Question

The period of the function
$f\left( x \right) = \dfrac{{\sin 8x\cos x - \sin 6x\cos 3x}}{{\cos 2x\cos x - \sin 3x\sin 4x}}$ is
$\left( a \right)\pi$
$\left( b \right)2\pi$
$\left( c \right)\dfrac{\pi }{2}$
$\left( d \right)$ None of these

Answer 1

Hint : In this particular question, simplify the given trigonometric function using the basic trigonometric identities such as, 2sin A cos B = sin (A + B) + sin (A – B), 2cos A cos B = cos (A + B) + cos (A – B) and 2sin A sin B = cos (B – A) – cos (B + A), then use the concept that period of tan (kx) = $\dfrac{\pi }{k}$, so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given function
$f\left( x \right) = \dfrac{{\sin 8x\cos x - \sin 6x\cos 3x}}{{\cos 2x\cos x - \sin 3x\sin 4x}}$
Now as we know that, 2sin A cos B = sin (A + B) + sin (A – B), 2cos A cos B = cos (A + B) + cos (A – B) and
2sin A sin B = cos (B – A) – cos (B + A) so use this property in the above equation we have,
$f\left( x \right) = \dfrac{{\dfrac{1}{2}\left( {\sin \left( {8x + x} \right) + \sin \left( {8x - x} \right)} \right) - \left[ {\dfrac{1}{2}\left( {\sin \left( {6x + 3x} \right) + \sin \left( {6x - 3x} \right)} \right)} \right]}}{{\dfrac{1}{2}\left( {\cos \left( {2x + x} \right) + \cos \left( {2x - x} \right)} \right) - \left[ {\dfrac{1}{2}\left( {\cos \left( {4x - 3x} \right) - \cos \left( {4x + 3x} \right)} \right)} \right]}}$
Now simplify this we have,
$f\left( x \right) = \dfrac{{\sin \left( {9x} \right) + \sin \left( {7x} \right) - \sin \left( {9x} \right) - \sin \left( {3x} \right)}}{{\cos \left( {3x} \right) + \cos \left( x \right) - \cos \left( x \right) + \cos \left( {7x} \right)}}$
Now cancel out the same terms having positive and negative signs from numerator as well as from denominator we have,
$f\left( x \right) = \dfrac{{\sin \left( {7x} \right) - \sin \left( {3x} \right)}}{{\cos \left( {3x} \right) + \cos \left( {7x} \right)}}$
Now as we know that sin C – sin D = $2\sin \left( {\dfrac{{C - D}}{2}} \right)\cos \left( {\dfrac{{C + D}}{2}} \right)$
And cos C + cos D = $2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ so use these properties in the above equation we have.
$f\left( x \right) = \dfrac{{2\sin \left( {\dfrac{{7x - 3x}}{2}} \right)\cos \left( {\dfrac{{7x + 3x}}{2}} \right)}}{{2\cos \left( {\dfrac{{3x + 7x}}{2}} \right)\cos \left( {\dfrac{{3x - 7x}}{2}} \right)}}$
Now simplify we get
$f\left( x \right) = \dfrac{{2\sin \left( {2x} \right)\cos \left( {5x} \right)}}{{2\cos \left( {5x} \right)\cos \left( { - 2x} \right)}}$
Now cancel out the same terms from numerator and denominator we have,
$f\left( x \right) = \dfrac{{\sin \left( {2x} \right)}}{{\cos \left( { - 2x} \right)}}$
Now as we know that cos (-x) = cos x, so use these properties in the above equation we have.
$f\left( x \right) = \dfrac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}$
Now as we know that (sin x/cos x) = tan x, so use these properties in the above equation we have.
$f\left( x \right) = \tan 2x$
Now as we know that the period of tan (kx) = $\dfrac{\pi }{k}$ so the period of tan 2x = $\dfrac{\pi }{2}$
So this is the required period of the given trigonometric function.
Hence option (C) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric identities, whatever are used is all written above after that simplification is the key then after simplification we get only one terms which is tan 2x, so the period of the tan 2x is $\dfrac{\pi }{2}$, so it is the required answer.