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Question: Prove that \[\cos A\cos 2A\cos 4A\cos 8A = \dfrac{{\sin 16A}}{{16\sin A}}\]...

Prove that cosAcos2Acos4Acos8A=sin16A16sinA\cos A\cos 2A\cos 4A\cos 8A = \dfrac{{\sin 16A}}{{16\sin A}}

Explanation

Solution

Here, we will use the half-angle formulas in this question to prove that the LHS is equal to the RHS. We will multiply and divide the given LHS by twice the sine of the angle. Then we will use the half-angle formula of sine to simplify it further. We will then multiply and divide by 2 and simplify it using the half-angle formula of sine. These two steps will be followed repeatedly to get the required RHS.

Formula Used:
We will use the formula 2sinAcosA=sin2A2\sin A\cos A = \sin 2A.

Complete step-by-step answer:
We have to prove cosAcos2Acos4Acos8A=sin16A16sinA\cos A\cos 2A\cos 4A\cos 8A = \dfrac{{\sin 16A}}{{16\sin A}}.
We will first take into consideration the left-hand side of the equation and solve it further.
LHS =cosAcos2Acos4Acos8A = \cos A\cos 2A\cos 4A\cos 8A
Now, we will multiply and divide the LHS by 2sinA2\sin A. Therefore, we get
\Rightarrow LHS =12sinA[(2sinAcosA)cos2Acos4Acos8A] = \dfrac{1}{{2\sin A}}\left[ {\left( {2\sin A\cos A} \right)\cos 2A\cos 4A\cos 8A} \right]
Here, using the half angle formula 2sinAcosA=sin2A2\sin A\cos A = \sin 2A, we get,
\Rightarrow LHS =12sinA[sin2Acos2Acos4Acos8A] = \dfrac{1}{{2\sin A}}\left[ {\sin 2A\cos 2A\cos 4A\cos 8A} \right]
Now, again multiplying and dividing the LHS by 2, we get,
\Rightarrow LHS =14sinA[(2sin2Acos2A)cos4Acos8A] = \dfrac{1}{{4\sin A}}\left[ {\left( {2\sin 2A\cos 2A} \right)\cos 4A\cos 8A} \right]
Again, using the half angle formula2sinAcosA=sin2A2\sin A\cos A = \sin 2A, we get
\Rightarrow LHS =14sinA[sin4Acos4Acos8A] = \dfrac{1}{{4\sin A}}\left[ {\sin 4A\cos 4A\cos 8A} \right]
Now, again multiplying and dividing the LHS by 2, we get,
\Rightarrow LHS =18sinA[(2sin4Acos4A)cos8A] = \dfrac{1}{{8\sin A}}\left[ {\left( {2\sin 4A\cos 4A} \right)\cos 8A} \right]
Again, using the half angle formula2sinAcosA=sin2A2\sin A\cos A = \sin 2A, we get
\Rightarrow LHS =18sinA[sin8Acos8A] = \dfrac{1}{{8\sin A}}\left[ {\sin 8A\cos 8A} \right]
Now, again multiplying and dividing the LHS by 2, we get,
\Rightarrow LHS =116sinA[2sin8Acos8A] = \dfrac{1}{{16\sin A}}\left[ {2\sin 8A\cos 8A} \right]
Again, using the half angle formula2sinAcosA=sin2A2\sin A\cos A = \sin 2A, we get
\Rightarrow LHS =sin16A16sinA = \dfrac{{\sin 16A}}{{16\sin A}}
Hence, LHS == RHS
Hence, proved

Note: Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, cartographers (to make maps) use trigonometry. It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine, and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’, and ‘tan’.