Question

If $k = \sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}$, then the numerical value of $k$ is
A) $\dfrac{1}{4}$
B) $\dfrac{1}{8}$
C) $\dfrac{1}{16}$
D) None of the above

Answer 1

We are provided with trigonometric functions and we are asked to find the numerical value by solving those functions. In this trigonometric problem the trigonometric functions are in multiplication. In order to find the numerical value of $\sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}$ convert them into the terms of degrees. In trigonometry, pi ($\pi$) means ${180^0}$ to represent the angle in radians. Therefore, $\sin \pi = {180^0}$.

Trigonometric values to be remembered:
$\sin {0^0} = 0$, $\sin {30^0} = \dfrac{1}{2}$, $\sin {45^0} = \dfrac{1}{{\sqrt 2 }}$, $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ and $\sin {90^0} = 1$.
$\cos {0^0} = 1$, $\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$, $\cos {45^0} = \dfrac{1}{{\sqrt 2 }}$, $\cos {60^0} = \dfrac{1}{2}$ and $\cos {90^0} = 0$.

Complete step by step solution:
We are given the problem,
$k = \sin \dfrac{\pi }{{18}}\sin \dfrac{{5\pi }}{{18}}\sin \dfrac{{7\pi }}{{18}}$
We already mentioned that $\sin \pi = {180^0}$, by substituting this,
$= \sin \dfrac{{{{180}^0}}}{{18}}\sin \dfrac{{5({{180}^0})}}{{18}}\sin \dfrac{{7({{180}^0})}}{{18}}$
By dividing $\dfrac{{180}}{{18}}$ we will get $10$,
$= \sin {10^0}\sin 5({10^0})\sin 7({10^0})$
Multiplying the values inside the bracket,
$= \sin {10^0}\sin {50^0}\sin {70^0}$
For further simplification let us multiply and divide the first two terms by $2$,
$= \dfrac{1}{2}\left[ {2(\sin {{10}^0}\sin {{50}^0})} \right]\sin {70^0}$
We have the trigonometry formula: $2\sin A\sin B = \cos (A - B) - \cos (A + B)$, by applying this formula,
$= \dfrac{1}{2}\left[ {\cos ({{10}^0} - {{50}^0}) - \cos ({{10}^0} + {{50}^0})} \right]\sin {70^0}$
$= \dfrac{1}{2}\left[ {\cos ( - {{40}^0}) - \cos {{60}^0}} \right]\sin {70^0}$
$\cos ( - \theta ) = \cos \theta$, therefore $\cos ( - {40^0}) = \cos {40^0}$,
By applying this,
$= \dfrac{1}{2}\left[ {\cos {{40}^0} - \cos {{60}^0}} \right]\sin {70^0}$
We know that the value of $\cos {60^0} = \dfrac{1}{2}$, by substituting this,
$= \dfrac{1}{2}\left[ {\cos {{40}^0} - \dfrac{1}{2}} \right]\sin {70^0}$
Now multiplying $\sin {70^0}$ inside the bracket,
$= \dfrac{1}{2}\left[ {\cos {{40}^0}\sin {{70}^0} - \dfrac{1}{2}\sin {{70}^0}} \right]$
Multiply and divide $\cos {40^0}\sin {70^0}$ by $2$ for further simplification,
$= \dfrac{1}{2}\left[ {\dfrac{1}{2}\left( {2(\cos {{40}^0}\sin {{70}^0})} \right) - \dfrac{1}{2}\sin {{70}^0}} \right]$
Here $\dfrac{1}{2}$ is common for both the terms inside the bracket, so let's take it as common outside,
$= \dfrac{1}{2} \times \dfrac{1}{2}\left[ {2\cos {{40}^0}\sin {{70}^0} - \sin {{70}^0}} \right]$
We have another formula that, $2\cos A\sin B = \sin (A + B) - \sin (A - B)$, by applying this,
$= \dfrac{1}{4}\left[ {\sin ({{40}^0} + {{70}^0}) - \sin ({{40}^0} - {{70}^0}) - \sin {{70}^0}} \right]$
$= \dfrac{1}{4}\left[ {\sin {{110}^0} - \sin ( - {{30}^0}) - \sin {{70}^0}} \right]$
And the value of $\sin ( - \theta ) = - \sin \theta$, hence $\sin ( - {30^0}) = - \sin ({30^0})$, apply this,
$= \dfrac{1}{4}\left[ {\sin {{110}^0} - ( - \sin {{30}^0}) - \sin {{70}^0}} \right]$
Now the $- \times -$ will become $+$, because $- \times - = +$ then,
$= \dfrac{1}{4}\left[ {\sin {{110}^0} + \sin {{30}^0} - \sin {{70}^0}} \right]$
We know that $\sin {30^0} = \dfrac{1}{2}$,
$= \dfrac{1}{4}\left[ {\sin {{110}^0} + \dfrac{1}{2} - \sin {{70}^0}} \right]$
We can also write $\sin (x) = \sin (180 - x)$, therefore, $\sin ({110^0}) = \sin ({180^0} - {70^0}) = \sin ({70^0})$
$= \dfrac{1}{4}\left[ {\sin {{70}^0} + \dfrac{1}{2} - \sin {{70}^0}} \right]$
Now $+ \sin {70^0}$ and $- \sin {70^0}$ will cancel each other,
$= \dfrac{1}{4}\left[ {\dfrac{1}{2}} \right]$
$= \dfrac{1}{8}$
$\therefore k = \dfrac{1}{8}$
Hence, option (B) $\dfrac{1}{8}$ is correct.

Note:
One should study all the formulas and values based on trigonometry thoroughly. Then only one can get any idea about how to solve these types of questions. Here in two steps we multiply and divide the terms by $2$. Remember that any number multiplying and dividing by the same number will not make any changes in the value, but still we follow this step to provide some clue and thus to carry forward the problems. Thus we change the values in the form of formulas and solve them easily.