5 sin(x+arctan(4/3)) = 3 sin x + 4 cos x
(arctan is the inverse tangent)
I'm really having trouble trying to prove this. I'm mainly stuck with what to do with the inverse tangent.
let $\displaystyle \theta = \arctan\left(\frac{4}{3}\right)$ ... using right triangle ratios, $\displaystyle \sin{\theta} = \frac{4}{5}$ and $\displaystyle \cos{\theta} = \frac{3}{5}$
sum identity for sine ...
$\displaystyle 5 \sin(x + \theta) = 5[\sin{x}\cos{\theta} + \cos{x}\sin{\theta}]$
substitute in the values for $\displaystyle \cos{\theta}$ and $\displaystyle \sin{\theta}$ and finish it.