Using the sine rule , prove the cosine rule.

Explanation

By sine Rule, we have
`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)=k("say")`
`:.a=k.sinA,b=k.sinB,c=k.sinC`
Now consider `(b^(2)+c^(2)-a^(2))/(2bc)`
`=(k^(2).sin^(2)B+k^(2).sin^(2)C-k^(2)sin^(2)A)/(2.ksinB.k.sinC)`
`=(k^(2)(sin^(2)B+sin^(2)C-sin^(2)A))/(2.ksinB.sinC)`
`=(sin^(2)B+sin^(2)C-sin^(2)A)/(2.sinB.sinC)`
`=(2sin^(2)B+2sin^(2)C-2sin^(2)A)/(4sinB.sinC)`
[Multiplying numerator and denominator by 2.]
`=(2sin^(2)B+(1-cos2c)-(1-cos2A))/(4.sinB.sinC)`
`=(2sin^(2)B+(cos2A-cos2C))/(4.sinB.sinC)`
`=(2sin^(2)B+2sin(A+C).sin(C-A))/(4.sinB.sinC)`
`=(2sin^(2)B+2sinB.sin(C-A))/(4sinB.sinC)`
`[{:("":'A+C,=pi-B),("":.sin(A+C),=sin(pi-B)),(,=sinB):}]`
`=(2sinB(sinB+sin(C-A)))/(4sinB.sinC)`
`=(sinB+sin(C-A))/(2sinC)`
`=(sin(C+A)+sin(C-A))/(2sinC)`
`[:'sinB=sin(A+C)]`
`=(2sinC.cosA)/(2sinC)=cosA`
Hence, `(b^(2)+c^(2)-a^(2))/(2bc)=cosA`
Similarly, we can prove `cosB=(c^(2)+a^(2)-b^(2))/(2ca)`
`cosC=(a^(2)+b^(2)-c^(2))/(2ab)`
Hence, the cosine rule is proved.