Question

If $x+y+z={{180}^{\circ }}$, then $\cos 2x+\cos 2y-\cos 2z$ is equal to
A. $4\sin x\sin y\sin z$
B. $1-4\sin x\sin y\cos z$
C. $4\sin x\sin y\sin z-1$
D. $\cos x\cos y\cos z$

Answer 1

Hint: We first take the sum of angles for the trigonometric ratios. We also use the multiple angle formula of $\cos 2x=1-2{{\sin }^{2}}x$. We convert them to their multiple forms. We take $-2\sin x$ common and find the required solution.

Complete step by step answer:
It is given that $x+y+z=\pi $. We get $y+z=\pi -x$ and $x=\pi -\left( y+z \right)$.
We are going to use the formulas of sum of angles for the trigonometric ratios. We have $\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$.
We use the representation of $A=2y;B=2z$ and get
$\begin{align}
  & \cos 2y-\cos 2z \\
 & =2\sin \left( \dfrac{2y+2z}{2} \right)\sin \left( \dfrac{2z-2y}{2} \right) \\
 & =2\sin \left( y+z \right)\sin \left( z-y \right) \\
\end{align}$
We change the angle and get
$\begin{align}
  & 2\sin \left( y+z \right)\sin \left( z-y \right) \\
 & =2\sin \left( \pi -x \right)\sin \left( z-y \right) \\
 & =2\sin x\sin \left( z-y \right) \\
\end{align}$
We also use the multiple angle formula of $\cos 2x=1-2{{\sin }^{2}}x$.
Therefore,
$\begin{align}
  & \cos 2x+\cos 2y-\cos 2z \\
 & =1-2{{\sin }^{2}}x+2\sin \left( y+z \right)\sin \left( z-y \right) \\
 & =1-2{{\sin }^{2}}x+2\sin x\sin \left( z-y \right) \\
\end{align}$
We take $-2\sin x$ common and get
$\begin{align}
  & \cos 2x+\cos 2y-\cos 2z \\
 & =1-2\sin x\left[ \sin x-\sin \left( z-y \right) \right] \\
\end{align}$
We again change the $\sin x$ in the bracket. $\sin x=\sin \left[ \pi -\left( y+z \right) \right]=\sin \left( y+z \right)$.
$\begin{align}
  & \cos 2x+\cos 2y-\cos 2z \\
 & =1-2\sin x\left[ \sin \left( y+z \right)-\sin \left( z-y \right) \right] \\
\end{align}$
We now use $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$.
We use the representation of $A=y+z;B=z-y$ and get
$\begin{align}
  & \sin \left( y+z \right)-\sin \left( z-y \right) \\
 & =2\cos \left( \dfrac{y+z+z-y}{2} \right)\sin \left( \dfrac{y+z-z+y}{2} \right) \\
 & =2\cos z\sin y \\
\end{align}$
Therefore,
$\begin{align}
  & \cos 2x+\cos 2y-\cos 2z \\
 & =1-2\sin x\left[ \sin \left( y+z \right)-\sin \left( z-y \right) \right] \\
 & =1-4\sin x\sin y\cos z \\
\end{align}$
Therefore, the correct option is option (B).

Note:
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $0\le a\le \pi $.