Question

In a given triangle $\Delta ABC$ prove that ${\left( {a - b} \right)^2}{\cos ^2}\left( {c/2} \right) + {\left( {a + b} \right)^2}{\sin ^2}\left( {c/2} \right) = {c^2}$.

Answer 1

Hint: Before attempting this question one should have prior knowledge about the trigonometric identities and remember to use $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$, using this information will help you to approach closer towards the solution of the problem.

Complete step-by-step solution -
According to the given information we have ${\left( {a - b} \right)^2}{\cos ^2}\left( {c/2} \right) + {\left( {a + b} \right)^2}{\sin ^2}\left( {c/2} \right) = {c^2}$
Let take as L.H.S and \[{c^2}\] as R.H.S
Simplifying L.H.S using the trigonometric identities
\[{\left( {a - b} \right)^2}{\cos ^2}\left( {c/2} \right) + {\left( {a + b} \right)^2}{\sin ^2}\left( {c/2} \right)\]
Using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ and ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the above equation we get
\[\left( {{a^2} + {b^2} - 2ab} \right){\cos ^2}\left( {c/2} \right) + \left( {{a^2} + {b^2} + 2ab} \right){\sin ^2}\left( {c/2} \right)\]
$ \Rightarrow {a^2}{\cos ^2}\left( {c/2} \right) + {b^2}{\cos ^2}\left( {c/2} \right) - 2ab{\cos ^2}\left( {c/2} \right) + {a^2}{\sin ^2}\left( {c/2} \right) + {b^2}{\sin ^2}\left( {c/2} \right) + 2ab{\sin ^2}\left( {c/2} \right) $
$ \Rightarrow {a^2}\left( {{{\cos }^2}\left( {c/2} \right) + {{\sin }^2}\left( {c/2} \right)} \right) + {b^2}\left( {{{\sin }^2}\left( {c/2} \right) + {{\cos }^2}\left( {c/2} \right)} \right) - 2ab\left( {{{\cos }^2}\left( {c/2} \right) - {{\sin }^2}\left( {c/2} \right)} \right) $
Using the trigonometric identities \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] and \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \] in the above equation we get
\[{a^2} + {b^2} - 2ab\cos C\]
Now since we know that $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$therefore
\[{a^2} + {b^2} - 2ab\left( {\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right)\]
$ \Rightarrow $\[{a^2} + {b^2} - {a^2} - {b^2} + {c^2}\]
$ \Rightarrow $\[{c^2}\]
Since L.H.S is equal to R.H.S
Therefore ${\left( {a - b} \right)^2}{\cos ^2}\left( {c/2} \right) + {\left( {a + b} \right)^2}{\sin ^2}\left( {c/2} \right) = {c^2}$ equation hence proved.

Note: In the above solution we used the term “trigonometric identities” which can be explained as formulas which contains trigonometric function such as$\sin \theta ,\cos \theta ,\tan \theta $, etc. and are valid for each value of the variables which occur when both sides of equality are defined. The example of some trigonometric identities are$\cos \theta = \dfrac{1}{\theta }$, $\cot \theta = \dfrac{1}{{\tan \theta }}$, etc.