Math, asked by neerajvermag11, 1 year ago

prove that:- (cosA+cosB)^2+(sinA+sinB)^2=4cos^2A-B/2

Answers

Answered by lovelymansianiit
2
=> (cos^2A + cos^2B + 2cosAcosB) + (sin^2A + sin^2B - 2sinAsinB)

=> cos^2A + cos^2B + sin^2A + sin^2B + 2cosAcosB - 2sinAsinB

=> cos^2A + sin^2A + cos^2B + sin^2B + 2(cosA*cosB - sinA*sinB) 

=> 1 + 1 + 2(cosA*cosB - sinA*sinB) 


=> 2 + 2(cosA*cosB - sinA*sinB) 

=> 2 (1 + (cosA*cosB + sinA*sinB))

=> 2 * (1 + cos(A-B)) 

{Because: cosA*cosB - sinA*sinB = cos(A+B)}

=> 2 * 2cos^2 ((A+B)/2)

=> 4cos^2 (A+B)/2
Answered by siddhartharao77
2
The answer is explained in the attachment.


Hope it helps!
Attachments:

siddhartharao77: :-)
Similar questions