The value of cos2⁡A3−4cos2⁡A2+sin2⁡A3−4sin2⁡A2 is equal to 

The value of cos2A34cos2A2+sin2A34sin2A2 is equal to

 

  1. A
  2. B
  3. C
  4. D

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    Solution:

    cos2A34cos2A2+sin2A34sin2A2=3cosA4cos3A2+3sinA4sin3A2=(cos3A)2+(sin3A)2=1

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