Question

cos^A/2+cos^B/2-cos^C/2=2cosA/2×cosB/2×sinC/2

Answer 1

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2 answers · Mathematics 

 Best Answer

Let us assume that A+B+C=180 degrees ( pie radian) 
L.H.S= cos square A/2 +cos square B/2 +cos square C/2 
First take the cos square A/2 + cos square B/2 ( from L.H.S) 
By using the formula cos square A/2= (1+ cos A)/2 in cos square A/2 +cos square B/2, we get, 
(1+cos A)/2 +( 1+cos B)/2 [ PUTTING THIS IN L.H.S] 
THEN, 
(1+cos A)/2 + (1+ cos B)/2+ cos square C/2 
taking 1 and 1/2 as common from (1+cos A)/2 + (1+ cos B)/2, then 
=1+1/2 (cos A +cos B) + cos square C/2 
now we can use cos C + cos D= 2 cos(C+D)/2. cos (C-D)/2 formula in the above statement, THEN, 
=1+ 1/2[2cos (A+B)/2 . cos (A-B)/2 ] + cos square C/2 
as we know A+B+C= 180 degrees or pie radian , 
=1+ cos (180/2 - C/2) cos (A-B)/2 + cos square C/2 
=1+ sin C/2 . cos (A-B)/2 + 1- sin square C/2 
=2+ sin C/2 [ cos(A-B)/2 - sin C/2] 
=2+ sin C/2 { cos (A-B)/2 - sin [ 180/2-(A+B)/2 ]} 
=2+ sin square C/2 [ cos (A-B)/2 - cos (A+B)/2] 
=2+ sin C/2 . 2sin A/2 sin B/2 sin C/2 
= 2+ 2 sin A/2 sin B/2 sin C/2