Trigonometric exercise asked in mathematics teacher group LeveL 3

Exercise:

Show that, if $\cos \left( B-C \right)=\frac{2\sin B\sin C}{{{\sin }^{2}}A}$ then $\vartriangle ABC$ is right

Solution: we know that the sum of angle in any triangle is $180$

hence $A+B+C=\pi $ so $\sin \left( B+C \right)=\sin \left( \pi -A \right)=\sin A$

so $\cos \left( B-C \right)\sin \left( B+C \right)=\frac{2\sin B\sin C}{\sin A}$

Apply the trigonometric identities from product to sum we get:

$\cos \left( B-C \right)\sin \left( B+C \right)=\frac{1}{2}\left( \sin \left( B-C+B+C \right)-\sin \left( B-C-B-C \right) \right)$

                              $=\frac{1}{2}\left( \sin 2B-\sin \left( -2C \right) \right)=\frac{1}{2}\left( \sin 2B+\sin 2C \right)$

Hence $\frac{1}{2}\left( \sin 2B+\sin 2C \right)=\frac{2\sin B\sin C}{\sin A}\Leftrightarrow \frac{\sin 2B+\sin 2C}{2\sin B\sin C}=\frac{2}{\sin A}$

So $\frac{2\sin B\cos B}{2\sin B\sin C}+\frac{2\sin C\cos C}{2\sin B\sin C}=\frac{2}{\operatorname{Sin}A}\Leftrightarrow \frac{\cos B}{\sin C}+\frac{\cos C}{\sin B}=\frac{2}{\sin A}$

$\Leftrightarrow \cos B\frac{\sin A}{\sin C}+\cos C\frac{\sin A}{\sin B}=2$

Take $AB=a\,\,,\,\,BC=b\,\,\And \,\,AC=c$ then by Using cosine law :

${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos B\Leftrightarrow \cos B=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$

${{a}^{2}}={{c}^{2}}+{{b}^{2}}-2cb\cos C\Leftrightarrow \cos C=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$

Also $\frac{\sin A}{\sin C}=\frac{\frac{h}{a}}{\frac{h}{b}}=\frac{b}{a}$ and $\frac{\sin A}{\sin B}=\frac{\frac{h*}{c}}{\frac{h*}{b}}=\frac{b}{c}$

So $\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\times \frac{b}{a}+\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\times \frac{b}{c}=2$ $\Leftrightarrow \frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2{{a}^{2}}}+\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2{{c}^{2}}}=2$

$\Leftrightarrow \frac{{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)+{{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}{2{{a}^{2}}{{c}^{2}}}=2$ $\Leftrightarrow {{c}^{2}}{{a}^{2}}+{{c}^{2}}{{b}^{2}}-{{c}^{4}}+{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}-{{a}^{4}}=4{{a}^{2}}{{c}^{2}}$

$\Leftrightarrow {{c}^{2}}{{a}^{2}}+{{c}^{2}}{{b}^{2}}+{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}={{a}^{4}}+{{c}^{4}}+2{{a}^{2}}{{c}^{2}}+2{{a}^{2}}{{c}^{2}}={{\left( {{a}^{2}}+{{c}^{2}} \right)}^{2}}+2{{a}^{2}}{{c}^{2}}$

$\Leftrightarrow {{b}^{2}}\left( {{a}^{2}}+{{c}^{2}} \right)+2{{a}^{2}}{{c}^{2}}={{\left( {{a}^{2}}+{{c}^{2}} \right)}^{2}}+2{{a}^{2}}{{c}^{2}}\Leftrightarrow {{b}^{2}}\left( {{a}^{2}}+{{c}^{2}} \right)={{\left( {{a}^{2}}+{{c}^{2}} \right)}^{2}}$

$\Leftrightarrow {{b}^{2}}={{a}^{2}}+{{c}^{2}}$ hence the triangle is right at $A$ and satisfied the Pythagoras theorem