Question

If in a triangle $\frac{\sin A}{\sin C}=\frac{\sin \left(A-B\right)}{\sin \left(B-C\right)}$, then

• A. ${\sin }^{2}A,{\sin }^{2}B,{\sin }^{2}C$ are in GP
• B. $\sin 2A,\sin 2B,\sin 2C$ are in AP
• C. $\cos 2A,\cos 2B,\cos 2C$ are in AP
• D. $a\sin A,b\sin B,c\sin C$ are in GP

Answer 1

The correct option is C

$\cos 2A,\cos 2B,\cos 2C$ are in AP

Explanation for the correct option:

Find the correct relation.

Cross multiply the given equation $\frac{\sin A}{\sin C}=\frac{\sin \left(A-B\right)}{\sin \left(B-C\right)}$ and use trigonometric identities to simplify it.

$$\begin{gathered} \sin A\sin \left(B-C\right)=\sin \left(A-B\right)\sin C \\ \Rightarrow \sin \left(180°-\left(B+C\right)\right)\sin \left(B-C\right)=\sin \left(A-B\right)\sin \left(180°-\left(A+B\right)\right)\left[A+B+C=180°\right] \\ \Rightarrow \sin \left(B+C\right)\sin \left(B-C\right)=\sin \left(A-B\right)\sin \left(A+B\right) \\ \Rightarrow {\sin }^{2}B-{\sin }^{2}C={\sin }^{2}A-{\sin }^{2}B\left[sin\left(A+B\right)sin\left(A-B\right)=si{n}^{2}A-si{n}^{2}B\right] \\ \Rightarrow 2{\sin }^{2}B={\sin }^{2}A+{\sin }^{2}C \\ \Rightarrow -2\times 2{\sin }^{2}B=-2{\sin }^{2}A-2{\sin }^{2}C \\ \Rightarrow 2\left(1-2{\sin }^{2}B\right)=\left(1-2{\sin }^{2}A\right)+\left(1-2{\sin }^{2}C\right) \\ \Rightarrow 2\cos 2B=\cos 2A+\cos 2C\left[cos2\theta =1-2si{n}^2\theta \right] \end{gathered}$$

So for the terms $\cos 2A,\cos 2B,\cos 2C$ it is found that $2\cos 2B=\cos 2A+\cos 2C$. And thus $\cos 2A,\cos 2B,\cos 2C$ are in AP.

Hence, the correct option is C.