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Question: Period of \[\cos x\cos \left( 60^{\circ}-x\right) \cos \left( 60^{\circ}+x\right) \] is A) \[\dfra...

Period of cos⁡xcos⁡(60∘−x)cos⁡(60∘+x)\cos x\cos \left( 60^{\circ}-x\right) \cos \left( 60^{\circ}+x\right) is
A) π2\dfrac{\pi }{2}
B) π3\dfrac{\pi }{3}
C) 2π3\dfrac{2\pi }{3}
D) π\pi

Explanation

Solution

Hint: In this question it is given that we have to find the period of cos⁡xcos⁡(60∘−x)cos⁡(60∘+x)\cos x\cos \left( 60^{\circ}-x\right) \cos \left( 60^{\circ}+x\right) . So in order to get the solution we have to apply some formulas in the appropriate steps in order to get the solution, which are,
2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos \left( A+B\right) +\cos \left( A-B\right) .........(1)
cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2\cos^{2} \theta -1.......................(2)
4cos⁡3θ−3cos⁡θ=cos⁡3θ4\cos^{3} \theta -3\cos \theta =\cos 3\theta........(3)

Complete step-by-step solution:
Let, f(x)=cos⁡xcos⁡(60∘−x)cos⁡(60∘+x)f\left( x\right) =\cos x\cos \left( 60^{\circ}-x\right) \cos \left( 60^{\circ}+x\right)
=12cos⁡x[2cos⁡(60∘−x)cos⁡(60∘+x)]=\dfrac{1}{2} \cos x\left[ 2\cos \left( 60^{\circ}-x\right) \cos \left( 60^{\circ}+x\right) \right]
=12cos⁡x[2cos⁡(60∘+x)cos⁡(60∘−x)]=\dfrac{1}{2} \cos x\left[ 2\cos \left( 60^{\circ}+x\right) \cos \left( 60^{\circ}-x\right) \right]
Now applying formula (1) , where A
f\left( x\right) =\dfrac{1}{2} \cos x\left[ \cos \left\\{ \left( 60^{\circ}+x\right) +\left( 60^{\circ}-x\right) \right\\} +\cos \left\\{ \left( 60^{\circ}+x\right) -\left( 60^{\circ}-x\right) \right\\} \right]
=12cos⁡x[cos⁡(60∘+x+60∘−x)+cos⁡(60∘+x−60∘+x)]=\dfrac{1}{2} \cos x\left[ \cos \left( 60^{\circ}+x+60^{\circ}-x\right) +\cos \left( 60^{\circ}+x-60^{\circ}+x\right) \right]
=12cos⁡x[cos⁡120∘+cos⁡2x]=\dfrac{1}{2} \cos x\left[ \cos 120^{\circ}+\cos 2x\right]
=12cos⁡x(−12+cos⁡2x)=\dfrac{1}{2} \cos x\left( -\dfrac{1}{2} +\cos 2x\right) [ since, cos⁡120∘=−12\cos 120^{\circ} =\dfrac{-1}{2}]
=12cos⁡x(−12+2cos⁡2x−1)=\dfrac{1}{2} \cos x\left( -\dfrac{1}{2} +2\cos^{2} x-1\right) [ by formula (2)]
=12cos⁡x(2cos⁡2x−32)=\dfrac{1}{2} \cos x\left( 2\cos^{2} x-\dfrac{3}{2} \right)
=cos⁡x(cos⁡2x−34)=\cos x\left( \cos^{2} x-\dfrac{3}{4} \right)
=cos⁡x(4cos⁡2x−34)=\cos x\left( \dfrac{4\cos^{2} x-3}{4} \right)
=(4cos⁡3x−3cos⁡x4)=\left( \dfrac{4\cos^{3} x-3\cos x}{4} \right)
=cos⁡3x4=\dfrac{\cos 3x}{4} [ by using formula (3)]

As we know that if f(x)=cos⁡kxf\left( x\right) =\cos kx then period is 2π∣k∣\dfrac{2\pi }{\left\vert k\right\vert }
So here k=3, therefore, the period of cos⁡3x\cos 3x is 2π3\dfrac{2\pi }{3}.
Note: So while solving you need to have the basic idea about the period of a trigonometric function, i.e, the distance between the repetition of any function is called the period of the function or we can say that the length of one complete cycle is called a period.