Question

If $A+B+C=\pi ,$then$\cos 2A+\cos 2B+\cos 2C$ is

• A. $1+4\cos A\cos B\sin C$
• B. $-1+4\sin A\sin B\cos C$
• C. $-1–4\cos A\cos B\cos C$
• D. None of these

Answer 1

The correct option is C

$-1–4\cos A\cos B\cos C$

Step 1: Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{B}$:

$\begin{array}{rcl}A+B+C& =& 180°\dots .\left(1\right)\\ cos2A+cos2B& =& 2cos\frac{(2A+2B)}{2}cos\frac{(2A–2B)}{2}\\ & =& 2cos(A+B)cos(A–B)\\ & =& 2cos(180°–C)cos(A–B)(u\sin g(1\left)\right)\\ & =& -2cosCcos(A–B)\end{array}$

Step 2: Find the value of $\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{C}$:

$\begin{array}{rcl}cos2A+cos2B+cos2C& =& -2cosCcos(A–B)+cos2C\\ & =& -2cosCcos(A–B)+(2co{s}^{2}C–1)(Usingtheformulacos2x=2co{s}^{2}x–1)\\ & =& -2cosC\left[cos\right(A–B)–cosC]–1\\ & =& -2cosC\left[cos\right(A–B)+cos(A+B\left)\right]–1\\ & & \because cosC=cos[180°–(A+B\left)\right]\\ & =& -2cosC\left[2cosAcosB\right]–1\\ & =& -1–4cosAcosBcosC\end{array}$

Hence, option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is the correct option.