Question

Prove that:
$\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x$

Answer 1

Hint: For a question like this we approach the solution by simplifying anyone the side and proving it equal to the other side, here also we will simplify the left-hand side using some of the trigonometric formulas like
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We simplify in such a manner that it results in the equivalent value to the other side expression

Complete step by step Answer:

Given data: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x$
Taking the left-hand side
$ \Rightarrow \sin x + \sin 3x + \sin 5x + \sin 7x$
On rearranging we get,
$ \Rightarrow \sin x + \sin 7x + \sin 3x + \sin 5x$
Using the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ in the first two and last two terms, we get,
$ \Rightarrow 2\sin (4x)\cos 3x + 2\sin (4x)\cos x$
Taking 2sin(4x) common from both the terms
$ \Rightarrow 2\sin (4x)\left[ {\cos 3x + \cos x} \right]$
Now using the formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$, we get,
$ \Rightarrow 2\sin (4x)2\cos (2x)\cos (x)$
$ \Rightarrow 4\cos x\cos 2x\sin 4x$, which is equal to the right-hand side in the given equation
Since, Left-hand side=right-hand side
We have proved the given equation

Note: An alternative method for the solution of the given question can be
 This time we’ll simplify the term in the right-hand side and will prove it equal to the term in the left-hand side
$ \Rightarrow 4\cos x\cos 2x\sin 4x$
$ \Rightarrow 4\cos \left( {\dfrac{{3x - x}}{2}} \right)\cos \left( {\dfrac{{3x + x}}{2}} \right)\sin 4x$
Using the formula $2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) = \cos A + \cos B$
$ \Rightarrow 2[\cos 3x + \cos x]\sin 4x$
$ \Rightarrow 2\sin 4x\cos 3x + 2\cos x\sin 4x$
$ \Rightarrow 2\sin \left( {\dfrac{{7x + x}}{2}} \right)\cos \left( {\dfrac{{7x - x}}{2}} \right) + 2\cos \left( {\dfrac{{5x - 3x}}{2}} \right)\sin \left( {\dfrac{{5x + 3x}}{2}} \right)$
Using the formula $2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) = \sin A + \sin B$
$ \Rightarrow \sin 7x + \sin x + \sin 3x + \sin 5x$, which is equal to the left-hand side in the given equation
Since, Left-hand side=right-hand side
We have proved the given equation