prove that (cos a-cosb)^2+(sina-sinb)^2=4 sin^2(a-b/2)
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Prove (cosA+cosB)^2+(sinA-sinB)^2 = 4cos^2(A+B)/2
(cosA+cosB)^2 + (sinA-sinB)^2
=> (cos^2A + cos^2B + 2cosAcosB) + (sin^2A + sin^2B - 2sinAsinB)
=> cos^2A + cos^2B + sin^2A + sin^2B + 2cosAcosB - 2sinAsinB
=> cos^2A + sin^2A + cos^2B + sin^2B + 2(cosA*cosB - sinA*sinB)
=> 1 + 1 + 2(cosA*cosB - sinA*sinB)
=> 2 + 2(cosA*cosB - sinA*sinB)
=> 2 (1 + (cosA*cosB + sinA*sinB))
=> 2 * (1 + cos(A-B))
{Because: cosA*cosB - sinA*sinB = cos(A+B)}
=> 2 * 2cos^2 ((A+B)/2)
=> 4cos^2 (A+B)/2nd 3
(cosA+cosB)^2 + (sinA-sinB)^2
=> (cos^2A + cos^2B + 2cosAcosB) + (sin^2A + sin^2B - 2sinAsinB)
=> cos^2A + cos^2B + sin^2A + sin^2B + 2cosAcosB - 2sinAsinB
=> cos^2A + sin^2A + cos^2B + sin^2B + 2(cosA*cosB - sinA*sinB)
=> 1 + 1 + 2(cosA*cosB - sinA*sinB)
=> 2 + 2(cosA*cosB - sinA*sinB)
=> 2 (1 + (cosA*cosB + sinA*sinB))
=> 2 * (1 + cos(A-B))
{Because: cosA*cosB - sinA*sinB = cos(A+B)}
=> 2 * 2cos^2 ((A+B)/2)
=> 4cos^2 (A+B)/2nd 3
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