Question

Prove that:
$\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x$

Answer 1

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
$\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
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: $\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x$
Taking the numerator of the expression in the left-hand side
$\Rightarrow \sin x - \sin 3x + \sin 5x - \sin 7x$
On rearranging we get,
$\Rightarrow \;\left[ {\sin x - \sin 3x} \right]{\text{ }} + {\text{ }}\left[ {\sin 5x - \sin 7x} \right]$
Using the formula $\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$
$\Rightarrow 2\sin \left( {\dfrac{{x - 3x}}{2}} \right)\cos \left( {\dfrac{{x + 3x}}{2}} \right) + 2\sin \left( {\dfrac{{5x - 7x}}{2}} \right)\cos \left( {\dfrac{{5x + 7x}}{2}} \right)$
On simplification, we get
$\Rightarrow 2\sin \left( { - x} \right)\cos \left( {2x} \right) + 2\sin \left( { - x} \right)\cos \left( {6x} \right)$
Taking $-sinx$ common from both the terms
$\Rightarrow - 2\sin x\left( {\cos 2x + \cos 6x} \right)$
Using the formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\Rightarrow - 2\sin x\left( {2\cos \left( {\dfrac{{2x + 6x}}{2}} \right)\cos \left( {\dfrac{{6x - 2x}}{2}} \right)} \right)$
On simplification we get,
$\Rightarrow - 4\sin x\cos 4x\cos 2x$
Now, Taking the denominator of the expression in the left-hand side
$\Rightarrow (\cos x - \cos 3x) - (\cos 5x - \cos 7x)$
Using the formula$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$

$\Rightarrow - 2\sin \left( {\dfrac{{x - 3x}}{2}} \right)\sin \left( {\dfrac{{x + 3x}}{2}} \right) + 2\sin \left( {\dfrac{{5x - 7x}}{2}} \right)\sin \left( {\dfrac{{5x + 7x}}{2}} \right)$
On simplification we get,
$\Rightarrow - 2\sin \left( { - x} \right)\sin \left( {2x} \right) + 2\sin \left( { - x} \right)\sin \left( {6x} \right)$
Using $\sin ( - x) = - \sin x$, we get
$\Rightarrow 2\sin \left( x \right)\sin \left( {2x} \right) - 2\sin \left( x \right)\sin \left( {6x} \right)$
Taking $2\sin x$ common from both the terms, we get
$\Rightarrow 2\sin x\left( {\sin 2x - \sin 6x} \right)$
Now using $\sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)$, we get
$\Rightarrow 2\sin x\left( {2\sin \left( {\dfrac{{2x - 6x}}{2}} \right)\cos \left( {\dfrac{{2x + 6x}}{2}} \right)} \right)$
On simplification we get,
$\Rightarrow 2\sin x\left( { - 2\sin 2x\cos 4x} \right)$
$\Rightarrow - 4\sin x\sin 2x\cos 4x$
Therefore the left-hand expression will be $\dfrac{{numerator}}{{deno\min ator}}$
i.e. $\dfrac{{ - 4\sin x\cos 4x\cos 2x}}{{ - 4\sin x\sin 2x\cos 4x}}$
Dividing $- 4\sin x\cos 4x$ from the numerator and the denominator, we get
$= \dfrac{{\cos 2x}}{{\sin 2x}}$
Using $\dfrac{{\cos A}}{{\sin A}} = \cot A$, we get
$= \cot 2x$, which equal to the right-hand side expression
Since the left-hand side and right=hand side are equal, the equation has been proved.

Note: We also prove that $\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x$, by substituting the value of $\dfrac{\pi }{6}$
Substituting $x = \dfrac{\pi }{6}$
Left-hand side$= \dfrac{{\sin \left( {\dfrac{\pi }{6}} \right) - \sin \left( {3\dfrac{\pi }{6}} \right) + \sin \left( {5\dfrac{\pi }{6}} \right) - \sin \left( {7\dfrac{\pi }{6}} \right)}}{{\cos \left( {\dfrac{\pi }{6}} \right) - \cos \left( {3\dfrac{\pi }{6}} \right) - \cos \left( {5\dfrac{\pi }{6}} \right) + \cos \left( {7\dfrac{\pi }{6}} \right)}}$
$= \dfrac{{\dfrac{1}{2} - 0 + \dfrac{1}{2} - \left( { - \dfrac{1}{2}} \right)}}{{\dfrac{{\sqrt 3 }}{2} - 0 - \left( { - \dfrac{{\sqrt 3 }}{2}} \right) - \dfrac{{\sqrt 3 }}{2}}}$
On simplifying
$= \dfrac{1}{{\sqrt 3 }}$
Right-hand side$= \cot 2\left( {\dfrac{\pi }{6}} \right)$
$= \dfrac{1}{{\sqrt 3 }}$ , $\cot \dfrac{\pi }{3} = \dfrac{1}{{\sqrt 3 }}$
Since, Left-hand side$=$Right-hand side
We have proved the given equation, but do not attempt this type of solution for the descriptive type question, this substitution method is just a way to check or verify.