# Show that bcosB+ccosC=acos(b-c) in triangle?

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We know that;

`sinx-siny = 2sin((x-y)/2)*cos((x+y)/2)`

From sin law for a triangle;

`sinA/a = sinB/b = sinC/c = k` where k is a constant and other letters have standard definitions.

`sinA/a = sinB/b = sinC/c = k `

`a = sinA/k`

But for a triangle` A+B+C = 180`

Therefore `A = 180-(B+C)`

`a = sin(180-(B+C))/k`

`a = (sin(B+C))/k`

`acos(B-C)`

`= (sinA/k)*cos(B-C)`

`= (sin(B+C))/k*cos(B-C)`

`= 1/2(sin(2B)-sin(-2C))/k`

`= (1/2)(1/k)(2cosBsinB+2sinCcosC)`

`= (sinB/k)cosB+(sinC/k)cosC`

`= bcosB+c cosC`

*Therefore `bcosB+c cosC = acos(B-C)` represent a triangle.*

**Sources:**

You should use the following notations for the sides of triangle such that:

a represents the side opposed to angle A

b represents the side opposed to angle B

c represents the side opposed to angle C

Using the law of sines yields:

`a/sin A = 2R, b/sin B = 2R, c/sin C = 2R`

You should substitute `2Rsin B` for b and `2R sin C` for c such that:

`2Rsin B cosB + 2R sin C cos C = 2R sin Acos (B-C)`

Factoring out R yields:

`R(2sin B cos B + 2 sin C cos C) = 2R sin Acos (B-C)`

Reducing by R yields:

`sin 2B + sin 2C = 2sin Acos (B-C)`

You need to convert the sum of sines into a product such that:

`2 sin ((2B+2C)/2) cos ((2B-2C)/2) = 2sin Acos (B-C)`

`2 sin (B+C) cos (B-C) = 2sin Acos (B-C)`

Notice that A,B,C represents the angles of triangle, hence, `A+B+C = 180^o` =>`B+C = 180^o - A` such that:

`sin (B+C) = sin(180^o - A) = sin180^o cosA - sin A*cos 180^o`

Since `sin180^o = 0` and `cos180^o = -1` yields:

`sin (B+C) = sin A`

Hence, substituting `sin (B+C)` for `sin A` to the right side yields:

`2 sin (B+C) cos (B-C) = 2 sin (B+C) cos (B-C)`

**Hence, using the law of sines in triangle ABC yields that the given identity holds.**