Question

In $\Delta ABC,$ the value of $(a-b{)}^2{\cos }^2\frac{C}{2}+(a+b{)}^2{\sin }^2\frac{C}{2}=$

• A. $a^2+b^2+c^2$
• B. $a^2$
• C. $b^2$
• D. $c^2$

Answer 1

The correct option is D $c^2$

$=(a-b{)}^2{\cos }^2\frac{C}{2}+(a+b{)}^2{\sin }^2\frac{C}{2}$
$=(a^2+b^2-2ab){\cos }^2\frac{C}{2}+(a^2+b^2+2ab){\sin }^2\frac{C}{2}$ $=(a^2+b^2)({\cos }^2\frac{C}{2}+{\sin }^2\frac{C}{2})-2ab({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2})$
$=(a^2+b^2)-2ab(2{\cos }^2\frac{C}{2}-1)$
$=(a^2+b^2)-2ab(2\frac{s(s-c)}{ab}-1)$
$=(a^2+b^2)-2ab(\frac{2\times \frac{(a+b+c)}{2}(\frac{a+b+c}{2}-c)}{ab}-1)$
$=(a^2+b^2)-2ab\left(\frac{(a+b+c)(a+b-c)-2ab}{2ab}\right)$
$=(a^2+b^2)-2ab\left(\frac{a^2+b^2+2ab-c^2-2ab}{2ab}\right)$
$=(a^2+b^2)-(a^2+b^2-c^2)$
$=c^2$