Question

The minimum value of $ {\cos ^3}x + {\cos ^3}\left( {{{120}^ \circ } + x} \right) + {\cos ^3}\left( {{{120}^ \circ } - x} \right) $ is

Answer 1

Hint: In the given trigonometric expression we have three cube powered cosine functions. Cube powered cosine functions are found in the formula of $ \cos 3\theta $ . So first write the formula of $ \cos 3\theta $ and write all the cube powered cosines in terms of $ \cos 3\theta $ . After this, you will get an idea on how to proceed further. Solve the further referring to the below mentioned formulas.
Formulas used:
1. $ \cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $
2. $ \cos \left( {{{360}^ \circ } + \theta } \right) = \cos \theta $
3. $ \cos \left( { - \theta } \right) = \cos \theta $
4. $ \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
5. $ 2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right) $

Complete step by step solution:
We are given to the minimum value of $ {\cos ^3}x + {\cos ^3}\left( {{{120}^ \circ } + x} \right) + {\cos ^3}\left( {{{120}^ \circ } - x} \right) $
We know that the value $ \cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $
This means $ \cos 3x = 4{\cos ^3}x - 3\cos x $
Therefore, $ {\cos ^3}x = \dfrac{1}{4}\left( {\cos 3x + 3\cos x} \right) $
On replacing $ {\cos ^3}x $ in $ {\cos ^3}x + {\cos ^3}\left( {{{120}^ \circ } + x} \right) + {\cos ^3}\left( {{{120}^ \circ } - x} \right) $ with its above obtained formula, we get
 $ \Rightarrow \dfrac{1}{4}\left( {\cos 3x + 3\cos x} \right) + \dfrac{1}{4}\left[ {\cos 3\left( {{{120}^ \circ } + x} \right) + 3\cos \left( {{{120}^ \circ } + x} \right)} \right] + \dfrac{1}{4}\left[ {\cos 3\left( {{{120}^ \circ } - x} \right) + 3\cos \left( {{{120}^ \circ } - x} \right)} \right] $
 $ \Rightarrow \dfrac{1}{4}\left[ {\cos 3x + 3\cos x + \cos \left( {{{360}^ \circ } + 3x} \right) + 3\cos \left( {{{120}^ \circ } + x} \right) + \cos \left( {{{360}^ \circ } - 3x} \right) + 3\cos \left( {{{120}^ \circ } - x} \right)} \right] $
We know that $ \cos \left( {{{360}^ \circ } + \theta } \right) = \cos \theta $ , so $ \cos \left( {{{360}^ \circ } + 3x} \right) = \cos 3x $ and $ \cos \left( {{{360}^ \circ } - 3x} \right) = \cos \left( { - 3x} \right) $
Substituting the above value we get
 $ \Rightarrow \dfrac{1}{4}\left[ {\cos 3x + 3\cos x + \cos 3x + 3\cos \left( {{{120}^ \circ } + x} \right) + \cos \left( { - 3x} \right) + 3\cos \left( {{{120}^ \circ } - x} \right)} \right] $
We know that $ \cos \left( { - \theta } \right) = \cos \theta $ , so $ \cos \left( { - 3x} \right) = \cos 3x $
This gives,
 $ \Rightarrow \dfrac{1}{4}\left[ {\cos 3x + 3\cos x + \cos 3x + 3\cos \left( {{{120}^ \circ } + x} \right) + \cos 3x + 3\cos \left( {{{120}^ \circ } - x} \right)} \right] $
 $ \Rightarrow \dfrac{1}{4}\left[ {3\cos 3x + 3\cos x + 3\cos \left( {{{120}^ \circ } + x} \right) + 3\cos \left( {{{120}^ \circ } - x} \right)} \right] $
Taking out 3 common, we get
 $ \Rightarrow \dfrac{{1 \times 3}}{4}\left[ {\cos 3x + \cos x + \cos \left( {{{120}^ \circ } + x} \right) + \cos \left( {{{120}^ \circ } - x} \right)} \right] $
 $ \Rightarrow \dfrac{3}{4}\left[ {\cos 3x + \cos x + \cos \left( {{{120}^ \circ } + x} \right) + \cos \left( {{{120}^ \circ } - x} \right)} \right] $
As we can see $ \cos 3x + \cos x $ is in the form of $ \cos A + \cos B $ , where A is 3x and B is x.
We know that $ \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
Therefore, $ \cos 3x + \cos x = 2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right) = 2\cos 2x\cos x $
In $ \cos \left( {{{120}^ \circ } + x} \right) + \cos \left( {{{120}^ \circ } - x} \right) $ , A is $ \left( {{{120}^ \circ } + x} \right) $ and B is $ \left( {{{120}^ \circ } - x} \right) $
 $ \cos \left( {{{120}^ \circ } + x} \right) + \cos \left( {{{120}^ \circ } - x} \right) = 2\cos \left( {\dfrac{{{{120}^ \circ } + x + {{120}^ \circ } - x}}{2}} \right)\cos \left( {\dfrac{{{{120}^ \circ } + x - \left( {{{120}^ \circ } - x} \right)}}{2}} \right) = 2\cos {120^ \circ }\cos x $
Substituting the obtained values, we get
 $ \Rightarrow \dfrac{3}{4}\left( {2\cos 2x\cos x + 2\cos {{120}^ \circ }\cos x} \right) $
 $ \cos {120^ \circ } = - \dfrac{1}{2} $
 $ \Rightarrow \dfrac{3}{4}\left( {2\cos 2x\cos x + 2\left( {\dfrac{{ - 1}}{2}} \right)\cos x} \right) = \dfrac{3}{4}\left( {2\cos 2x\cos x - \cos x} \right) $
 We know that $ 2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right) $
This gives us $ 2\cos 2x\cos x = \cos \left( {2x + x} \right) + \cos \left( {2x - x} \right) = \cos 3x + \cos x $
 $ \Rightarrow \dfrac{3}{4}\left( {\cos 3x + \cos x - \cos x} \right) = \dfrac{3}{4}\left( {\cos 3x} \right) $
Cosine function ranges from -1 to +1. So the minimum value of $ \cos \theta $ is -1, this means the minimum value of $ \cos 3x $ is -1.
Therefore, the minimum value of $ \dfrac{3}{4}\left( {\cos 3x} \right) $ is $ \Rightarrow \dfrac{3}{4}\left( { - 1} \right) = \dfrac{{ - 3}}{4} $
The minimum value of $ {\cos ^3}x + {\cos ^3}\left( {{{120}^ \circ } + x} \right) + {\cos ^3}\left( {{{120}^ \circ } - x} \right) $ is $ \dfrac{{ - 3}}{4} $
So, the correct answer is “ $ \dfrac{{ - 3}}{4} $”.

Note: Sine, Cosine and tangent functions are periodic functions; which means their values get repeated after a certain interval. Sine values and cosine values repeat after every 360 degrees (2π radians) whereas tangent values get repeated after every 180 degrees (π radians). Be careful while writing the formulas. A little replacement of sine with cosine will change the complete answer. So please be careful.