# Harmonic Identities Help

#### andyt

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)

#### HallsofIvy

MHF Helper
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 $$\displaystyle \alpha= cos^{-1}\left(\frac{2\sqrt{13}}{13}\right)$$.

1 person

#### andyt

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)

#### romsek

MHF Helper
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

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

$R\sin(\beta) = 4$

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

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

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

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

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

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

2 people

#### andyt

Morning Romsek

Rsin(3θ+β)≡Acos(3θ)-Bsin(3θ)

What equation did you use to get to: R cos (β) = -6 and R sin (β) = 4

Thanks
Andy

Last edited:

#### romsek

MHF Helper
Morning Romsek

Rsin(3θ+β)≡Acos(3θ)-Bsin(3θ)

What equation did you use to get to: R cos (β) = -6 and R sin (β) = 4

Thanks
Andy
I equated coefficients of $\cos(3\theta)$ and $\sin(3\theta)$

Thanks romsek