cosA+cosB+cosC=1+4sinA/2. sinB/2. sinC/2
Answers
Step-by-step explanation:
A + B + C = π ...... (1)
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L.H.S.
= ( cos A + cos B ) + cos C
= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C
= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C
= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }
= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)
= 1 + 4 sin(A/2) sin(B/2) sin(C/2)
= R.H.S. ............................. Q.E.D.
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In step (2), we used the Factorization formula
cos x - cos y = 2 sin [ (x+y)/2 ] · sin [ (y-x)/2 ]
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