Q. Let $P Q R$ be a triangle of area $\Delta$ with $a=2, b=7 / 2$, and $c=5 / 2$, where $a, b$, and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$, and $R$, respectively. Then $\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}$ equals
Trigonometric Functions
Solution:
$\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}$
$=\frac{1-\cos P}{1+\cos P}=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}$
$=\tan ^{2} \frac{P}{2}$
$=\frac{(s-b)(s-c)}{s(s-a)}$
$=\frac{((s-b)(s-c))^{2}}{\Delta^{2}}$
$=\frac{\left(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\right)^{2}}{\Delta^{2}}$
$=\left(\frac{3}{4 \Delta}\right)^{2}$
