I want to show that (sin2kx / 2sinx) + cos(2k-1)x = sin(2k+2)x / 2sinx
I presume I could convert the cos expression to sin first. This is more difficult than the trigonometry I normally do.
Printable View
I want to show that (sin2kx / 2sinx) + cos(2k-1)x = sin(2k+2)x / 2sinx
I presume I could convert the cos expression to sin first. This is more difficult than the trigonometry I normally do.
$\displaystyle \sin{[(2k + 2)x]} = \sin{(2kx + 2x)}$
$\displaystyle = \sin(2kx)\cos(2x) + \cos(2kx)\sin(2x)$
$\displaystyle = \sin(2kx)[1 - 2\sin^2{x}] + 2\sin{x}\cos{x}\cos{(2kx)}$
$\displaystyle = \sin(2kx) - 2\sin(2kx)\sin^2{x} + 2\sin{x}\cos{x}\cos(2kx)$
$\displaystyle = \sin(2kx) - 2\sin{x}[\sin(2kx)\sin{x} + \cos(2kx)\cos{x}]$
$\displaystyle = \sin(2kx) - 2\sin{x}\cos{(2kx - x)}$
$\displaystyle = \sin(2kx) - 2\sin{x}\cos[(2k - 1)x]$.
Therefore
$\displaystyle \displaystyle{\frac{\sin{[(2k + 2)x]}}{2\sin{x}} = \frac{\sin(2kx) - 2\sin{x}\cos[(2k - 1)x]}{2\sin{x}}}$
$\displaystyle \displaystyle{ = \frac{\sin(2kx)}{2\sin{x}} - \cos[(2k-1)x]}$
Thanks, that was more work than I thought it would be. You have made a mistake at the 5th line and further. The last + should be -.
