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Show that bcosB+ccosC=acos(b-c) in triangle?
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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.
Posted by sciencesolve on October 2, 2012 at 5:43 PM (Answer #1)
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`
`= bcosB+c cosC`
Therefore `bcosB+c cosC = acos(B-C)` represent a triangle.
Posted by jeew-m on October 2, 2012 at 6:00 PM (Answer #2)
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