Question

Prove the following: $\cos A\cos 2A\cos 4A\cos 8A=\dfrac{\sin 16A}{16\sin A}$

Answer 1

Hint: In this question we have been given a trigonometric expression which we have to prove the left-hand side equal to the right-hand side. We will first consider the left-hand side of the expression and then solve it to get the terms on the right-hand side. We will use the property of double angles in trigonometry that $\sin 2A=2\sin A\cos A$ and substitute the value required to get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\cos A\cos 2A\cos 4A\cos 8A=\dfrac{\sin 16A}{16\sin A}$
Consider the left-hand side of the expression, we get:
$= \cos A\cos 2A\cos 4A\cos 8A$
Now we will rearrange the terms in the expression. On multiplying and dividing the expression by $2\sin A$, we get:
$= \dfrac{2\sin A\times \cos A\cos 2A\cos 4A\cos 8A}{2\sin A}$
On multiplying the terms, we get:
$= \dfrac{2\sin A\cos A\cos 2A\cos 4A\cos 8A}{2\sin A}$
Now we can see that the numerator has the term $2\sin A\cos A$ in it therefore, on using the formula $\sin 2A=2\sin A\cos A$ and substituting, we get:
$= \dfrac{\sin 2A\cos 2A\cos 4A\cos 8A}{2\sin A}$
On multiplying and dividing the term by $2$, we get:
$= \dfrac{2\sin 2A\cos 2A\cos 4A\cos 8A}{2\times 2\sin A}$
On simplifying, we get:
$= \dfrac{2\sin 2A\cos 2A\cos 4A\cos 8A}{4\sin A}$
Now we can see that the numerator has the term $2\sin 2A\cos 2A$ in it therefore, we can use the double angle formula, in this case the angle is doubled so we have the formula as $\sin 4A=2\sin 2A\cos 2A$. On substituting, we get:
$= \dfrac{\sin 4A\cos 4A\cos 8A}{4\sin A}$
On multiplying and dividing the term by $2$, we get:
$= \dfrac{2\times \sin 4A\cos 4A\cos 8A}{2\times 4\sin A}$
On simplifying, we get:
$= \dfrac{2\sin 4A\cos 4A\cos 8A}{8\sin A}$
Now we can see that the numerator has the term $2\sin 4A\cos 4A$ in it therefore, we can use the double angle formula, in this case the angle is doubled so we have the formula as $\sin 8A=2\sin 4A\cos 4A$. On substituting, we get:
$= \dfrac{\sin 8A\cos 8A}{8\sin A}$
On multiplying and dividing the term by $2$, we get:
$= \dfrac{2\times \sin 8A\cos 8A}{2\times 8\sin A}$
On simplifying, we get:
$= \dfrac{2\sin 8A\cos 8A}{16\sin A}$
Now we can see that the numerator has the term $2\sin 8A\cos 8A$ in it therefore, we can use the double angle formula, in this case the angle is doubled so we have the formula as $\sin 16A=2\sin 8A\cos 8A$. On substituting, we get:
$= \dfrac{\sin 16A}{16\sin A}$, which is the right-hand side, hence proved.

Note: It is to be remembered that in these types of questions where there is proving of the right-hand side and the left-hand side required, we can take either the left-hand side and prove it to be equal to the right-hand side, or we can take the right-hand side and prove it to be equal to left-hand side. It is to be noted that multiplying and dividing by the same term does not affect any terms value.