Given that $\displaystyle sin(A+B) = 2sin(A-B)$, express $\displaystyle tan A$ in terms of $\displaystyle tan B$
$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}$.
Given that $\displaystyle \sin{(A + B)} = 2\sin{(A - B)}$
$\displaystyle \sin{A}\cos{B} - \cos{A}\sin{B} = 2(\sin{A}\cos{B} + \cos{A}\sin{B})$
$\displaystyle \sin{A}\cos{B} - \cos{A}\sin{B} = 2\sin{A}\cos{B} + 2\cos{A}\sin{B}$
$\displaystyle -3\cos{A}\sin{B} = \sin{A}\cos{B}$
$\displaystyle \frac{-3\sin{B}}{\cos{B}} = \frac{\sin{A}}{\cos{A}}$
$\displaystyle -3\tan{B} = \tan{A}$.
$\displaystyle \sin{4x} = 2\sin{2x}\cos{2x}$
$\displaystyle = 2\cdot 2\sin{x}\cos{x}(\cos^2{x} - \sin^2{x})$
$\displaystyle = 4\cos{x}\sqrt{1 - \cos^2{x}}[\cos^2{x} - (1 - \cos^2{x})]$
$\displaystyle = 4\cos{x}(2\cos^2{x} - 1)\sqrt{1 - \cos^2{x}}$
$\displaystyle = 4p(2p^2 - 1)\sqrt{1 - p^2}$.
Dear Punch,
Do you know that,
$\displaystyle Sin2A=2SinACosA$
$\displaystyle SinA=\sqrt{1-Cos^2A}$
These are the two equations I used. In case you do not know them I think you should refer List of trigonometric identities - Wikipedia, the free encyclopedia When going from 3 to 4 I used the second identity.
Hope this helps.