Question

A+B+C=180then prove that cos2A+cos2B-cos2C=1-4sinAsinBsinC

Answer 1

cos2a + cos2b + cos2c 

= cos2a + [ cos(2b) + cos(2c) ] 
= cos2a + 2*cos(b + c)*cos(b - c) 
= (2cos²a - 1) + 2cos(180 - a)*cos(b - c) 
= 2cos²(a) - 1 - 2cos(a)*cos(b - c) 
= - 1 + 2cos(a) * [cos(a) - cos(b - c)] 
= - 1 + 2cos(a) [ cos {180 - (b + c) } - cos(b - c) ] 
= - 1 + 2cos(a) [ - cos (b + c) - cos(b - c) ] 
= - 1 - 2cos(a) [ cos (b + c) - cos(b - c) ] 
= - 1 - 2cos(a) [ 2*cos(b)*cos(c) ] 
= - 1 - 4 cosa cosb cosc

Answer 2

hiii!!! folk..

A+B+C=180°(given)

=)from lhs
cos2A+,cos2B-cos2C

=)2cos(2A+2B)/2*Cos(2A-2B/2)-cos2C

=)2cos(A+B)*cos(A-B)+2cos^2C+1. √【cos2C=2cos,^2C+1】

=)2cos(π-C)*cos(A-B)+2cos,^2+1

=)1-2cosC/2(cosA-B)+cos【π-(A+B)】


=)1-2CosC/2(cosA-B)-cosC(A+B)

=)1-2cosC(2sin(A-B+A+B/2)*sin(,A+B-A+B)/2

=)1-2cosC(2sunA*sinB)

=)1-4sinA*sinB*sinC=Rhs.prooved.


hope it help you.

@rajukumar☺