Question

Prove that $\cos {20^ \circ }\cos {40^ \circ }\cos {80^ \circ } = \dfrac{1}{8}$.

Answer 1

Hint: The given problem requires us to simplify and find the value of the given trigonometric expression. The question requires us to have thorough knowledge of trigonometric functions, formulae and identities. The question describes the wide ranging applications of trigonometric identities and formulae. We must know the double angle formula for sine: $\sin 2x = 2\sin x\cos x$ and that the sine and cosine functions are complementary to each other. Hence, $\sin x = \cos \left( {{{90}^ \circ } - x} \right)$.

Complete step-by-step answer:
In the given question, we are required to prove the value of product $\cos {20^ \circ }\cos {40^ \circ }\cos {80^ \circ }$ using the basic concepts of trigonometry and identities.
So, L.H.S.$ = \cos {20^ \circ }\cos {40^ \circ }\cos {80^ \circ }$
Now, we know that sine and cosine functions are complementary functions. So, we have, $\sin x = \cos \left( {{{90}^ \circ } - x} \right)$.
$ \Rightarrow \cos {20^ \circ }\cos {40^ \circ }\cos \left( {{{90}^ \circ } - {{80}^ \circ }} \right)$
So, using the trigonometric formula $\sin x = \cos \left( {{{90}^ \circ } - x} \right)$, we get,
$ \Rightarrow \cos {20^ \circ }\cos {40^ \circ }\sin {10^ \circ }$
Now, multiplying and dividing the numerator and denominator by $2\cos {10^ \circ }$, we get,
$ \Rightarrow \dfrac{{2\sin {{10}^ \circ }\cos {{10}^ \circ }}}{{2\cos {{10}^ \circ }}} \times \cos {20^ \circ }\cos {40^ \circ }$
Now, using the double angle formula for sine $\sin 2x = 2\sin x\cos x$ in the numerator, we get,
$ \Rightarrow \dfrac{{\sin {{20}^ \circ } \times \cos {{20}^ \circ } \times \cos {{40}^ \circ }}}{{2\cos {{10}^ \circ }}}$
Now, we multiply both the numerator and denominator by \[2\] .
$ \Rightarrow \dfrac{{2 \times \sin {{20}^ \circ } \times \cos {{20}^ \circ } \times \cos {{40}^ \circ }}}{{4\cos {{10}^ \circ }}}$
Now, again using the trigonometric formula $\sin 2x = 2\sin x\cos x$, we get,
$ \Rightarrow \dfrac{{\sin {{40}^ \circ } \times \cos {{40}^ \circ }}}{{4\cos {{10}^ \circ }}}$
Again, we multiply both the numerator and denominator by \[2\].
$ \Rightarrow \dfrac{{2\sin {{40}^ \circ } \times \cos {{40}^ \circ }}}{{8\cos {{10}^ \circ }}}$
So, again condensing he trigonometric expression by using the trigonometric formula $\sin 2x = 2\sin x\cos x$, we get,
\[ \Rightarrow \dfrac{{\sin {{80}^ \circ }}}{{8\cos {{10}^ \circ }}}\]
Now, we know that cosine and sine are complementary functions. So, using the formula $\sin x = \cos \left( {{{90}^ \circ } - x} \right)$, we get,
\[ \Rightarrow \dfrac{{\sin {{80}^ \circ }}}{{8\cos {{\left( {90 - 80} \right)}^ \circ }}}\]
\[ \Rightarrow \dfrac{{\sin {{80}^ \circ }}}{{8\sin {{80}^ \circ }}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{1}{8}\]
Also, R.H.S$ = \dfrac{1}{8}$
As LHS $ = $ RHS.
So, $\cos {20^ \circ }\cos {40^ \circ }\cos {80^ \circ } = \dfrac{1}{8}$.

Note: Basic trigonometric identities include ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$. These identities are of vital importance for solving any question involving trigonometric functions and identities. All the trigonometric ratios can be converted into each other using the simple trigonometric identities listed above. The given problem involves the use of trigonometric formulae and identities. Such questions require thorough knowledge of trigonometric conversions and ratios. Algebraic operations and rules like transposition rules come into significant use while solving such problems.