The third Harmonic of a sound is given by 4 cos x -6 sin
Using the Harmonic Identity A cos x – B sin ≡ R cos (x + α)
R is given in Cos and alpha.
Yet I need to answer in the form R sin (3 θ + β)
One thing you need is that $\displaystyle cos(p+ q)= cos(p)cos(q)- sin(p)sin(q)$. From that, Rcos(x+ a)= Rcos(a)cos(x)- Rsin(a)sin(x). So you want $\displaystyle \alpha$ and R such that $\displaystyle R cos(\alpha)= 4$ and $\displaystyle R sin(\alpha)= 6$. Then $\displaystyle R^2cos^2(\alpha)+ R^2sin^2(\alpha)= R^2= 4^2+ (-6)^2= 16+ 36= 52$. $\displaystyle R= \sqrt{52}= 2\sqrt{13}$. Then $\displaystyle R cos(\alpha)= 2\sqrt{13} cos(\alpha)= 4$. $\displaystyle cos(\alpha)= \frac{2}{\sqrt{13}}= \frac{2\sqrt{13}}{13}$ so $\displaystyle \alpha= cos^{-1}\left(\frac{2\sqrt{13}}{13}\right)$.
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
$R\cos(\beta) = -6$
$R\sin(\beta) = 4$
$\tan(\beta) = -\dfrac 2 3$
$\beta = \arctan\left(-\dfrac 2 3\right)$
$\beta = \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)$
$R^2 = (-6)^2 + (4)^2 = 52$
$R = \sqrt{52}=2\sqrt{13}$
$4 cos (3 \theta) - 6 sin (3 \theta) = 2\sqrt{13}\sin\left(3\theta + \pi -\tan ^{-1}\left(\dfrac{2}{3}\right)\right)$