cosA+cosB−cosC=−1+4 cos A/2 cos B/2...
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(cos A+cos B-cos C=-1+4 cos frac{A}{2} cos frac{B}{2} sin frac{C}{2} )

JEE/Engineering Exams
Maths
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cosA+cosB−cosC=−1+4 cos A/2 cos B/2 sin C/2
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( mathrm{LHS}[mathrm{cosA}+mathrm{cosB}]-mathrm{cos} C )
( =2 cos (A+B / 2) cdot cos (A-B / 2)-cos C )
it is given ( mathrm{A}+mathrm{B}+mathrm{C}= ) pi or ( 180 mathrm{eqn} )
(i) ( mathrm{so} ) ( =2 cos (p mathrm{i} / 2-C / 2) cdot cos (A-B / 2)-cos C )
as we know that ( cos (90-A)=sin A )
( =2 sin C / 2 cdot cos (A-B / 2)-1+2 sin ^{wedge} 2 C / 2 )
( =2 sin C / 2[cos (A-B / 2)+sin C / 2]-1 )
( =2 sin C / 2[cos (A-B / 2)+cos (A+B / 2)]-1 )
by eqn(i) ( =2 sin C / 2[2 cos A / 2 cdot cos B / 2]-1 )
( =4 cos A / 2 cdot cos B / 2 cdot sin C / 2-1=R H S )

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