Question

Prove that \[\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{{16}}\]

Answer 1

Hint:
Here, we will first use the product of the cosine formula to find the product of the first two terms. Then by using the trigonometric ratio value we will simplify it further. We will use the product of the cosine formula again to prove the given trigonometric ratio.

Formula Used:
We will use the following formula:
1) The product of the cosine of the angles is given by the formula \[\cos A\cos B = \dfrac{1}{2}\left[ {\cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \right]\]
2) Trigonometric Ratio: \[\cos \left( {180^\circ - 80^\circ } \right) = - \cos 80^\circ \]

Complete step by step solution:
We have to prove that \[\cos 20\cos 40\cos 60\cos 80 = \dfrac{1}{{16}}\]
Now we will start simplifying the equation from the left hand side. Therefore,
\[\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \left( {\cos 20^\circ \cdot \cos 40^\circ } \right) \cdot \cos 60^\circ \cdot \cos 80^\circ \]
By using the product of the cosine formula \[\cos A\cos B = \dfrac{1}{2}\left[ {\cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \right]\], we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{2}\left( {\cos \left( {20^\circ + 40^\circ } \right) + \cos \left( {20^\circ - 40^\circ } \right)} \right) \cdot \cos 60^\circ \cdot \cos 80^\circ \]
Adding and subtracting the terms inside the bracket, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{2}\left( {\cos \left( {60^\circ } \right) + \cos \left( { - 20^\circ } \right)} \right) \cdot \cos 60^\circ \cdot \cos 80^\circ \]
Now substituting \[\cos 60^\circ = \dfrac{1}{2}\] in the above equation, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{2}\left( {\cos \left( {60^\circ } \right) + \cos \left( { - 20^\circ } \right)} \right) \times \dfrac{1}{2} \times \cos 80^\circ \]

Multiplying the terms, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4}\left( {\cos \left( {60^\circ } \right) + \cos \left( { - 20^\circ } \right)} \right)\cos 80^\circ \]
By multiplying \[\cos 80^\circ \] to the terms inside the bracket, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4}\left( {\cos \left( {60^\circ } \right) \cdot \cos \left( {80^\circ } \right) + \cos \left( { - 20^\circ } \right) \cdot \cos \left( {80^\circ } \right)} \right)\]
Again substituting \[\cos 60^\circ = \dfrac{1}{2}\] in the above equation, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4}\left( {\dfrac{1}{2} \cdot \cos \left( {80^\circ } \right) + \cos \left( { - 20^\circ } \right) \cdot \co s \left( {80^\circ } \right)} \right)\]
Since cosine function is an even function, so we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4}\left( {\dfrac{1}{2} \cdot \cos \left( {80^\circ } \right) + \cos \left( {20^\circ } \right) \cdot \cos \left( {80^\circ } \right)} \right)\]
By using the product of the cosine formula \[\cos A\cos B = \dfrac{1}{2}\left[ {\cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \right]\], we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4}\left( {\dfrac{1}{2} \cdot \cos \left( {80^\circ } \right) + \dfrac{1}{2}\cos \left( {20^\circ + 80^\circ } \right) + \cos \left( {20^\circ - 80^\circ } \right)} \right)\]
Adding and subtracting the terms, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{4} \cdot \dfrac{1}{2}\left( {\cos \left( {80^\circ } \right) + \cos \left( {100^\circ } \right) + \cos \left( { - 60^\circ } \right)} \right)\]
We know that \[\cos 100^\circ = \cos \left( {180^\circ - 80^\circ } \right)\]
By substituting the value of \[\cos 100^\circ \], we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{8}\left( {\cos \left( {80^\circ } \right) + \cos \left( {180^\circ - 80^\circ } \right) + \cos \left( { - 60^\circ } \right)} \right)\]
By substituting \[\cos \left( {180^\circ - 80^\circ } \right) = - \cos 80^\circ \], we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{8}\left( {\cos \left( {80^\circ } \right) - \cos \left( {80^\circ } \right) + \cos \left( { - 60^\circ } \right)} \right)\]
Since cosine function is an even function, so we get
 \[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{8}\left( {\cos 60^\circ } \right)\]
By substituting \[\cos 60^\circ = \dfrac{1}{2}\], we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{8} \cdot \dfrac{1}{2}\]
Multiplying the terms, we get
\[ \Rightarrow \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{{16}}\]

Therefore, \[\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{{16}}\] and hence proved.

Note:
We know that we have many trigonometric identities which are related to all the other trigonometric equations. We should note that sine and tangent are odd functions since both the functions are symmetric about the origin. Cosine is an even function since the function is symmetric about y axis. So, we take the arguments in the negative sign for odd functions and positive signs for even functions.