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