Question

sin^2B +4cos(A+B)-sinAsinB+ cos2(A+B)=cos2A​

Answer 1

Step-by-step explanation:

alpha=A. and. beta = B.

2sin^2A+4cos(A+B).sinA.sinB+cos2(A+B).

we know that. cos2A=1–2sin^2A or putting 2sin^2A= 1-cos2A………..(1)

and cos2(A+B)= 2cos^2(A+B)-1 …………………(2)

=1-cos2A+4.cos(A+B).sinAsinB+2cos^2(A+B) -1.

= 2cos^2(A+B)+4.cos(A+B).sinAsinB-cos2A.

=2.cos(A+B)[cos(A+B)+2sinAsinB] -cos2A.

=2.cos(A+B).[cosAcosB-sinA.sinB+2sinAsinB]-cos2A.

=2.cos(A+B).[cosAcosB+sinAsinB]-cos2A.

=2.cos(A+B).cos(A-B) - cos2A.

=cos(A+B+A-B)+cos(A+B-A+B)-cos2A.

= cos2A +cos2B-cos2A.

= cos2B. or. cos 2 beta. Answer.