Question

If \[ABC\] is a triangle, then find the equivalent expression for \[{\left( {a - b} \right)^2}{\cos ^2}\dfrac{C}{2} + {\left( {a + b} \right)^2}{\sin ^2}\dfrac{C}{2}\].
A. \[{a^2}\]
B. \[{b^2}\]
C. \[{c^2}\]
D. None of these

Answer 1

Hint: We will apply the square identity in the given expression and simplify it. After that, we will apply cosine law.

Formula used:
Algebraical Identity:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Cosine laws:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
The given expression is \[{\left( {a - b} \right)^2}{\cos ^2}\dfrac{C}{2} + {\left( {a + b} \right)^2}{\sin ^2}\dfrac{C}{2}\].
Apply the algebraical identity
 \[ = \left( {{a^2} + {b^2} - 2ab} \right){\cos ^2}\dfrac{C}{2} + \left( {{a^2} + {b^2} + 2ab} \right){\sin ^2}\dfrac{C}{2}\]
\[ = \left( {{a^2} + {b^2}} \right){\cos ^2}\dfrac{C}{2} - 2ab{\cos ^2}\dfrac{C}{2} + \left( {{a^2} + {b^2}} \right){\sin ^2}\dfrac{C}{2} + 2ab{\sin ^2}\dfrac{C}{2}\]
\[ = \left( {{a^2} + {b^2}} \right)\left( {{{\cos }^2}\dfrac{C}{2} + {{\sin }^2}\dfrac{C}{2}} \right) - 2ab{\cos ^2}\dfrac{C}{2} + 2ab{\sin ^2}\dfrac{C}{2}\]
Now applying trigonometry identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]:
\[ = \left( {{a^2} + {b^2}} \right) \cdot 1 - 2ab{\cos ^2}\dfrac{C}{2} + 2ab{\sin ^2}\dfrac{C}{2}\]
\[ = \left( {{a^2} + {b^2}} \right) \cdot 1 + 2ab\left( {{{\sin }^2}\dfrac{C}{2} - {{\cos }^2}\dfrac{C}{2}} \right)\]
Apply the formula \[{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta \]
\[ = \left( {{a^2} + {b^2}} \right) \cdot 1 + 2ab\left( { - \cos C} \right)\]
\[ = {a^2} + {b^2} - 2ab\cos C\]
Applying cosine law \[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
\[ = {c^2}\]
Hence option C is the correct option.

Additional information:
There are some few laws on triangle. The sine laws are
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]

Note: Students often do a mistake to apply the formula for \[\cos 2\theta \]. They apply a wrong formula that is \[{\sin ^2}\theta - {\cos ^2}\theta = \cos 2\theta \]. The correct formula is \[{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta \].