Question

Prove that ${\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = \dfrac{3}{4}\cos 3\theta $.

Answer 1

Hint: Here, we will proceed by using the formula i.e., when $\left( {a + b + c = 0} \right)$ then ${a^3} + {b^3} + {c^3} = 3abc$. Then, we will use the formulas $\cos {\text{A}} + \cos {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ and $\cos \left( {{{180}^0} + \alpha } \right) = - \cos \alpha $ in order to obtain the required equation which needs to be proved.

Complete Step-by-Step solution:
To prove: ${\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = \dfrac{3}{4}\cos 3\theta {\text{ }} \to {\text{(1)}}$
As we know that ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right)$
When $\left( {a + b + c = 0} \right)$, the above formula becomes
$
  {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)\left( {{a^2} + {b^2} + {c^2}} \right) \\
   \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \\
   \Rightarrow {a^3} + {b^3} + {c^3} = 3abc{\text{ }} \to {\text{(2)}} \\
 $
Also we know that $\cos {\text{A}} + \cos {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right){\text{ }} \to {\text{(3)}}$
Now, let us consider $\cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta + \left[ {\cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right)} \right]$
By using the formula given in equation (3) in the above equation, we get
$
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta + \left[ {2\cos \left( {\dfrac{{\left( {{{120}^0} + \theta } \right) + \left( {{{240}^0} + \theta } \right)}}{2}} \right)\cos \left( {\dfrac{{\left( {{{120}^0} + \theta } \right) - \left( {{{240}^0} + \theta } \right)}}{2}} \right)} \right] \\
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta + 2\cos \left( {\dfrac{{{{360}^0} + 2\theta }}{2}} \right)\cos \left( {\dfrac{{ - {{120}^0}}}{2}} \right) \\
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta + 2\cos \left( {{{180}^0} + \theta } \right)\cos \left( { - {{60}^0}} \right) \\
 $
Using the formulas $\cos \left( {{{180}^0} + \alpha } \right) = - \cos \alpha $ and $\cos \left( { - \alpha } \right) = \cos \alpha $ in the above equation, we get
$ \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta - 2\left( {\cos \theta } \right)\left( {\cos {{60}^0}} \right){\text{ }} \to {\text{(4)}}$
According to the general trigonometric table, $\cos {60^0} = \dfrac{1}{2}$
By using $\cos {60^0} = \dfrac{1}{2}$, equation (4) becomes
$
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta - 2\left( {\cos \theta } \right)\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = \cos \theta - \cos \theta \\
   \Rightarrow \cos \theta + \cos \left( {{{120}^0} + \theta } \right) + \cos \left( {{{240}^0} + \theta } \right) = 0{\text{ }} \to {\text{(5)}} \\
 $
Using the formula given by equation (2), we get
$
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {{{180}^0} - {{60}^0} + \theta } \right)} \right]\left[ {\cos \left( {{{180}^0} + {{60}^0} + \theta } \right)} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {{{180}^0} + \left( {\theta - {{60}^0}} \right)} \right)} \right]\left[ {\cos \left( {{{180}^0} + \left( {\theta + {{60}^0}} \right)} \right)} \right] \\
 $
Using the formulas $\cos \left( {{{180}^0} + \alpha } \right) = - \cos \alpha $ in the above equation, we get
\[
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ { - \cos \left( {\theta - {{60}^0}} \right)} \right]\left[ { - \cos \left( {\theta + {{60}^0}} \right)} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {\theta - {{60}^0}} \right)} \right]\left[ {\cos \left( {\theta + {{60}^0}} \right)} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( { - \left( {{{60}^0} - \theta } \right)} \right)} \right]\left[ {\cos \left( {{{60}^0} + \theta } \right)} \right] \\
 \]
Using the formula $\cos \left( { - \alpha } \right) = \cos \alpha $, the above equation becomes
\[
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {{{60}^0} - \theta } \right)} \right]\left[ {\cos \left( {{{60}^0} + \theta } \right)} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\cos \left( {{{60}^0} + \theta } \right)\cos \left( {{{60}^0} - \theta } \right)} \right] \\
 \]
Using the formula $\cos \left( {{\text{A}} + {\text{B}}} \right)\cos \left( {{\text{A}} - {\text{B}}} \right) = {\left( {\cos {\text{A}}} \right)^2} - {\left( {\sin {\text{A}}} \right)^2}$ in the above equation, we get
\[ \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {{{\left( {\cos {{60}^0}} \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]
Using the formula $
  {\left( {\sin \alpha } \right)^2} + {\left( {\cos \alpha } \right)^2} = 1 \\
   \Rightarrow {\left( {\sin \alpha } \right)^2} = 1 - {\left( {\cos \alpha } \right)^2} \\
 $ in the above equation, we get
\[ \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {{{\left( {\cos {{60}^0}} \right)}^2} - \left( {1 - {{\left( {\cos \theta } \right)}^2}} \right)} \right]\]
Using $\cos {60^0} = \dfrac{1}{2}$ in the above equation, we get
\[
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {{{\left( {\dfrac{1}{2}} \right)}^2} - 1 + {{\left( {\cos \theta } \right)}^2}} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\dfrac{1}{4} - 1 + {{\left( {\cos \theta } \right)}^2}} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\dfrac{{1 - 4 + 4{{\left( {\cos \theta } \right)}^2}}}{4}} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\cos \theta } \right]\left[ {\dfrac{{ - 3 + 4{{\left( {\cos \theta } \right)}^2}}}{4}} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = 3\left[ {\dfrac{{ - 3\left( {\cos \theta } \right) + 4{{\left( {\cos \theta } \right)}^3}}}{4}} \right] \\
   \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = \dfrac{3}{4}\left[ {4{{\left( {\cos \theta } \right)}^3} - 3\cos \theta } \right] \\
 \]
Using the formula \[4{\left( {\cos \alpha } \right)^3} - 3\cos \alpha = \cos 3\alpha \] in the above equation, we get
\[ \Rightarrow {\left[ {\cos \theta } \right]^3} + {\left[ {\cos \left( {{{120}^0} + \theta } \right)} \right]^3} + {\left[ {\cos \left( {{{240}^0} + \theta } \right)} \right]^3} = \dfrac{3}{4}\cos 3\theta \]
Clearly, the above equation obtained is the same as the equation which was needed to be proved i.e., equation (1). Hence, we have proved the required equation.

Note: In this particular problem, the equation which needs to be proved is actually the sum of cubes of three trigonometric functions. Here, if we sum these three trigonometric functions we will get to know that their sum comes to be zero which resembles with the formula ${a^3} + {b^3} + {c^3} = 3abc$ where $\left( {a + b + c = 0} \right)$.