Prove

that

2 cos (A + 45°) cos (B-45°) = cos 2A

Step-by-step explanation:

2 cos (A+45) cos (B-45)

=2[cos A cos 45-sin A sin 45][cos B cos 45+sin B sin 45]

=2[cos A * 1/√2-sin A * 1/√2][cos B*1/√2+sin B*1/√2]

=2[1/√2 (cos A-sin A) ][1/√2 (cos B-sin B]

=2*1/√2*1/√2 (cos A-sin A) (cos B+sin B)

=cosA cos B+cos A sin B-sin A cos B-sin A sin B

=cos A cos B-sin A sin B - (sin A cos B-cos A sin B)

=cos (A+B) - sin (A-B)

i think there should be A in place of B.

then

=cos (A+A) - sin (A-A)

=cos 2A-sin 0

=cos 2A

as sin 0=0