Show that area K of a triangle ABC is K a2 sin 2B b2 sin 2A

Show that area K, of a triangle ABC is K= a^2 sin 2B + b2 sin 2A/4 = b^2 sin 2C + c^2 sin 2B/4 = c^2 sin 2A + a^2 sin 2C/4

Solution

Consider the second equation (b²Sin2C + c²Sin2B) / 4

Substituting the value for Sin2C = 2 SinACosA, we get

(2b²SinC CosC + 2c²SinB CosB) /4----------------------------------------------------------(i)
Now from Cosine rule we have,
CosC = (a² + b² - c²⁾ / 2ab and CosB = (a² + c² - b²⁾ / 2ac

Substituting CosC and CosB in eq (i), we get
(b(a² + b² - c²)SinC + c( a² + c² -b²) SinB) / 4a

Also from Sine Rule, we have

SinC / c = Sin A / a ; So SinC = cSinA / a

Similarly SinB = bSinA / a

Substituting the same in above equation, we get

{ b (a² + b² - c²) (cSinA / a) + c(a² + c² - b²) (bSinA/a)} / 4a

Simplying we get,

( bcSinA (a²+ b² -c²⁾ + bcSinA (a² +c² -b²)) / 4a²

= 2a²bcSinA / 4a2
= bcSinA/2-------------- which is the formula for the area of the traingle.

Similarly the other two equations can also be proved by same method,