Question

P.T

Don't spaM............!

Answer 1

Answer:

$cos A + cos B = 2cos((A+B)/2) cos((A-B)/2)\\\\cos(A+B+C) + cos(-A+B+C) = 2*cos((A+B+C-A+B+C)/2) cos((A+B+C-(-A+B+C))/2)\\ = 2cosA cos(B+C)\\\\cos(A-B+C)+cos(A+B-C) = 2*cos(((A-B+C)+(A+B-C))/2) cos(((A-B+C)-(A+B-C))/2)\\= 2cosAcos(C-B)\\\\cos(A+B+C) + cos(-A+B+C) + cos(A-B+C)+cos(A+B-C) = 2cosAcos(C+B) + 2cosAcos(C-B)\\=>2cosA[cos(C+B) + cos(C-B)]\\=>2cosA[2cos((C+B+C-B)/2)cos((C+B-C+B)/2)=(2cosA)2(cosC)(cosB)$

=4 cos A cos B cos C

Step-by-step explanation:

Proved

Hope it helps :-)