Question

Hi Prove that: sin8xcosx - cos8xsin6x / cos2xcosx - sin3xsinx = tan2x

Answer 1

Formulas required: 1. 2 sinA cosB = sin (A+B) + sin (A-B) 2. 2 cosA cosB = cos (A+B) + cos (A-B) 3. 2 sinA sinB = cos (A-B) - cos (A+B) 4. sinx / cosx = tanx 5. sinC - sinD = 2 cos ( (C+D) / 2 ) sin ( (C-D) / 2 ) 6. cosC + cosD = 2 cos ( (C+D) / 2 ) cos ( (C-D) / 2 ) (sin8xcosx - sin6xcos3x) / (cos2xcosx - sin3xsin4x) = 2 (sin8xcosx - sin6xcos3x) / 2 (cos2xcosx - sin3xsin4x) = (2sin8xcosx - 2sin6xcos3x) / (2cos2xcosx - 2sin3xsin4x) = (sin9x + sin7x - (sin 9x + sin3x)) / (cos3x + cosx - (cosx - cos7x)) = (sin7x - sin3x) / (cos7x + cos3x) = 2 cos5x sin2x / 2 cos5x cos2x = sin2x / cos2x = tan2x