Question

Using Cosine rule prove the Sine rule.​

Answer 1

Step-by-step explanation:

Law of cosines––––––––––––––

cosA=b²+c²−a² / 2bc

cosB=a²+c²−b² / 2ac

cosC=a²+b²−c² / 2ab

(sinA/a)²=sin2²A/a²

=1−cos²A/a²={1−(b²+c²−a²/2bc)²} / a²

=(2bc)²−(b²+c²−a²)² / a²(2bc)²

=(2bc+b²+c²−a²)(2bc−b²−c²+a2) / (2abc)²

={(b+c)²−a²}{a²−(b²+c²−2bc)} / (2abc)2

={(b+c)²−a²}{a²−(b−c)²} / (2abc)²

=(b+c+a)(b+c−a)(a+b−c)(a−b+c) / (2abc)² .........(1)

(sinB/b)²=sin²B/b²

=1−cos²B / b²={1−(a²+c²−b²/2ac)²} / b²

=(2ac)²−(a²+c²−b²)² / b²(2ac)²

=(2ac+a²+c²−b²)(2ac−a²−c²+b²) / (2abc)²

={(a+c)²−b²}{b²−(a²+c²−2ac)} / (2abc)²

={(a+c)²−b²}{b²−(a−c)²} / (2abc)²

=(a+c+b)(a+c−b)(b+a−c)(b−a+c) / (2abc)²

=(b+c+a)(b+c−a)(a+b−c)(a−b+c) / (2abc)²by re-arrangement.......(2)

by (1) and (2)

(sinA/a)²=(sinB/b)²

⟹sinA/a=sinB/b

the expression(b+c+a)(b+c−a)(a+b−c)(a−b+c) / (2abc)²

is cyclic in nature. When we replace a by b, b by c and c by a we get the same expression.

∴it can be proved in a similar way that

(sinC/c)²=(b+c+a)(b+c−a)(a+b−c)(a−b+c) / (2abc)²

∴sinA/a=sinB/b=sinC/c