tangential angle of bifoliate

Question

Consider the curve given by polar equation$$r=f(\theta)={8\cos\theta\sin^2\theta\over3+\cos(4\theta)}$$for $\theta$ in $[0,\pi]$.

By Mathworld's equation (9) the tangential angle is given by $$\phi(θ) = \tan^{-1}\left(\frac{4\cos(θ)\sin(θ)[3+\cos(4θ)]}{6+25\cos(2θ)+2\cos(4θ)-\cos(6θ)}\right)\tag1$$

My computation:

By Wikipedia's formula $$\tanψ = f(θ)/f'(θ)$$ the tangential angle $\phi=ψ+θ$. $$\tan\phi=\frac{\tan(ψ)+\tan(θ)}{1-\tan(ψ)\tan(θ)}=-\frac{{(2 \, \cos^{4}(θ) + 2 \, \cos^{2}(θ) - 1)} \sin(θ)}{2 \, {(2 \, \cos(θ) \sin^{2} (θ)- \cos(θ))}}$$ This is different from (1). Which step is wrong? Thanks.

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my computation in sage:

var('θ')
r = 8*cos(θ)*sin(θ)^2/(3+cos(4*θ))
tanψ = r/r.diff(θ)

print(((tanψ+tan(θ))/(1-tanψ*tan(θ))).full_simplify())

Answer 1

Actually, Mathworld's equation (9) gives the polar tangential angle rather than the tangential angle. Indeed, we have \begin{align*} \frac{f(\theta)}{f'(\theta)} = \frac{4\cos \theta \sin \theta [3 + \cos 4\theta]}{6 + 25\cos 2\theta + 2\cos 4\theta - \cos 6\theta}. \tag{1} \end{align*}

Proof of (1).

We have $$f'(\theta) = \frac{- 8\sin^3\theta + 16\cos^2\theta \sin \theta}{3 + \cos 4\theta} + \frac{32\cos \theta \sin^2\theta \sin 4\theta}{[3 + \cos 4\theta]^2}.$$

(1) is written as \begin{align*} &\frac{8\cos \theta \sin^2\theta [3 + \cos 4\theta]}{(-8\sin^3 \theta + 16\cos^2\theta \sin\theta) [3 + \cos 4\theta] + 32\cos \theta \sin^2\theta \sin 4\theta}\\[6pt] ={}& \frac{4\cos \theta \sin \theta [3 + \cos 4\theta]}{6 + 25\cos 2\theta + 2\cos 4\theta - \cos 6\theta}. \end{align*} It suffices to prove that \begin{align*} &\frac{1}{(-4\sin^2 \theta + 8\cos^2\theta) [3 + \cos 4\theta] + 16\cos \theta \sin\theta \sin 4\theta}\\[6pt] ={}& \frac{1}{6 + 25\cos 2\theta + 2\cos 4\theta - \cos 6\theta} \end{align*} or (using $-4\sin^2 \theta + 8\cos^2\theta = 2(\cos^2 \theta + \sin^2 \theta) + 6(\cos^2\theta - \sin^2 \theta) = 2 + 6\cos 2\theta$ and $\sin 2\theta = 2\sin\theta \cos \theta$) $$ \frac{1}{(2 + 6\cos 2\theta) [3 + \cos 4\theta] + 8\sin 2\theta \sin 4\theta} = \frac{1}{6 + 25\cos 2\theta + 2\cos 4\theta - \cos 6\theta} $$ or $$6\cos 2\theta \cos 4\theta + 8\sin 2\theta \sin 4 \theta = 7\cos 2\theta - \cos 6\theta$$ or $$7(\cos 2\theta \cos 4\theta + \sin 2\theta \sin 4 \theta) - (\cos 2\theta \cos 4\theta - \sin 2\theta \sin 4 \theta) = 7\cos 2\theta - \cos 6\theta$$ which is true.

We are done.