Question

The value of $\sin 10^\circ \,\sin 30^\circ \,\sin 50^\circ \,\sin 70^\circ $is:
A) $\dfrac{1}{{36}}$
B) $\dfrac{1}{{32}}$
C) $\dfrac{1}{{18}}$
D) $\dfrac{1}{{16}}$

Answer 1

Hint:
In this question, we will try to use this formula sin2A = 2sinAcosA. But before we use this formula we need to multiply and divide it by $2\cos 10^\circ $ so that we can use the formula. Now at the place of $2\sin 10^\circ \cos 10^\circ $ we can write $\sin 20^\circ $. Then we put the value of $\sin 30^\circ $ that is $\dfrac{1}{2}$. Let’s see how we solve it further.

Complete step by step solution:
$\sin 10^\circ \,\sin 30^\circ \,\sin 50^\circ \,\sin 70^\circ $
Here, we will multiply and divide it by $2\cos 10^\circ $.
$\dfrac{{2\cos 10^\circ \sin 10^\circ \,\sin 30^\circ \,\sin 50^\circ \,\sin 70^\circ }}{{2\cos 10^\circ }}$
Here, we can use the formula sin2A = 2sinAcosA on $2\cos 10^\circ \sin 10^\circ $$ = \sin 20^\circ $
$\dfrac{{\sin 20^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ }}{{2\cos 10^\circ }}$
$ \Rightarrow \dfrac{{\sin 20^\circ \sin 50^\circ \sin 70^\circ }}{{4\cos 10^\circ }}$(the value of $\sin 30^\circ = \dfrac{1}{2}$)
$ \Rightarrow \dfrac{{\sin 20^\circ \sin 50^\circ \sin (90^\circ - 20^\circ )}}{{4\cos 10^\circ }}$(we know that $\sin (90^\circ - \theta ) = \cos \theta $)
$ \Rightarrow \dfrac{{\sin 20^\circ \sin 50^\circ \cos 20^\circ }}{{4\cos 10^\circ }}$
Multiplying with 2 in numerator and denominator.
$ \Rightarrow \dfrac{{2\sin 20^\circ \cos 20^\circ \sin 50^\circ }}{{8\cos 10^\circ }}$
We can see that after multiplying with 2, we can use this formula sin2A = 2sinAcosA on $2\sin 20^\circ \cos 20^\circ = \sin 40^\circ $
$ \Rightarrow \dfrac{{\sin 40^\circ \sin 50^\circ }}{{8\cos 10^\circ }}$
$ \Rightarrow \dfrac{{\sin 40^\circ \sin (90^\circ - 50^\circ )}}{{8\cos 10^\circ }}$(We know that $\sin (90^\circ - \theta ) = \cos \theta $)
$ \Rightarrow \dfrac{{\sin 40^\circ \cos 40}}{{8\cos 10^\circ }}$
Again, multiplying numerator and denominator with 2.
$ \Rightarrow \dfrac{{2\sin 40^\circ \cos 40^\circ }}{{16\cos 10^\circ }}$.
We can see that after multiplying with 2, we can use this formula sin2A = 2sinAcosA on $2\sin 40^\circ \cos 40^\circ = \sin 80^\circ $
$ \Rightarrow \dfrac{{\sin 80^\circ }}{{16\cos 10^\circ }}$
$ \Rightarrow \dfrac{{\sin (90^\circ - 10^\circ )}}{{16\cos 10^\circ }}$(We know that $\sin (90^\circ - \theta ) = \cos \theta $)
$ \Rightarrow \dfrac{{\cos 10^\circ }}{{16\cos 10^\circ }}$= $\dfrac{1}{{16}}$.
Thus, the value of $\sin 10^\circ \,\sin 30^\circ \,\sin 50^\circ \,\sin 70^\circ $is $\dfrac{1}{{16}}$.

Hence, the option D is the correct option.

Note:
Note: Students should appropriately use the formula. To convert $\sin 70^\circ $into $\cos 20^\circ $ by using formula $\sin (90^\circ - \theta ) = \cos \theta $. Then again, we multiplied numerator and denominator by 2 because we needed to use this formula sin2A = 2sinAcosA. You should be careful while solving this question.