i have some questions i really need help with
sin3 A = 3sin A - 4sin^3 A
(Hint use sin3 A = sin(2a + a)
and
sin3 x / sin x - cos3 x / cos x = 2
please help im stuck
um, use the hint! did you try? what did you get?
I will start you off:and
sin3 x / sin x - cos3 x / cos x = 2
please help im stuck
Consider the LHS:
$\displaystyle \frac {\sin 3x}{\sin x} - \frac {\cos 3x}{\cos x} = \frac {\sin 3x}{\sin x} \cdot \frac {2 \cos x}{2 \cos x} - \frac {\cos 3x}{\cos x} \cdot \frac {2 \sin x}{2 \sin x} $
$\displaystyle = \frac {2 \sin 3x \cos x}{2 \sin x \cos x} - \frac {2 \sin x \cos 3x}{2 \sin x \cos x}$
$\displaystyle = \frac {2 \sin 3x \cos x - 2 \sin x \cos 3x}{\sin 2x}$
now can you finish up?