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How can I prove that in any triangle this identity is true:

1+ cos A - cos (B-C)=(b^2+c^2)/4R^2

First of all, (b^2+c^2)/4R^2 could be written as: [(b^2)/4R^2] + [(c^2)/4R^2]

From the sine theorem we know that:

a/ sin A=b/sinB=c/sinC=2R, where R is the radius of the circle circumscribing the triangle

b/sinB=c/sinC=2R, so (b/sinB)^2=(c/sinC)^2=(2R)^2,

b^2/4R^2=(sin B)^2 and c^2/4R^2=(sin C)^2

So, [(b^2)/4R^2] + [(c^2)/4R^2]= (sin B)^2 + (sin C)^2

If A,B,C, are the angles of a triangle, then A+B+C=pi

(sin B)^2 + (sin C)^2= (sin B+sin C)^2 - 2sinBsinC

We'll try to work a bit in the left side of the equal:

so,  1+cosA-cos (B-C)=1+cosA-cosBcosC-sinBsinC

1+cosA-cosBcosC-sinBsinC=(sin B+sin C)^2 - 2sinBsinC

1+cosA-cosBcosC-sinBsinC + 2sinBsinC=(sin B+sin C)^2

1+cosA-cosBcosC+sinBsinC=(sin B+sin C)^2

 1+cosA- cos (B+C)=(sin B+sin C)^2

But, B+C=pi-A and cos (pi-A)=-cos A

1-2cos (B+C)=(sin B+sin C)^2

1-2cos 2(B+C)/2=4sin^2(B+C)/2*cos^2(B-C)/2

1-2+4sin^2(B+C)/2-4sin^2(B+C)/2*cos^2(B-C)/2=0

4sin^2(B+C)/2*sin^2(B-C)/2=1