Question

The value of $\sin \left( \dfrac{\pi }{16} \right)\sin \left( \dfrac{3\pi }{16} \right)\sin \left( \dfrac{5\pi }{16} \right)\sin \left( \dfrac{7\pi }{16} \right)$is:
A.$\dfrac{1}{16}$
B.$\dfrac{\sqrt{2}}{16}$
C.$\dfrac{1}{8}$
D.$\dfrac{\sqrt{2}}{8}$

Answer 1

Hint : In this particular problem first of all we have to divide and multiply by 4 to this equation then separate the term in the form of formula after that we have to use the formula to this equation. Formula like$2\sin \left( A \right)\sin \left( B \right)=\cos \left( A-B \right)-\cos \left( A+B \right)$. So, in this way we have to solve further and get the values.

Complete step-by-step answer :
According to the question, it is given that $\sin \left( \dfrac{\pi }{16} \right)\sin \left( \dfrac{3\pi }{16} \right)\sin \left( \dfrac{5\pi }{16} \right)\sin \left( \dfrac{7\pi }{16} \right)$
So, we have to find the values.
Before applying directly applying formula,
We need to make this above equation in the form of formula
To make it in the form of a formula , first we need to multiply and divide by 4 n this above equation.
That means,
$=\dfrac{1}{4}\times 4\times \sin \left( \dfrac{\pi }{16} \right)\sin \left( \dfrac{3\pi }{16} \right)\sin \left( \dfrac{5\pi }{16} \right)\sin \left( \dfrac{7\pi }{16} \right)$
From the above equation we grouped the product of first and last term and second and third term then simplifying further we get:
$=\dfrac{1}{4}\times \left[ 2\times \sin \left( \dfrac{7\pi }{16} \right)\sin \left( \dfrac{\pi }{16} \right) \right]\left[ 2\times \sin \left( \dfrac{5\pi }{16} \right)\sin \left( \dfrac{3\pi }{16} \right) \right]$
So, now it is in the form of formula,
We need to apply the formula in the above equation we get:
That is, $2\sin \left( A \right)\sin \left( B \right)=\cos \left( A-B \right)-\cos \left( A+B \right)$
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{7\pi }{16}-\dfrac{\pi }{16} \right)-\cos \left( \dfrac{7\pi }{16}+\dfrac{\pi }{16} \right) \right]\left[ \cos \left( \dfrac{5\pi }{16}-\dfrac{3\pi }{16} \right)-\cos \left( \dfrac{5\pi }{16}+\dfrac{3\pi }{16} \right) \right]$
By simplifying and solving this above equation we get:
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{6\pi }{16} \right)-\cos \left( \dfrac{8\pi }{16} \right) \right]\left[ \cos \left( \dfrac{2\pi }{16} \right)-\cos \left( \dfrac{8\pi }{16} \right) \right]$
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{6\pi }{16} \right)-\cos \left( \dfrac{\pi }{2} \right) \right]\left[ \cos \left( \dfrac{2\pi }{16} \right)-\cos \left( \dfrac{\pi }{2} \right) \right]$
As you can notice in this equation that $\cos \left( \dfrac{\pi }{2} \right)=0$
Substitute in above equation we get:
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{6\pi }{16} \right)-0 \right]\left[ \cos \left( \dfrac{2\pi }{16} \right)-0 \right]$
Hence, we get this
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{6\pi }{16} \right) \right]\left[ \cos \left( \dfrac{2\pi }{16} \right) \right]$
After simplifying we get:
$=\dfrac{1}{4}\times \left[ \cos \left( \dfrac{3\pi }{8} \right) \right]\left[ \cos \left( \dfrac{\pi }{8} \right) \right]$
So, again we have to repeat the same procedure
But here we have to multiply and divide by 2 on above equation we get
$=\dfrac{1}{4}\times \dfrac{1}{2}\times \left[ 2\times \cos \left( \dfrac{3\pi }{8} \right)\cos \left( \dfrac{\pi }{8} \right) \right]$
as $\dfrac{\pi}{2} - \dfrac{3\pi }{8} = \dfrac{\pi }{8}$ and $\dfrac{\pi}{2} - \dfrac{3\pi }{8} = \dfrac{\pi }{8}$
$=\dfrac{1}{4}\times \dfrac{1}{2}\times \left[ 2\times \sin \left( \dfrac{3\pi }{8} \right)\sin \left( \dfrac{\pi }{8} \right) \right]$
We need to apply the formula in the above equation we get:
That is, $2\sin \left( A \right)\sin \left( B \right)=\cos \left( A-B \right)-\cos \left( A+B \right)$
And after simplifying we get:
$=\dfrac{1}{8}\times \left[ \cos \left( \dfrac{3\pi }{8}-\dfrac{\pi }{8} \right)-\cos \left( \dfrac{3\pi }{8}+\dfrac{\pi }{8} \right) \right]$
After simplifying this we get:
$=\dfrac{1}{8}\times \left[ \cos \left( \dfrac{2\pi }{8} \right)-\cos \left( \dfrac{4\pi }{8} \right) \right]$
Further solving we get:
$=\dfrac{1}{8}\times \left[ \cos \left( \dfrac{\pi }{4} \right)-\cos \left( \dfrac{\pi }{2} \right) \right]$
As we know here that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$and $\cos \left( \dfrac{\pi }{2} \right)=0$substitute this value in above equation we get:
$=\dfrac{1}{8}\times \left[ \dfrac{1}{\sqrt{2}}-0 \right]$
Hence, it becomes
$=\dfrac{1}{8\sqrt{2}}$
But to get the desired option in the question we have to multiply by $\sqrt{2}$ on numerator as well as denominator we get:
$=\dfrac{\sqrt{2}}{8\sqrt{2}\times \sqrt{2}}$
After simplifying this we get:
$=\dfrac{\sqrt{2}}{16}$
Therefore, $\sin \left( \dfrac{\pi }{16} \right)\sin \left( \dfrac{3\pi }{16} \right)\sin \left( \dfrac{5\pi }{16} \right)\sin \left( \dfrac{7\pi }{16} \right)=\dfrac{\sqrt{2}}{16}$
So, the correct answer is “Option B”.

Note : In this particular problem keep in mind while solving first of all we need to multiply and divide by 4 on this equation to always apply the formula correctly. In case after simplification, if we want to split the angle to get the exact value of the trigonometric function, we must be aware of the sign’s in all four quadrants. That is in first quadrant all the functions are positive in second quadrant only sine and cosecant is positive remaining are negative, in third quadrant only tangent and cotangent function are positive and remaining are negative and in fourth quadrant only cosine and secant functions are positive remaining are negative.