Re: Trigonometric identity

On Oct 7, 8:31 am, Saysero <says...@xxxxxxxxx> wrote:
Is there an relatively easy way to show that

sinx - 2sin(2x) + sin(3x)
-------------------------------------- = tg(2x)
cosx - 2cos(2x) + cos(3x)

Thank you in advance.

sinx - 2sin(2x) + sin(3x) sin(2x)
------------------------- = tg(2x) = -------
cosx - 2cos(2x) + cos(3x) cos(2x)

so you have to prove
(sinx - 2sin(2x) + sin(3x))(cos(2x) =
(cosx - 2cos(2x) + cos(3x))(sin(2x)

The second term after expansion cancels out. So
(sinx + sin(3x))(cos(2x) = (cosx + cos(3x))(sin(2x)

After re-arranging the terms, the proof becomes
sin x cos(2x) - cos x sin(2x) =
sin(2x)cos(3x) - cos(2x)sin(3x)

Using the tri. indentity of
sin(A-B) = sin A cos B - cos A sin B

LHS = sin(x-2x) = sin(-x)
RHS = sin(2x-3x) = sin(-x) = LHS.

.