Trigonometric Identity?

Question

For any triangle ABC, show that

sin(A - B)/sin(A + B) = (a² - b²)/c²

Answer 1

sin(180-x) = sin(x)
C = 180 - (A+B)
sinC = sin(180 - (A+B)) = sin(A+B)

By the sine rule:
sinA/a = sinB/b = sinC/c
sinA/sinC = a/c
sinB/sinC = b/c

By the cosine rule:
a² = b² + c² - 2bc.cosA
cosA = (b² + c² - a²)/2bc
Similarly:
cosB = (a² + c² - b²)/2ac
cosC = (a² + b² - c²)/2ab

sin(A-B)/sin(A+B)
= sin(A-B) / sinC
= (sinAcosB - cosAsinB) / sinC
= (a/c)(a² + c² - b²)/2ac - (b/c)(b² + c² - a²)/2bc
= (a² + c² - b²)/2c² - (b² + c² - a²)/2c²
= (a² + c² - b² - b² - c² + a²) / 2c²
= (2a² - 2b²) / (2c²)
= (a² - b²) / c²

Answer 2

OK, according to properties, sin (A-B) is the same as sin A / B and sin (A+B) =sin A times the (sin B). so:::

sin A / sin B all divided by sin A sin A so:::

sin A/ sin B sin A sin A simplify:::

sin sin B/ sin A which is equal to sin (B-A)

from there, I don't know where to go, but hopefully this will help some.

Good Luck! God Bless!

Answer 3

sin(A -- B) / sin(A + B)
=sin(A + B)sin(A -- B) / sin^2(A + B)
= {sin^2A -- sin^2B} / sin^2C
[as A + B = 180 deg -- C]
= (a^2 -- b^2) / c^2
[as a/sinA = b/sinB = c/sinC]