Question

Find the numerical value of k if $k=\sin \dfrac{\pi }{18}\sin \dfrac{5\pi }{18}\sin \dfrac{7\pi }{18}$ .
(a) $\dfrac{1}{2}$
(b) $\dfrac{1}{4}$
(c) $\dfrac{1}{8}$
(d) $\dfrac{1}{18}$

Answer 1

Hint: First of all, we will convert all the angles in radian into degrees by using the formula, $\text{degree}=\dfrac{180}{\pi }\times \text{radian}$. Then, by using the formulas of trigonometry which can be given as, $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ and $2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)$, we will solve the problem and find the value of k.

Complete step-by-step solution -
In question we are given that, $k=\sin \dfrac{\pi }{18}\sin \dfrac{5\pi }{18}\sin \dfrac{7\pi }{18}$, and we are asked to find the value of k, so first of all we will convert radian into degree by using the formula,
$\text{degree}=\dfrac{180}{\pi }\times \text{radian}$
So, the radian to degree conversion for $\dfrac{\pi }{18}$, $\dfrac{5\pi }{18}$ and $\dfrac{7\pi }{18}$, can be given as,
$\dfrac{\pi }{18}=\dfrac{180}{\pi }\times \dfrac{\pi }{18}={{10}^{\circ }}$
$\dfrac{5\pi }{18}=\dfrac{180}{\pi }\times \dfrac{5\pi }{18}={{50}^{\circ }}$
$\dfrac{7\pi }{18}=\dfrac{180}{\pi }\times \dfrac{7\pi }{18}={{70}^{\circ }}$
Substituting these values in expression, we will get,
$k=\sin {{10}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}$ …………..(i)
Now, from expression we can see that two sines are in multiplication so to simplify further we can apply the formula of trigonometry which can be given as,
$2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ ……………(ii)
Now, on comparing expression (ii) with expression (i), we can say that $\alpha ={{10}^{\circ }}$ and $\beta ={{50}^{\circ }}$, on substituting these values in expression (ii) we will get,
$2\sin 10\sin 50=\cos \left( 10-50 \right)-\cos \left( 10+50 \right)$ …………….(iii)
Now, first of all we will multiply and divide 2 in expression (i), to bring it in the form of expression (ii), so, on multiplying and dividing expression (i) with 2, we will get,
$k=\dfrac{2}{2}\sin {{10}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}\Rightarrow k=\dfrac{1}{2}\left( 2\sin {{10}^{\circ }}\sin {{50}^{\circ }} \right)\sin {{70}^{\circ }}$
Now, on substituting the values of expression (iii) in expression (i) we will get,
$k=\dfrac{1}{2}\left( \cos \left( 10-50 \right)-\cos \left( 10+50 \right) \right)\sin {{70}^{\circ }}$
$\Rightarrow k=\dfrac{1}{2}\left( \cos \left( -40 \right)-\cos \left( 60 \right) \right)\sin {{70}^{\circ }}$
Now, we know that cos is even function so, $\cos \left( -40 \right)=\cos {{40}^{\circ }}$ and value of $\cos \left( 60 \right)=\dfrac{1}{2}$, so on substituting these values in above expression we will get,
$\Rightarrow k=\dfrac{1}{2}\left( \cos \left( 40 \right)-\dfrac{1}{2} \right)\sin {{70}^{\circ }}$
On further the expression simplifying we will get,
$\Rightarrow k=\dfrac{1}{2}\cos {{40}^{\circ }}\sin {{70}^{\circ }}-\dfrac{1}{4}\sin {{70}^{\circ }}$ ………………..(iv)
Now, on looking at the expression (iv), we can say that cos and sin are in multiplication, so, we can use the trigonometric formula which can be given as,
$2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)$ …………….(v)
Now, again doing same process as above, i.e. on comparing the expression (iv) with expression (v), we can say that $\alpha ={{40}^{\circ }}$ and $\beta ={{70}^{\circ }}$ , on substituting these values in expression (v) we will get,
$2\cos 40\sin 70=\sin \left( 40+70 \right)-\sin \left( 40-70 \right)$ ………………..(vi)
Now, again we will multiply and divide expression (iv) with 2 and then on substituting the value of expression (vi) we will get,
$k=\dfrac{1}{2}\left( \dfrac{2}{2}\cos {{40}^{\circ }}\sin {{70}^{\circ }} \right)-\dfrac{1}{2}\sin {{70}^{\circ }}\Rightarrow k=\dfrac{1}{4}\left( 2\cos {{40}^{\circ }}\sin {{70}^{\circ }} \right)-\dfrac{1}{4}\sin {{70}^{\circ }}$
$\Rightarrow k=\dfrac{1}{4}\left( \sin \left( 110 \right)-\sin \left( -30 \right) \right)-\dfrac{1}{4}\sin {{70}^{\circ }}$
Now, as sin is odd function the value of $\sin \left( -30 \right)=-\sin {{30}^{\circ }}$ and the value of $\sin \left( 30 \right)=\dfrac{1}{2}$, so on substituting these values in above expression we will get,
$\Rightarrow k=\dfrac{1}{4}\left( \sin {{110}^{\circ }}+\sin {{30}^{\circ }} \right)-\dfrac{1}{+}\sin {{70}^{\circ }}$
$\Rightarrow k=\dfrac{1}{4}\sin {{110}^{\circ }}+\dfrac{1}{4}\times \dfrac{1}{2}-\dfrac{1}{4}\sin {{70}^{\circ }}$
Now, we know that $\sin 110=\sin \left( 180-70 \right)=\sin 70$, so on substituting this in expression above we will get,
$\Rightarrow k=\dfrac{1}{4}\sin {{70}^{\circ }}+\dfrac{1}{4}\times \dfrac{1}{2}-\dfrac{1}{4}\sin {{70}^{\circ }}\Rightarrow k=\dfrac{1}{4}\times \dfrac{1}{2}=\dfrac{1}{8}$
Hence, the value of k is $\dfrac{1}{8}$.
Thus, option (c) is the correct answer.

Note: Student might use the formula of $2\cos \alpha \sin \beta$, as $2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right)$ instead of $2\cos \alpha \sin \beta =\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)$ and due to that the value will become, $2\cos \alpha \sin \beta =\sin \left( 40+70 \right)+\sin \left( 40-70 \right)=\sin {{110}^{\circ }}-\sin {{30}^{\circ }}$. When we substitute these value in expression (vi) we will get, $k=\dfrac{1}{4}\left( \sin \left( 110 \right)+\sin \left( -30 \right) \right)-\dfrac{1}{4}\sin {{70}^{\circ }}=\dfrac{1}{4}\sin {{110}^{\circ }}-\dfrac{1}{4}\times \dfrac{1}{2}-\dfrac{1}{4}\sin {{70}^{\circ }}$
And by simplifying it further we will get, $k=-\dfrac{1}{8}$, which is completely opposite to our answer and it is incorrect also. So, students must be careful while using the formulas.