I,m Really sorry HallsofIvy I have wrote the equation incorrectly
The third Harmonic of a sound is given by 4 cos (3 θ) - 6 sin (3 θ), Using the Harmonic Identity A cos x – B sin ≡ R cos (x + α)
R is given in Cos and alpha.
Yet I need to express this sound wave in the form R sin (3 θ + β)
$R \sin(3\theta+\beta) = R \sin(3\theta)\cos(\beta) + R\cos(3\theta)\sin(\beta)$
equating like terms we get
$R\cos(\beta) = -6$
$R\sin(\beta) = 4$
$\tan(\beta) = -\dfrac 2 3$
$\beta = \arctan\left(-\dfrac 2 3\right)$
$\beta = \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)$
$R^2 = (-6)^2 + (4)^2 = 52$
$R = \sqrt{52}=2\sqrt{13}$
$4 cos (3 \theta) - 6 sin (3 \theta) = 2\sqrt{13}\sin\left(3\theta + \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)\right)$