Question

(ii) cos 3A cos 2A + sin 4A sin A = cos A cos 2A​

Answer 1

Answer:

in the explanation

Step-by-step explanation:

1/2 (2cos 3A cos 2A + 2sin 4A sin A)

= 1/2[{cos(3A+2A) + cos(3A-2A)} +{cos (4A-A) - cos (4A+A)}

= 1/2 {cos 5A + cos A +cos 3A - cos 5A}

= 1/2 (cos A + cos3A)

= 1/2 {2cos(A+3A)/2 cos (3A-A)/2}

= cos 2A cos A