Question

The period of function $f(x) = \dfrac{{\sin 8x\cos x - \sin 6x\cos 3x}}{{\cos 2x\cos - \sin 3x\sin 4x}}$ is

1. $\pi$
2. $2\pi$
3. $\dfrac{\pi }{2}$
4. None of these

Answer 1

Here in this question we have to determine the period of a trigonometric function. Since the given trigonometric function is a difference or sum of a product of the trigonometric functions. So on considering the transformation formulas we are going to obtain the solution for the given question.

Complete step by step answer:
In trigonometry we have 6 trigonometric ratios namely, sine, cosine, tangent, cosecant, secant and cotangent. These ratios are abbreviated as sin, cos, tan, csc, sec and cot.
Now consider the given question
$\Rightarrow f(x) = \dfrac{{\sin 8x\cos x - \sin 6x\cos 3x}}{{\cos 2x\cos x - \sin 3x\sin 4x}}$
By the transformations formulas we have $\sin a\cos b = \dfrac{1}{2}\left[ {\sin (a + b) + \sin (a - b)} \right]$, $\cos a\cos b = \dfrac{1}{2}\left[ {\cos (a + b) + \cos (a - b)} \right]$ and $\sin a\sin b = - \dfrac{1}{2}\left[ {\cos (a + b) - \cos (a - b)} \right]$, where a and b represents the angle. Now the above function can be written as
$\Rightarrow f(x) = \dfrac{{\dfrac{1}{2}\left( {\sin (8x + x) + \sin (8x - x)} \right) - \dfrac{1}{2}\left( {\sin (6x + 3x) + \sin (6x - 3x)} \right)}}{{\dfrac{1}{2}\left( {\cos (2x + x) + \cos (2x - x)} \right) + \dfrac{1}{2}\left( {\cos (3x + 4x) - \cos (4x - 3x)} \right)}}$
On simplifying we have
$\Rightarrow f(x) = \dfrac{{\dfrac{1}{2}\left( {\sin 9x + \sin 7x} \right) - \dfrac{1}{2}\left( {\sin 9x + \sin 3x} \right)}}{{\dfrac{1}{2}\left( {\cos 3x + \cos x} \right) + \dfrac{1}{2}\left( {\cos 7x - \cos x} \right)}}$
Take $\dfrac{1}{2}$ as common in both numerator and denominator and we have
$\Rightarrow f(x) = \dfrac{{\dfrac{1}{2}\left[ {\left( {\sin 9x + \sin 7x} \right) - \left( {\sin 9x + \sin 3x} \right)} \right]}}{{\dfrac{1}{2}\left[ {\left( {\cos 3x + \cos x} \right) + \left( {\cos 7x - \cos x} \right)} \right]}}$
On cancelling the terms we have
$\Rightarrow f(x) = \dfrac{{\left( {\sin 9x + \sin 7x} \right) - \left( {\sin 9x + \sin 3x} \right)}}{{\left( {\cos 3x + \cos x} \right) + \left( {\cos 7x - \cos x} \right)}}$
On applying the sign conventions, the above function can be written as
$\Rightarrow f(x) = \dfrac{{\sin 9x + \sin 7x - \sin 9x - \sin 3x}}{{\cos 3x + \cos x + \cos 7x - \cos x}}$
On simplifying we have
$\Rightarrow f(x) = \dfrac{{\sin 7x - \sin 3x}}{{\cos 3x + \cos 7x}}$
By the transformations formulas we have $\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$. So applying these formulas to the above function we have
$\Rightarrow f(x) = \dfrac{{2\cos \left( {\dfrac{{7x + 3x}}{2}} \right)\sin \left( {\dfrac{{7x - 3x}}{2}} \right)}}{{2\cos \left( {\dfrac{{3x + 7x}}{2}} \right)\cos \left( {\dfrac{{3x - 7x}}{2}} \right)}}$
On simplifying we have
$\Rightarrow f(x) = \dfrac{{2\cos \left( {\dfrac{{10x}}{2}} \right)\sin \left( {\dfrac{{4x}}{2}} \right)}}{{2\cos \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right)}}$
On cancelling terms and on dividing we have
$\Rightarrow f(x) = \dfrac{{\cos 5x\sin 2x}}{{\cos 5x\cos 2x}}$
On cancelling the terms we have
$\Rightarrow f(x) = \dfrac{{\sin 2x}}{{\cos 2x}}$
As we know, the ratio of sine trigonometric ratio to the cosine trigonometric ratio is the tangent trigonometric ratio.
This can be written as
$\Rightarrow f(x) = \tan 2x$
As we know that the period for the tangent trigonometric ratio is ${\pi }$
Therefore the period of function $f(x) = \dfrac{{\sin 8x\cos x - \sin 6x\cos 3x}}{{\cos 2x\cos - \sin 3x\sin 4x}}$ is $\dfrac{\pi }{2}$
Here we had angle 2x so the period is $\dfrac{\pi }{2}$.

So, the correct answer is “Option 3”.

Note: The student must know about the transformation formulae. They are defined as follows:
$1.\,\,sinAcosB{\text{ }} = \left[ {sin\left( {A + B} \right){\text{ }} + {\text{ }}sin\left( {A{\text{ }}-{\text{ }}B} \right)} \right]$
$2.\,\,cosA{\text{ }}.sinB{\text{ }} = \left[ {sin\left( {A + B} \right){\text{ }}-\;sin\left( {A{\text{ }}-{\text{ }}B} \right)} \right]$
$3.\,\,cosAcosB{\text{ }} = \left[ {cos\left( {A + B} \right){\text{ }} + {\text{ }}cos\left( {A{\text{ }}-{\text{ }}B} \right)} \right]$

$$4.\,\,sinA{\text{ }}.sinB{\text{ }} = \left[ {cos\left( {A + B} \right){\text{ }}-{\text{ }}cos\left( {A{\text{ }}-{\text{ }}B} \right)} \right] \\\ 5.\;\,\,sinC{\text{ }} + {\text{ }}sinD{\text{ }} = {\text{ }}2sin\left( {\dfrac{{C + D}}{2}} \right)cos\left( {\dfrac{{C - D}}{2}} \right) \\\ 6.\;\,\,sinC{\text{ }}-\;sinD{\text{ }} = {\text{ }}2cos\left( {\dfrac{{C + D}}{2}} \right)\;sin\left( {\dfrac{{C - D}}{2}} \right) \\\ 7.\;\;cosC{\text{ }} + {\text{ }}cosD{\text{ }} = {\text{ }}2cos\left( {\dfrac{{C + D}}{2}} \right)cos\left( {\dfrac{{C - D}}{2}} \right) \\\ 8.\;\,\,cosC{\text{ }}-{\text{ }}cosD{\text{ }} = {\text{ }}-2sin\left( {\dfrac{{C + D}}{2}} \right)sin\left( {\dfrac{{C - D}}{2}} \right) \\\$$

These formulas are simplified forms of the sum and difference formulas of trigonometric ratios.