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

$R\cos(\beta) = -6$

$R\sin(\beta) = 4$

$\tan(\beta) = -\dfrac 2 3$

$\beta = \arctan\left(-\dfrac 2 3\right)$

$\beta = \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)$

$R^2 = (-6)^2 + (4)^2 = 52$

$R = \sqrt{52}=2\sqrt{13}$

$4 cos (3 \theta) - 6 sin (3 \theta) = 2\sqrt{13}\sin\left(3\theta + \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)\right)$