Question

Prove that:
sin A - 2sin³A / 2cos³A - cos A = tan A 

Answer 1

$\frac{sin\ A\ -\ 2\ sin^3\ A}{2\ cos^3\ A\ -\ cos\ A} = \frac{sin\ A(1\ -\ 2\ sin^2\ A)}{cos\ A(2\ cos^2\ A - 1)} \\ \\ = \frac{sin\ A}{cos\ A} * \frac{cos\ 2A}{cos\ 2A} = tan\ A,\ \ \ \ \ \ \ \ if A\ is\ not\ \pi/4$

Answer 2

Step-by-step explanation:

sinA−2sin

sinA−2sin 3

sinA−2sin 3 A

sinA−2sin 3 A

sinA−2sin 3 A =

sinA−2sin 3 A = cosA(2cos

sinA−2sin 3 A = cosA(2cos 2

sinA−2sin 3 A = cosA(2cos 2 A−1)

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A)

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A)

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A) =

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A) = cosAcos2A

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A) = cosAcos2AsinAcos2A

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A) = cosAcos2AsinAcos2A

sinA−2sin 3 A = cosA(2cos 2 A−1)sinA(1−2sin 2 A) = cosAcos2AsinAcos2A =tanA.