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Question: Trigonometric function: In \( \Delta ABC \) with usual notations. Prove that – \( {(a - b)^2}{\co...

Trigonometric function: In ΔABC\Delta ABC with usual notations. Prove that –
(a−b)2cos⁡2(c2)+(a+b)2sin⁡2(c2)=c2{(a - b)^2}{\cos ^2}\left( {\dfrac{c}{2}} \right) + {(a + b)^2}{\sin ^2}\left( {\dfrac{c}{2}} \right) = {c^2}

Explanation

Solution

Hint : Follow step by step approach; apply sine and cosine identity and the expansion of the whole square formula. Also, substitute the semi-perimeter values and then simplify the equations for the resultant answer.

Complete step-by-step answer :
Take the Left hand side of the given equation –
LHS =(a−b)2cos⁡2(c2)+(a+b)2sin⁡2(c2)= {(a - b)^2}{\cos ^2}\left( {\dfrac{c}{2}} \right) + {(a + b)^2}{\sin ^2}\left( {\dfrac{c}{2}} \right)
Open the brackets using the algebraic expansion for the square of the difference of two terms and the square of the addition of two terms and place it in the above equation –
LHS =(a2−2ab+b2)cos⁡2(c2)+(a2+2ab+b2)sin⁡2(c2)= ({a^2} - 2ab + {b^2}){\cos ^2}\left( {\dfrac{c}{2}} \right) + ({a^2} + 2ab + {b^2}){\sin ^2}\left( {\dfrac{c}{2}} \right)
Open the brackets and multiply the terms. When you multiply one positive term with positive it gives positive term but when you multiply one negative term with positive it gives negative term.
LHS =a2cos⁡2(c2)−2abcos⁡2(c2)+b2cos⁡2(c2)+a2sin⁡2(c2)+2absin⁡2(c2)+b2sin⁡2(c2)= {a^2}{\cos ^2}\left( {\dfrac{c}{2}} \right) - 2ab{\cos ^2}\left( {\dfrac{c}{2}} \right) + {b^2}{\cos ^2}\left( {\dfrac{c}{2}} \right) + {a^2}{\sin ^2}\left( {\dfrac{c}{2}} \right) + 2ab{\sin ^2}\left( {\dfrac{c}{2}} \right) + {b^2}{\sin ^2}\left( {\dfrac{c}{2}} \right)
Make the pair of like terms –
LHS=a2cos⁡2(c2)+a2sin⁡2(c2)‾−2abcos⁡2(c2)+2absin⁡2(c2)‾+b2cos⁡2(c2)+b2sin⁡2(c2)‾= \underline {{a^2}{{\cos }^2}\left( {\dfrac{c}{2}} \right) + {a^2}{{\sin }^2}\left( {\dfrac{c}{2}} \right)} - \underline {2ab{{\cos }^2}\left( {\dfrac{c}{2}} \right) + 2ab{{\sin }^2}\left( {\dfrac{c}{2}} \right)} + \underline {{b^2}{{\cos }^2}\left( {\dfrac{c}{2}} \right) + {b^2}{{\sin }^2}\left( {\dfrac{c}{2}} \right)}
Take out the common multiples from the above paired terms-
LHS =a2(cos⁡2(c2)+sin⁡2(c2))+2ab(sin⁡2(c2)−cos⁡2(c2))+b2(cos⁡2(c2)+sin⁡2(c2))= {a^2}\left( {{{\cos }^2}\left( {\dfrac{c}{2}} \right) + {{\sin }^2}\left( {\dfrac{c}{2}} \right)} \right) + 2ab\left( {{{\sin }^2}\left( {\dfrac{c}{2}} \right) - {{\cos }^2}\left( {\dfrac{c}{2}} \right)} \right) + {b^2}\left( {{{\cos }^2}\left( {\dfrac{c}{2}} \right) + {{\sin }^2}\left( {\dfrac{c}{2}} \right)} \right)
Using the identity - cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 in the above equation-
LHS =a2(1)+2ab(sin⁡2(c2)−cos⁡2(c2))+b2(1)= {a^2}\left( 1 \right) + 2ab\left( {{{\sin }^2}\left( {\dfrac{c}{2}} \right) - {{\cos }^2}\left( {\dfrac{c}{2}} \right)} \right) + {b^2}\left( 1 \right)
It can be re-written as –
LHS =a2+b2+2ab(sin⁡2(c2)−cos⁡2(c2))= {a^2} + {b^2} + 2ab\left( {{{\sin }^2}\left( {\dfrac{c}{2}} \right) - {{\cos }^2}\left( {\dfrac{c}{2}} \right)} \right)
