Question

Evaluate the value of $\dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}}$ .

Answer 1

Here, the given fractional value is $\dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}}$ and we need to find the value of the given fractional value.
Then, find the values of $\cos x - \cos 3x$ , $\sin 8x + \sin 2x$ , $\sin 5x - \sin x$ and $\cos 4x - \cos 6x$ by using the formulae given below
$\sin a + \sin b = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right) \\\ \sin a - \sin b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \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) \\\$
Thus, substitute the values in the given fraction and find the required answer.

Complete step-by-step answer:
Here, the given fractional value is $\dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}}$ .
Since, we know that,
$\sin a + \sin b = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right) \\\ \sin a - \sin b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \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) \\\$
So, we will simplify the term $\cos x - \cos 3x$ of the fraction as
$\cos x - \cos 3x = - 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\sin \left( {\dfrac{{x - 3x}}{2}} \right) \\\ = - 2\sin \left( {\dfrac{{4x}}{2}} \right)\sin \left( {\dfrac{{ - 2x}}{2}} \right) \\\ = - 2\sin 2x\sin \left( { - x} \right) \\\ = 2\sin 2x\sin x\left( {\because \sin \left( { - x} \right) = - \sin x} \right) \\\$
Similarly, repeating the above steps for $\sin 8x + \sin 2x$ , $\sin 5x - \sin x$ and $\cos 4x - \cos 6x$
So,
$\sin 8x + \sin 2x = 2\sin \left( {\dfrac{{8x + 2x}}{2}} \right)\cos \left( {\dfrac{{8x - 2x}}{2}} \right) \\\ = 2\sin \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{6x}}{2}} \right) \\\ = 2\sin 5x\cos 3x \\\$
$\sin 5x - \sin x = 2\cos \left( {\dfrac{{5x + x}}{2}} \right)\sin \left( {\dfrac{{5x - x}}{2}} \right) \\\ = 2\cos \left( {\dfrac{{6x}}{2}} \right)\sin \left( {\dfrac{{4x}}{2}} \right) \\\ = 2\cos 3x\sin 2x \\\$

$$\cos 4x - \cos 6x = - 2\sin \left( {\dfrac{{4x + 6x}}{2}} \right)\sin \left( {\dfrac{{4x - 6x}}{2}} \right) \\\ = - 2\sin \left( {\dfrac{{10x}}{2}} \right)\sin \left( {\dfrac{{ - 2x}}{2}} \right) \\\ = - 2\sin 5x\sin \left( { - x} \right) \\\ = 2\sin 5x\sin x\left( {\because \sin \left( { - x} \right) = - \sin x} \right) \\\$$

Thus, we will substitute the values of $\cos x - \cos 3x$ , $\sin 8x + \sin 2x$ , $\sin 5x - \sin x$ and $\cos 4x - \cos 6x$ in the given fractional value.
$\therefore \dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}} = \dfrac{{\left( {2\sin 2x\sin x} \right)\left( {2\sin 5x\cos 3x} \right)}}{{\left( {2\cos 3x\sin 2x} \right)\left( {2\sin 5x\sin x} \right)}}$
Here, from the above step it is clear that, every term of the fraction will get eliminated. So, the value of the given fractional value will be 0.
$\therefore \dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}} = 1$
Thus, the value of the given fractional value $\dfrac{{\left( {\cos x - \cos 3x} \right)\left( {\sin 8x + \sin 2x} \right)}}{{\left( {\sin 5x - \sin x} \right)\left( {\cos 4x - \cos 6x} \right)}}$ is 1.

Note: While solving questions like these in which many identities are used at once, we should carefully substitute the values in the question because there is a high chance that we mess up with the signs and a change of even one sign is enough to get us a completely wrong answer.