# Proof using Law of Sines

• Aug 26th 2011, 08:44 AM
scruz10
Proof using Law of Sines
For any triangle ABC, even if B or C is an obtuse angle, a = b cos C + c cos B. Use the Law of Sines to deduce the "addition formula"

sin (B + C) = sin B cos C + sin C cos B

I got most of it but can't figure out the end.
I replaced b with 2R sin B and c with 2R sin C

so
a = 2R sin B cos C + 2R sin C cos B
a = 2R (sin B cos C + sin C cos B)
a/2R = sin B cos C + sin C cos B
sin A = sin B cos C + sin C cos B

I don't know where to go from here
• Aug 26th 2011, 09:22 AM
arangu1508
Re: Proof using Law of Sines
I Think in a triangle, A + B + C = 180 degrees. Therefore, A = 180 - (B + C). Subtituting in the equation Sin A = Sin B Cos C + Sin C cos B, we get Sin [180 - (B + C)] = Sin B Cos C + Sin C cos B. Since in the Second Quadrant, Sine is positive and Sine remains Sine, Sin (B + C) = Sin B Cos C + Sin C cos B. Hence proved. I do not know how to express mathematically. Hope it is right. Good Souls comment on this.
• Aug 26th 2011, 11:56 AM
scruz10
Re: Proof using Law of Sines
Sin A = Sin (180 - [B + C]) = Sin (180 - B - C) = Sin (B + C)

Does this sound right?
• Aug 27th 2011, 12:29 AM
arangu1508
Re: Proof using Law of Sines
Quote:

Originally Posted by scruz10
Sin A = Sin (180 - [B + C]) = Sin (180 - B - C) = Sin (B + C)

Does this sound right?

s(Evilgrin)