If A+B+C= π than prove that sinA+sinB-sinC=4.sin(1/2)A.sin(1/2)B.cos(1/2)C

Steps to prove sinA+sinB-sinC=4.sin(1/2)A.sin(1/2)B.cos(1/2)C

L.H.S

=sinA+sinB-sinC

=2sin(A+B)/2.cos(A-B)/2-2sinC/2.cosC/2

=2cosC/2.cos(A-B)/2-2sinC/2.cosC/2 [sin(A+B)/2=cosC/2]

=2cosC/2{cos(A-B)/2-sinC/2}

=2cosC/2{cos(A-B)/2-cos(A+B)/2}  [sinC/2=cos(A+B)/2]

=2cosC/2.2.sinA/2.cosA/2

=4.sin(1/2)A.sin(1/2)B.cos(1/2)C

=R.H.S

Hence sinA+sinB-sinC=4.sin(1/2)A.sin(1/2)B.cos(1/2)C is proved.

Detail Information:-

Similar Questions For You:-

•  If A+B+C=π that prove that tan(B+C)/2=cotA/2
•  If α,β are acute angles and cos2α=(3cos2β-1)/(3-cos2β) than prove that tanα=√2tanβ
• If cotθ=cos(x+y) and cotΦ=cos(x-y) show that tan(θ-Φ)=2.sinx.siny/(cos^2x+cos^2y)
• If acos(x+α)=bcos(x-α) show that (a+b)tanx=(a-b)cotα