Question

Prove that: sin8xcosx - sin6xcos3x / cos2xcosx - sin3xsin4x = tan2x

Answer 1

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