Question

In triangle $ABC$ if $\overline{AB}=\frac{\overline{u}}{\left|\overline{u}\right|}-\frac{\overline{v}}{\left|\overline{v}\right|}$ and $\overline{AC}=\frac{2\overline{u}}{\left|\overline{u}\right|}$ where $\left|\overline{u}\right|\ne \left|\overline{v}\right|$ then

• A. $1+\sum \cos 2A=0$
• B. $\sum \cos 2A=0$
• C. $2+\sum \cos 2A=0$
• D. $\sum \cos 4A=0$

Answer 1

The correct option is A $1+\sum \cos 2A=0$

$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$

$=\hat{u}+\hat{v}$

Let angle between $\hat{u}\&\hat{v}$ is $\alpha$

$So,\cos A=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|}=\frac{2{\left|\hat{u}\right|}^2-2\hat{u}.\hat{v}}{\sqrt{{\left|\hat{u}\right|}^2+{\left|\hat{v}\right|}^2-2\left|\hat{u}\right|\left|\hat{u}\right|\cos \alpha .}2\left|\hat{u}\right|}$

$=\frac{2-\mathrm{2.1.}\cos \alpha }{\sqrt{2-2\cos \alpha }.2}=\frac{2-2\cos \alpha }{\sqrt{2-2\cos \alpha }.2}$

$So,\cos 2A=2{\cos }^{2}A-1=-2\cos \alpha$

$\cos \left(\pi -\beta \right)=\frac{\overrightarrow{AB}.\overrightarrow{BC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{BC}\right|}=\frac{{\left|\hat{u}\right|}^2-{\left|\hat{v}\right|}^2}{\left|\hat{u}+\hat{v}\right|\left|\hat{u}-\hat{v}\right|}=0$

$\Rightarrow -\cos B=0\Rightarrow B=\frac{\pi }{2}$

So,

$\cos 2B=2{\cos }^{2}B-1=0-1=-1$

$\cos \left(\pi -C\right)=\frac{\left(\overrightarrow{BC}\right).\left(\overrightarrow{AC}\right)}{\left|\overrightarrow{BC}\left|\overrightarrow{AC}\right|\right|}$

$-\cos C=\frac{\left(\hat{u}+\hat{v}\right)\left(2\hat{u}\right)}{\sqrt{{\left|\hat{u}\right|}^2+{\left|\hat{v}\right|}^2+2\left|\hat{u}\right|\left|\hat{v}\right|\cos \alpha .2\left|\hat{u}\right|}}$

$=\frac{2{\left|\hat{u}\right|}^2+\hat{u}.\hat{v}}{\sqrt{2+2\cos \alpha }.2}=\frac{2+2\cos \alpha }{\sqrt{2+2\cos \alpha }.2}$

$\therefore \cos C=\frac{-\left(2+2\cos \alpha \right)}{\sqrt{2+2\cos \alpha }.2}$

$So,\cos 2C=2{\cos }^{2}C-1=2.\frac{{\left(2+2\cos \alpha \right)}^2}{\left(2+2\cos \alpha \right).4}-1$

$So,\cos 2A+\cos 2B+\cos 2C=-2\cos \alpha -1+2\cos \alpha$

$\Rightarrow \sum \cos 2A=-1$

$\Rightarrow 1+\sum \cos 2A=0$

Hence, the option $\left(A\right)$ is the correct answer.