Again place the identity, cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta in the above equation-
LHS =a2+b2+2ab(sin⁡2(c2)−[1−sin⁡2(c2)])= {a^2} + {b^2} + 2ab\left( {{{\sin }^2}\left( {\dfrac{c}{2}} \right) - \left[ {1 - {{\sin }^2}\left( {\dfrac{c}{2}} \right)} \right]} \right)
Open the brackets; there is negative sign outside the bracket, so sign of the terms inside the bracket will change. Positive term will become negative and vice-versa.
LHS =a2+b2+2ab(sin⁡2(c2)‾−1+sin⁡2(c2)‾)= {a^2} + {b^2} + 2ab\left( {\underline {{{\sin }^2}\left( {\dfrac{c}{2}} \right)} - 1 + \underline {{{\sin }^2}\left( {\dfrac{c}{2}} \right)} } \right)
Add the like terms and simplify the above equation –
LHS =a2+b2+2ab(2sin⁡2(c2)−1)= {a^2} + {b^2} + 2ab\left( {2{{\sin }^2}\left( {\dfrac{c}{2}} \right) - 1} \right)
But, sin⁡c2=(s−a)(s−b)ab\sin \dfrac{c}{2} = \sqrt {\dfrac{{(s - a)(s - b)}}{{ab}}} taking square on both the sides of the equation.
⇒sin⁡2c2=(s−a)(s−b)ab\Rightarrow {\sin ^2}\dfrac{c}{2} = \dfrac{{(s - a)(s - b)}}{{ab}} place this relation in the above LHS
LHS =a2+b2+2ab((2(s−a)(s−b)ab)−1)= {a^2} + {b^2} + 2ab\left( {\left( {\dfrac{{2(s - a)(s - b)}}{{ab}}} \right) - 1} \right)
Take LCM and simplify –
LHS =a2+b2+2ab((2(s−a)(s−b)−abab))= {a^2} + {b^2} + 2ab\left( {\left( {\dfrac{{2(s - a)(s - b) - ab}}{{ab}}} \right)} \right)
Also, semi-perimeter s=a+b+c2s = \dfrac{{a + b + c}}{2} place it in the above equation –
LHS =a2+b2+2ab((2(a+b+c2−a)(a+b+c2−b)ab−1))= {a^2} + {b^2} + 2ab\left( {\left( {\dfrac{{2(\dfrac{{a + b + c}}{2} - a)(\dfrac{{a + b + c}}{2} - b)}}{{ab}} - 1} \right)} \right)
Simplify the above equation –
LHS =a2+b2+2ab((2(a+b+c−2a2)(a+b+c−2b2)ab−1))= {a^2} + {b^2} + 2ab\left( {\left( {\dfrac{{2(\dfrac{{a + b + c - 2a}}{2})(\dfrac{{a + b + c - 2b}}{2})}}{{ab}} - 1} \right)} \right)
Common multiples from the numerator and the denominator cancel each other.
LHS =a2+b2+2ab((b+c−a)(a+c−b)2ab−1)= {a^2} + {b^2} + 2ab\left( {\dfrac{{(b + c - a)(a + c - b)}}{{2ab}} - 1} \right)
Now multiply the term outside the bracket.
LHS =a2+b2+((b+c−a)(a+c−b)−2ab)= {a^2} + {b^2} + \left( {(b + c - a)(a + c - b) - 2ab} \right)
The above equation can be rearranged as –
LHS =a2+b2+((c+b−a)(c+a−b)−2ab)= {a^2} + {b^2} + \left( {(c + b - a)(c + a - b) - 2ab} \right)
It is the difference of two squares formulas-
LHS =a2+b2‾+c2−(b−a)2−2ab‾= \underline {{a^2} + {b^2}} + {c^2} - {(b - a)^2}\underline { - 2ab}
Again the three terms makes the perfect square-
LHS =(b−a)2+c2−(b−a)2= {(b - a)^2} + {c^2} - {(b - a)^2}
Equal values with opposite signs cancel each other.
LHS =+c2== + {c^2} = RHS.
LHS=RHSLHS = RHS
Hence, proved.

Note : Use algebraic expansion formula correctly and be careful while opening the brackets and the sign of the terms. When there is a positive sign outside the bracket then there is no change in sign of the terms inside the bracket when you open but when there is negative sign outside the bracket then the sign of the terms inside the bracket changes when opened. Positive terms become negative and vice-versa.