If $A+B+C=\pi $ , prove that $\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$ .
Answer
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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:
$\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$
Now, before we proceed we should know the following three formulas:
$\begin{align}
& \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)..............................\left( 1 \right) \\
& A+B+C=\pi \\
& \Rightarrow A+B=\pi -C \\
& \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C.......................................\left( 2 \right) \\
& \cos 2\theta =2{{\cos }^{2}}\theta -1...................................................................\left( 3 \right) \\
\end{align}$
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 $\cos 2A+\cos 2B+\cos 2C$ so, using the formula from equation (1).
Then,
$\begin{align}
& \cos 2A+\cos 2B+\cos 2C \\
& \Rightarrow 2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)+\cos 2C \\
& \Rightarrow 2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C \\
\end{align}$
Now, using the formula from equation (2) and equation (3) in the above equation. Then,
$\begin{align}
& 2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C \\
& \Rightarrow -2\cos C\cos \left( A-B \right)+2{{\cos }^{2}}C-1 \\
& \Rightarrow -1-2\cos C\left( -\cos C+\cos \left( A-B \right) \right) \\
& \Rightarrow -1-2\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\
\end{align}$
Now, using the formula from the equation (1) in the above equation. Then,
$\begin{align}
& -1-2\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\
& \Rightarrow -1-2\cos C\left( 2\cos \left( \dfrac{A+B+A-B}{2} \right)\cos \left(
\dfrac{A+B-A+B}{2} \right) \right) \\
& \Rightarrow -1-2\cos C\left( 2\cos A\cos B \right) \\
& \Rightarrow -1-4\cos A\cos B\cos C \\
\end{align}$
Now, from the above result, we can say that $\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$.
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 formula $\cos C+\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.
Complete step-by-step answer:
Given:
It is given that if $A+B+C=\pi $ and we have to prove the following equation:
$\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$
Now, before we proceed we should know the following three formulas:
$\begin{align}
& \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)..............................\left( 1 \right) \\
& A+B+C=\pi \\
& \Rightarrow A+B=\pi -C \\
& \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C.......................................\left( 2 \right) \\
& \cos 2\theta =2{{\cos }^{2}}\theta -1...................................................................\left( 3 \right) \\
\end{align}$
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 $\cos 2A+\cos 2B+\cos 2C$ so, using the formula from equation (1).
Then,
$\begin{align}
& \cos 2A+\cos 2B+\cos 2C \\
& \Rightarrow 2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)+\cos 2C \\
& \Rightarrow 2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C \\
\end{align}$
Now, using the formula from equation (2) and equation (3) in the above equation. Then,
$\begin{align}
& 2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C \\
& \Rightarrow -2\cos C\cos \left( A-B \right)+2{{\cos }^{2}}C-1 \\
& \Rightarrow -1-2\cos C\left( -\cos C+\cos \left( A-B \right) \right) \\
& \Rightarrow -1-2\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\
\end{align}$
Now, using the formula from the equation (1) in the above equation. Then,
$\begin{align}
& -1-2\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\
& \Rightarrow -1-2\cos C\left( 2\cos \left( \dfrac{A+B+A-B}{2} \right)\cos \left(
\dfrac{A+B-A+B}{2} \right) \right) \\
& \Rightarrow -1-2\cos C\left( 2\cos A\cos B \right) \\
& \Rightarrow -1-4\cos A\cos B\cos C \\
\end{align}$
Now, from the above result, we can say that $\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C$.
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 formula $\cos C+\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.
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