Question

The value of $\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)\cos \left( {\dfrac{{16\pi }}{{15}}} \right)$ ?
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{8}$
D. $\dfrac{1}{{16}}$

Answer 1

Hint:The given question requires evaluating the product of multiple cosines of different angles that are in geometric progression with each other. The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\cos (\pi + x) = - \cos x$ and $\sin 2x = 2\sin x\cos x$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step by step answer:
In the given problem, we have, $\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)\cos \left( {\dfrac{{16\pi }}{{15}}} \right)$.
So, we have to simplify the expression using the trigonometric formulae.
We know the trigonometric formula $\cos (\pi + x) = - \cos x$. Using the same in the problem, we get,
$ \Rightarrow \cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)\cos \left( {\pi + \dfrac{\pi }{{15}}} \right)$
$ \Rightarrow \left[ { - \cos \left( {\dfrac{\pi }{{15}}} \right)} \right]\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)$
Now, we multiply the numerator and denominator by $2\sin \left( {\dfrac{\pi }{{15}}} \right)$.
\[ \Rightarrow \dfrac{{\left[ { - 2\sin \left( {\dfrac{\pi }{{15}}} \right)\cos \left( {\dfrac{\pi }{{15}}} \right)} \right]\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{2\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]

Using the trigonometric formulae $\sin 2x = 2\sin x\cos x$, we get,
\[ \Rightarrow \dfrac{{\left[ { - \sin \left( {\dfrac{{2\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{2\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
Now, we multiply the numerator and denominator by $2$.
\[ \Rightarrow \dfrac{{\left[ { - 2\sin \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{2\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{2 \times 2\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
Using the trigonometric formulae $\sin 2x = 2\sin x\cos x$, we get,
\[ \Rightarrow \dfrac{{\left[ { - \sin \left( {\dfrac{{4\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{4\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]

So, we multiply the numerator and denominator by $2$.
\[ \Rightarrow \dfrac{{\left[ { - 2\sin \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{2 \times 4\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
Using the trigonometric formulae $\sin 2x = 2\sin x\cos x$, we get,
\[ \Rightarrow \dfrac{{\left[ { - \sin \left( {\dfrac{{8\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{8\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
So, we multiply the numerator and denominator by $2$.
\[ \Rightarrow \dfrac{{\left[ { - 2\sin \left( {\dfrac{{8\pi }}{{15}}} \right)} \right]\cos \left( {\dfrac{{8\pi }}{{15}}} \right)}}{{2 \times 8\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
Using the trigonometric formulae $\sin 2x = 2\sin x\cos x$, we get,
\[ \Rightarrow \dfrac{{\left[ { - \sin \left( {\dfrac{{16\pi }}{{15}}} \right)} \right]}}{{16\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]

Now, we know the trigonometric formula $\sin \left( {\pi + \theta } \right) = - \sin \theta $. So, we get,
\[ \Rightarrow \dfrac{{\left[ { - \sin \left( {\pi + \dfrac{\pi }{{15}}} \right)} \right]}}{{16\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{\pi }{{15}}} \right)}}{{16\sin \left( {\dfrac{\pi }{{15}}} \right)}}\]
Cancelling the common factors in numerator and denominator,
\[ \Rightarrow \dfrac{1}{{16}}\]
So, the value of $\cos \left( {\dfrac{{2\pi }}{{15}}} \right)\cos \left( {\dfrac{{4\pi }}{{15}}} \right)\cos \left( {\dfrac{{8\pi }}{{15}}} \right)\cos \left( {\dfrac{{16\pi }}{{15}}} \right)$ is \[\dfrac{1}{{16}}\].

Hence, option D is the correct answer.

Additional information: Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths. There are $6$trigonometric functions, namely: $\sin (x)$,$\cos (x)$,$\tan (x)$,$\cos ec(x)$,$\sec (x)$and \[\cot \left( x \right)\] . Also, $\cos ec(x)$ ,$\sec (x)$and \[\cot \left( x \right)\]are the reciprocals of $\sin (x)$,$\cos (x)$and$\tan (x)$ respectively.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $\sin 2x = 2\sin x\cos x$ and $\cos \left( {x + \pi } \right) = - \cos x$. We can convert any given trigonometric function using the trigonometric identities and formulae. One should know the double angle formula of sine to tackle the problem. We should take care of the calculations while solving such questions.