ . Therefore, on equating both, we get
$
\dfrac{1}{m} = \dfrac{1}{{64}} \\
\Rightarrow m = 64 \;
$
Hence, the value of m obtained is 64.
Note: The values of sine function and cosine function repeat after 360 degrees or 2pi radians whereas the values of tan function repeats after pi radians. Cosine function of negative angle can be written as positive cosine but sine function of negative angle is negative sine. Be careful with this. Do not confuse the formula of $ \sin \left( {2\theta } \right) $ with the formula of $ \cos \left( {2\theta } \right) $ , which is $ {\cos ^2}\theta - {\sin ^2}\theta $ .
Formulas used:
1) $ \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) $
2) $ \sin \left( {\pi - \theta } \right) = \sin \theta $
3) $ \cos \left( { - \theta } \right) = \cos \theta $
4) $ 2\sin \theta \cos \theta = \sin \left( {2\theta } \right) $
Complete step-by-step answer:
The given trigonometric equation is $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} = \dfrac{1}{m} $
Considering the LHS:
$ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $
$ \sin \dfrac{{7\pi }}{{14}} $ is equal to $ \sin \dfrac{\pi }{2} $ and the value of $ \sin \dfrac{\pi }{2} $ is 1.
On substituting the above value in the given equation, we get
$
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}} \times \left( 1 \right) \times \sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} \\
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} \\
$
And we already know that $ \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) $
Therefore, we can write sine functions in terms of cosine functions.
$
\sin \dfrac{{3\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{{14}}} \right) = \cos \dfrac{{2\pi }}{7} \\
\sin \dfrac{{5\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{5\pi }}{{14}}} \right) = \cos \dfrac{\pi }{7} \\
\sin \dfrac{{9\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{9\pi }}{{14}}} \right) = \cos \dfrac{{ - \pi }}{7} = \cos \dfrac{\pi }{7} \;
\sin \dfrac{{11\pi }}{{14}} = \cos \left( {\dfrac{\pi }{2} - \dfrac{{11\pi }}{{14}}} \right) = \cos \dfrac{{ - 2\pi }}{7} = \cos \dfrac{{2\pi }}{7} \\
\sin \dfrac{{13\pi }}{{14}} = \sin \left( {\pi - \dfrac{{13\pi }}{{14}}} \right) = \sin \dfrac{\pi }{{14}} \;
$
On substituting the above obtained values in $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $ , we get
$
\Rightarrow \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} = \sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\sin \dfrac{\pi }{{14}} \\
\Rightarrow {\left( {\sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} \;
$
To get a form $ 2\sin \theta \cos \theta = \sin \left( {2\theta } \right) $ , we have to multiply and divide with $ \cos \theta $
So, first multiply and divide with $ 2\cos \dfrac{\pi }{{14}} $
$
\Rightarrow {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\cos \dfrac{\pi }{{14}}\sin \dfrac{\pi }{{14}}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} \\
2\sin \theta \cos \theta = \sin 2\theta\Rightarrow 2\cos \dfrac{\pi }{{14}}\sin \dfrac{\pi }{{14}} = \sin \left( {2 \times \dfrac{\pi }{{14}}} \right) = \sin \dfrac{\pi }{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}\cos \dfrac{\pi }{7}} \right)^2} = {\left( {\dfrac{1}{{\left( {2\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
$
Multiply and divide with 2.
$
\Rightarrow {\left( {\dfrac{1}{{\left( {2 \times 2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7} = \sin \left( {2 \times \dfrac{\pi }{7}} \right) = \sin \dfrac{{2\pi }}{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {4\cos \dfrac{\pi }{{14}}} \right)}} \times \sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
$
Multiply and divide with 2 again.
$
\Rightarrow {\left( {\dfrac{1}{{\left( {4 \times 2\cos \dfrac{\pi }{{14}}} \right)}} \times 2\sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7}} \right)^2} \\
2\sin \dfrac{{2\pi }}{7}\cos \dfrac{{2\pi }}{7} = \sin \left( {2 \times \dfrac{{2\pi }}{7}} \right) = \sin \dfrac{{4\pi }}{7} \\
\Rightarrow {\left( {\dfrac{1}{{\left( {8\cos \dfrac{\pi }{{14}}} \right)}}\sin \dfrac{{4\pi }}{7}} \right)^2} \;
$
$ \sin \dfrac{{4\pi }}{7} $ can be written as $ \cos \left( {\dfrac{\pi }{2} - \dfrac{{4\pi }}{7}} \right) = \cos \dfrac{{ - \pi }}{{14}} = \cos \dfrac{\pi }{{14}} $
Therefore, substituting the obtained value of $ \sin \dfrac{{4\pi }}{7} $ , we get
$
\left( {\dfrac{1}{{8\cos \left( {\dfrac{\pi }{{14}}} \right)}} \times \cos \dfrac{\pi }{{14}}} \right) \\
\Rightarrow {\left( {\dfrac{1}{8}} \right)^2} = \dfrac{1}{{64}} \;
$
Therefore, the value of $ \sin \dfrac{\pi }{{14}}\sin \dfrac{{3\pi }}{{14}}\sin \dfrac{{5\pi }}{{14}}\sin \dfrac{{7\pi }}{{14}}\sin \dfrac{{9\pi }}{{14}}\sin \dfrac{{11\pi }}{{14}}\sin \dfrac{{13\pi }}{{14}} $ is $ \dfrac{1}{{64}} $
But the given value is $ \dfrac{1}{m} $ . Therefore, on equating both, we get
$
\dfrac{1}{m} = \dfrac{1}{{64}} \\
\Rightarrow m = 64 \;
$
Hence, the value of m obtained is 64.
Note: The values of sine function and cosine function repeat after 360 degrees or 2pi radians whereas the values of tan function repeats after pi radians. Cosine function of negative angle can be written as positive cosine but sine function of negative angle is negative sine. Be careful with this. Do not confuse the formula of $ \sin \left( {2\theta } \right) $ with the formula of $ \cos \left( {2\theta } \right) $ , which is $ {\cos ^2}\theta - {\sin ^2}\theta $ .
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