Question

If $A+B+C=\pi $, prove that $\cos 4A+\cos 4B+\cos 4C=-1+4\cos 2A\cos 2B\cos 2C$.

Answer 1

Hint: For solving this question we will use some trigonometric formula like formula for $\cos C+\cos D$ and $\cos 2\theta $ for simplifying the term written on the left-hand side. After that, we will prove it equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

It is given that if $A+B+C=\pi $ and we have to prove the following equation:

$\cos 4A+\cos 4B+\cos 4C=-1+4\cos 2A\cos 2B\cos 2C$

Now, before we proceed we should know the following three formulas:

$\begin{align}

  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)..........................\left( 1 \right) \\

 & \cos 2\theta =2{{\cos }^{2}}\theta -1..............................................................\left( 2 \right) \\

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow 2\left( A+B \right)=2\pi -2C \\

 & \Rightarrow \cos \left( 2\left( A+B \right) \right)=\cos \left( 2\pi -2C \right)=\cos

2C...........................\left( 3 \right) \\

\end{align}$

Now, we will be using the above three formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to $\cos 4A+\cos 4B+\cos 4C$ so, using the formula from equation (1).

Then,

$\begin{align}

  & \cos 4A+\cos 4B+\cos 4C \\

 & \Rightarrow 2\cos \left( \dfrac{4A+4B}{2} \right)\cos \left( \dfrac{4A-4B}{2} \right)+\cos 4C \\

 & \Rightarrow 2\cos \left( 2\left( A+B \right) \right)\cos \left( 2\left( A-B \right) \right)+\cos 4C \\

\end{align}$

Now, using the formula from equation (3) and equation (2) in the above equation. Then,

$\begin{align}

  & 2\cos \left( 2\left( A+B \right) \right)\cos \left( 2\left( A-B \right) \right)+\cos 4C \\

 & \Rightarrow 2\cos 2C\cos \left( 2\left( A-B \right) \right)+2{{\cos }^{2}}2C-1 \\

 & \Rightarrow 2\cos 2C\left( \cos \left( 2\left( A-B \right) \right)+\cos 2C \right)-1 \\

 & \Rightarrow 2\cos 2C\left( \cos \left( 2\left( A+B \right) \right)+\cos \left( 2\left( A-B \right) \right) \right)-1 \\

\end{align}$

Now, using formula from equation (1) in the above equation. Then,

$\begin{align}

  & 2\cos 2C\left( \cos \left( 2\left( A+B \right) \right)+\cos \left( 2\left( A-B \right) \right) \right)-1 \\

 & \Rightarrow 2\cos 2C\left( 2\cos \left( \dfrac{2A+2B+2A-2B}{2} \right)\cos \left( \dfrac{2A+2B-2A+2B}{2} \right) \right)-1 \\

 & \Rightarrow 2\cos 2C\left( 2\cos 2A\cos 2B \right)-1 \\

 & \Rightarrow -1+4\cos 2A\cos 2B\cos 2C \\

\end{align}$

Now, we can say that $\cos 4A+\cos 4B+\cos 4C=-1+4\cos 2A\cos 2B\cos 2C$ .

Thus, $L.H.S=R.H.S$.

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification easier, we should also try to make use of trigonometric results like $\cos \left( \pi -\theta \right)=-\cos \theta $ for making equations that will help us further in the solution. Moreover, the formulas like $\cos C+\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.