Question

Prove that \[4 \cdot \cos x \cdot \cos \left( {\dfrac{\pi }{3} + x} \right) \cdot \cos \left( {\dfrac{\pi }{3} - x} \right) = \cos 3x\].

Answer 1

Hint: To prove this type of question we first take the left hand side part of the problem and we show the right hand side of the problem. Here in this problem we use the formula \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\]. We need to remember the concept of trigonometric transformation formulae and trigonometric ratios and where the signs of cosine and sine will be positive and negative.

Complete Step by Step Solution:
The objective of the problem is to prove \[4 \cdot \cos x \cdot \cos \left( {\dfrac{\pi }{3} + x} \right) \cdot \cos \left( {\dfrac{\pi }{3} - x} \right) = \cos 3x\]
For this we considering left hand side part of the given problem
LHS=\[4 \cdot \cos x \cdot \cos \left( {\dfrac{\pi }{3} + x} \right) \cdot \cos \left( {\dfrac{\pi }{3} - x} \right)\]
We are splitting four as two into two , we get
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {2\cos \left( {\dfrac{\pi }{3} + x} \right) \cdot \cos \left( {\dfrac{\pi }{3} - x} \right)} \right]\]
Here we are using the formula \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\]
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {\dfrac{\pi }{3} + x + \dfrac{\pi }{3} - x} \right) + \cos \left( {\dfrac{\pi }{3} + x - \dfrac{\pi }{3} + x} \right)} \right]\]
On solving above equation we get
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {\dfrac{{2\pi }}{3}} \right) + \cos \left( {2x} \right)} \right]\]
Converting \[\dfrac{{2\pi }}{3}\] by substituting \[\pi \] as 180. We get
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {\dfrac{{2 \times 180}}{3}} \right) + \cos \left( {2x} \right)} \right]\]
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {\dfrac{{360}}{3}} \right) + \cos \left( {2x} \right)} \right]\]
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {{{120}^\circ }} \right) + \cos \left( {2x} \right)} \right]\]
Taking 120 as \[\left( {{{180}^\circ } - {{60}^\circ }} \right)\] we get
\[ \Rightarrow 2 \cdot \cos x \cdot \left[ {\cos \left( {{{180}^\circ } - {{60}^\circ }} \right) + \cos \left( {2x} \right)} \right]\]
 We know that \[\cos \left( {{{180}^\circ } - x} \right) = - \cos x\]
We get , \[2 \cdot \cos x \cdot \left[ { - \cos \left( {{{60}^\circ }} \right) + \cos \left( {2x} \right)} \right]\]
On splitting the bracketed terms with multiplying 2cosx which is outside the bracket we get,
\[ \Rightarrow 2\cos x \cdot \cos 2x - 2\cos x \cdot \cos {60^\circ }\]
We know that the value of cos60 degrees is half , we get
\[ \Rightarrow 2\cos x \cdot \cos 2x - 2\cos x \cdot \dfrac{1}{2}\]
On solving above equation we get ,
\[ \Rightarrow 2\cos x \cdot \cos 2x - \cos x\]
Here again we are using the formula \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\]
\[
   \Rightarrow \cos \left( {x + 2x} \right) + \cos \left( {x - 2x} \right) - \cos x \\
   \Rightarrow \cos 3x + \cos \left( { - x} \right) - \cos x \\
 \]
Since $\cos \left( { - x} \right) = \cos x$ we get
\[
   \Rightarrow \cos 3x + \cos \left( { - x} \right) - \cos x \\
   \Rightarrow \cos 3x + \cos x - \cos x \\
   \Rightarrow \cos 3x = RHS \\
 \]
Therefore, the left hand side part is equal to the right hand side part.

Thus \[4 \cdot \cos x \cdot \cos \left( {\dfrac{\pi }{3} + x} \right) \cdot \cos \left( {\dfrac{\pi }{3} - x} \right) = \cos 3x\]

Note:
The sine trigonometric function is positive in first and second quadrants in the coordinate axis and in the remaining it is negative only. And for the tan trigonometric function it is positive in the first and third quadrants of the coordinate system. For this type of problem we should remember the trigonometric ratios and formulae. And one should be careful about the algebraic signs and trigonometric signs.