Solutions for Triangles

Square the equation and replace the cos squared term using the identity

$\displaystyle \cos^{2}(A/2)=(1/2)(1+\cos(A)).$

Use the cosine rule to replace the cosine term with (b^2 + c^2 - a^2)/(2bc), and simplify.

That should get you to 1 - a^2/((b + c)^2)), so finally cos^2 = 1 - sin^2.

Solutions for Triangles

$\begin{array}{|rcll|} \hline \mathbf{x^2+a^2} &=& \mathbf{(b+c)^2} \\ x^2+a^2 &=& b^2+2bc+c^2 \\ x^2&=& b^2+2bc+c^2-a^2 \\ x &=& \sqrt{b^2+2bc+c^2-a^2} \\ x &=& \sqrt{2bc+ \mathbf{b^2+c^2-a^2}} \\\\ && \boxed{a^2=b^2+c^2-2bc\cos(A)\\ \mathbf{b^2+c^2-a^2 = 2bc\cos(A)} } \\\\ x &=& \sqrt{2bc+ 2bc\cos(A)} \\ x &=& \sqrt{2bc\Big(1+ \mathbf{\cos(A)}\Big)} \\\\ && \boxed{\cos(A)=\cos^2\left(\dfrac{A}{2}\right)-\sin^2\left(\dfrac{A}{2}\right) \\ \cos(A)=\cos^2\left(\dfrac{A}{2}\right)-\left(1-\cos^2\left(\dfrac{A}{2}\right) \right) \\ \mathbf{\cos(A)=2\cos^2\left(\dfrac{A}{2}\right)-1} } \\\\ x &=& \sqrt{2bc\left(1+ 2\cos^2\left(\dfrac{A}{2}\right)-1\right)} \\ x &=& \sqrt{2bc\left( 2\cos^2\left(\dfrac{A}{2}\right) \right)} \\ x &=& \sqrt{4bc \cos^2\left(\dfrac{A}{2}\right) } \\ \mathbf{x} &=& \mathbf{2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin(\theta) &=& \dfrac{x}{b+c} \quad | \quad \mathbf{x=2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)} \\\\ \mathbf{\sin(\theta)} &=& \mathbf{ \dfrac{2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)}{b+c} } \\\\ \hline \\ \cos(\theta) &=& \dfrac{a}{b+c} \\\\ \mathbf{(b+c)\cos(\theta)} &=& \mathbf{a} \\ \hline \end{array}$

Thanks for your response!! Its a great joy to watch you solve such problems...very grateful ..& great respect for your super knowledge!!

Regards

Old Timer