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Question: If \(x+y+z={{180}^{\circ }}\), then \(\cos 2x+\cos 2y-\cos 2z\) is equal to A. \(4\sin x\sin y\si...

If x+y+z=180∘x+y+z={{180}^{\circ }}, then cos⁡2x+cos⁡2y−cos⁡2z\cos 2x+\cos 2y-\cos 2z is equal to
A. 4sin⁡xsin⁡ysin⁡z4\sin x\sin y\sin z
B. 1−4sin⁡xsin⁡ycos⁡z1-4\sin x\sin y\cos z
C. 4sin⁡xsin⁡ysin⁡z−14\sin x\sin y\sin z-1
D. cos⁡xcos⁡ycos⁡z\cos x\cos y\cos z

Explanation

Solution

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

Complete step by step answer:
It is given that x+y+z=πx+y+z=\pi . We get y+z=π−xy+z=\pi -x and x=π−(y+z)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=2sin⁡(A+B2)sin⁡(B−A2)\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=2zA=2y;B=2z and get
cos⁡2y−cos⁡2z =2sin⁡(2y+2z2)sin⁡(2z−2y2) =2sin⁡(y+z)sin⁡(z−y) \begin{aligned} & \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{aligned}
We change the angle and get
2sin⁡(y+z)sin⁡(z−y) =2sin⁡(π−x)sin⁡(z−y) =2sin⁡xsin⁡(z−y) \begin{aligned} & 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{aligned}
We also use the multiple angle formula of cos⁡2x=1−2sin⁡2x\cos 2x=1-2{{\sin }^{2}}x.
Therefore,
cos⁡2x+cos⁡2y−cos⁡2z =1−2sin⁡2x+2sin⁡(y+z)sin⁡(z−y) =1−2sin⁡2x+2sin⁡xsin⁡(z−y) \begin{aligned} & \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{aligned}
We take −2sin⁡x-2\sin x common and get
cos⁡2x+cos⁡2y−cos⁡2z =1−2sin⁡x[sin⁡x−sin⁡(z−y)] \begin{aligned} & \cos 2x+\cos 2y-\cos 2z \\\ & =1-2\sin x\left[ \sin x-\sin \left( z-y \right) \right] \\\ \end{aligned}
We again change the sin⁡x\sin x in the bracket. sin⁡x=sin⁡[π−(y+z)]=sin⁡(y+z)\sin x=\sin \left[ \pi -\left( y+z \right) \right]=\sin \left( y+z \right).
cos⁡2x+cos⁡2y−cos⁡2z =1−2sin⁡x[sin⁡(y+z)−sin⁡(z−y)] \begin{aligned} & \cos 2x+\cos 2y-\cos 2z \\\ & =1-2\sin x\left[ \sin \left( y+z \right)-\sin \left( z-y \right) \right] \\\ \end{aligned}
We now use sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\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−yA=y+z;B=z-y and get
sin⁡(y+z)−sin⁡(z−y) =2cos⁡(y+z+z−y2)sin⁡(y+z−z+y2) =2cos⁡zsin⁡y \begin{aligned} & \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{aligned}
Therefore,
cos⁡2x+cos⁡2y−cos⁡2z =1−2sin⁡x[sin⁡(y+z)−sin⁡(z−y)] =1−4sin⁡xsin⁡ycos⁡z \begin{aligned} & \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{aligned}
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 −∞≤x≤∞-\infty \le x\le \infty . In that case we have to use the formula x=nπ±ax=n\pi \pm a for cos⁡(x)=cos⁡a\cos \left( x \right)=\cos a where 0≤a≤π0\le a\le \pi .