Question

If $x+y+z=180°$, 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. $\cos A\cos B\cos C$

Answer 1

The correct option is B

$1-4\sin x\sin y\cos z$

Explanation for the correct option.

Step 1: Solve $\cos 2x+\cos 2y$.

$\begin{array}{rcl}\cos 2x+\cos 2y& =& 2\cos \frac{2x+2y}{2}\cos \frac{2x-2y}{2}\left[cosA+cosB=2cos\frac{A+B}{2}cos\frac{A-B}{2}\right]\\ & \Rightarrow & \cos 2x+\cos 2y=2\cos \left(x+y\right)\cos \left(x-y\right)...\left(1\right)\end{array}$

Step 2: Solve $\cos 2x+\cos 2y-\cos 2z$

$$\begin{gathered} \cos 2x+\cos 2y-\cos 2z \\ =2\cos \left(x+y\right)\cos \left(x-y\right)-\left(2{\cos }^{2}z-1\right)\left[by\cos 2A=2{\cos }^{2}A-1\right] \\ =2\cos \left(x+y\right)\cos \left(x-y\right)-2{\cos }^{2}z+1 \\ =2\cos \left(x+y\right)\cos \left(x-y\right)-2{\cos }^2\left[180°-\left(x+y\right)\right]+1\left[givenx+y+z=180°\right] \\ =2\cos \left(x+y\right)\cos \left(x-y\right)-2{\left[-\cos \left(x+y\right)\right]}^2+1\left[by\cos \left(180°-\theta \right)=-\cos \theta \right] \\ =2\cos \left(x+y\right)\cos \left(x-y\right)-2\left[{\cos }^2\left(x+y\right)\right]+1 \\ =2\cos \left(x+y\right)\left[\cos \left(x-y\right)-\cos \left(x+y\right)\right]+1 \\ =2\cos \left(x+y\right)\left[2\sin \frac{2x}{2}\sin \frac{2y}{2}\right]+1\mathbf{}\left[cosA-cosB=2sin\frac{A+B}{2}sin\frac{B-A}{2}\right] \\ =4\cos \left(x+y\right)\left(\sin x\sin y\right)+1 \\ =4\cos \left(180°-z\right)\left(\sin x\sin y\right)+1 \\ =-4\cos z\left(\sin x\sin y\right)+1 \\ =1-4\sin x\sin y\cos z \end{gathered}$$

Hence, option B is correct.