Question

The value of the expression $\cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {179^ \circ }$ is
(A) $0$
(B) $1$
(C) $\dfrac{1}{{\sqrt 2 }}$
(D) $ - 1$

Answer 1

Hint: In the given question, we are provided with a trigonometric expression whose value we have to find. The expression given to us involves the product of sines of all the angles from ${1^ \circ }$ to ${179^ \circ }$. We will have to expand the product a bit more by writing a few more terms in the middle of the expression so as to solve the problem.

Complete answer:
So, we have, $\cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {179^ \circ }$
Now, we are given the above trigonometric expression involving the product of sines of all angles from ${1^ \circ }$ to ${179^ \circ }$.
Now, to solve the trigonometric expression given to us in the question, we will expand a few more terms in the end of expression.
So, we have, \[\cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {177^ \circ }\cos {178^ \circ }\cos {179^ \circ }\]
Now, we know the trigonometric formula $\cos \left( {{{180}^ \circ } - \theta } \right) = - \cos \theta $. Hence, we get,
\[ \Rightarrow \cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {\left( {180 - 3} \right)^ \circ }\cos {\left( {180 - 2} \right)^ \circ }\cos {\left( {180 - 1} \right)^ \circ }\]
\[ \Rightarrow \cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\left[ { - \cos {3^ \circ }} \right]\left[ { - \cos {2^ \circ }} \right]\left[ { - \cos {1^ \circ }} \right]\]
So, we notice that the terms at the end of the expression are the negative of the terms at the beginning of the expression. So, somewhere in the middle of the expression, there would be a term that will not be paired with its negative term.
Now, writing some of the terms in the middle of the expression, we get,
\[ \Rightarrow \cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {89^ \circ }\cos {90^ \circ }\left[ { - \cos {{89}^ \circ }} \right].......\left[ { - \cos {3^ \circ }} \right]\left[ { - \cos {2^ \circ }} \right]\left[ { - \cos {1^ \circ }} \right]\]
So, the term \[\cos {90^ \circ }\] is not paired with any negative term. Also, we know that the value of \[\cos {90^ \circ }\] is zero. Hence, we get the value of product as,
\[ \Rightarrow \cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {89^ \circ }\left( 0 \right)\left[ { - \cos {{89}^ \circ }} \right].......\left[ { - \cos {3^ \circ }} \right]\left[ { - \cos {2^ \circ }} \right]\left[ { - \cos {1^ \circ }} \right]\]
\[ \Rightarrow 0\]
Hence, the value of the expression $\cos {1^ \circ }\cos {2^ \circ }\cos {3^ \circ }.......\cos {179^ \circ }$ is zero.
Hence, option (A) is the correct answer.

Note:
We must know the value of cosine of $90$ degrees in order to solve the given problem. One should remember some basic and simple trigonometric formula so as to tackle such problems. We must do the calculations with utmost care in order to be sure of the final answer. The problem can be solved easily if the terms in the middle of the expression are expanded directly.