Consequently,

and therefore $$\dd [y]{x} = \answer [given]{\frac {\cos (\theta ) + 4\sin (\theta )\cos (\theta )}{-\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta )}}.$$ When $\theta =\pi /4$,

Thus the equation of the line (in polarrectangular coordinates) tangent to the limaçon at $\theta =\pi /4$ is $$y=(-2\sqrt {2}-1)\big (x-(1+\sqrt {2}/2)\big )+1+\sqrt {2}/2 \approx -3.83 x+8.24.$$ $$\graph {r=1+2\sin (\theta ), y=(-2\sqrt {2}-1)(x-(1+\sqrt {2}/2))+1+\sqrt {2}/2}$$

• $\cos (\theta )=0$ or
• $1+4\sin (\theta )=0$.

Let’s examine the case of $\cos (\theta ) = 0$ first, and restrict to the situation where $\theta$ is between $0$ and $2\pi$. If $\theta$ is in the interval $[0,\pi ]$, then $\cos (\theta )=0$ when $\theta =\answer [given]{\pi /2}$. If $\theta$ is in the interval $[\pi ,2\pi ]$, then $\cos (\theta )=0$ when $\theta =\answer [given]{3\pi /2}$.

The corresponding points in rectangular coordinates are

meaning $(x,y) = (0,3)$, and

meaning $(x,y) = (0,1)$.

But the numerator was the product of $\cos (\theta )$ and another term, namely $1+4\sin (\theta )$ So the numerator is also zero when

At this point we have a choice to make: We can either apply the arcsin function and solve for $\theta$, or we can work with $\sin (\theta )$ directly. Perhaps the easier route is to work with $\sin (\theta )$ directly. To explore $\sin (\theta )$, we may choose an algebraic method employing the Pythagorean identity, or a geometric method looking at the unit circle with the Pythagorean theorem. Let’s take the geometric route to compute $\cos (\theta )$.

meaning $(x,y) = (\sqrt {15}/8, -1/8)$, and

meaning $(x,y) = (-\sqrt {15}/8, -1/8)$.

In summary, the graph of $r(\theta ) = 1+2\sin (\theta )$ on the interval $[0,2\pi ]$ has horizontal tangent lines at the points

• $(x,y) = (0,\answer [given]{3})$,
• $(x,y) = (0,1)$,
• $(x,y) = (\sqrt {15}/8, -1/8)$, and
• $(x,y) = (-\sqrt {15}/8, -1/8)$.

You can confirm this by looking at the graph below: $$\graph {r=1+2\sin (\theta ),y=3,y=1,y=-1/8}$$

We then set this equal to zero and note that the equation is quadratic equation in the variable $\sin (\theta )$. The quadratic formula gives that

Now we can find cosine using either geometrically via the unit circle or algebraically with the Pythagorean identity. This time, let’s take the more algebraic route with the Pythagorean identity. When $\sin (\theta ) =\frac {-1+\sqrt {33}}{8}$, then

So $$\cos (\theta ) = \pm \frac {\sqrt {(\answer [given]{15}+\sqrt {33})/2}}{4}$$ and when $\sin (\theta ) =\frac {-1-\sqrt {33}}{8}$,

So $$\cos (\theta ) = \pm \frac {\sqrt {(\answer [given]{15}-\sqrt {33})/2}}{4}.$$ So now we know that the graph of $r(\theta ) =1+2\sin (\theta )$ on $[0,2\pi ]$ has vertical tangent lines at the four points

we see that the graph of $r(\theta ) =1+2\sin (\theta )$ on $[0,2\pi ]$ has vertical tangent lines at

You can confirm this visually by looking at the graph below: $$\graph {r=1+2\sin (\theta ), x=\left (1+\frac {-1+\sqrt {33}}{4}\right )\frac {\sqrt {(15+\sqrt {33})/2}}{4}, x=\left (1+\frac {-1+\sqrt {33}}{4}\right )\frac {-\sqrt {(15+\sqrt {33})/2}}{4}, x=\left (1+\frac {-1-\sqrt {33}}{4}\right )\frac {\sqrt {(15-\sqrt {33})/2}}{4}, x=\left (1+\frac {-1-\sqrt {33}}{4}\right )\frac {-\sqrt {(15-\sqrt {33})/2}}{4}}$$

Thus the equations of the tangent lines, in polarrectangular coordinates, are

In rectangular form, the tangent lines are $y=\tan (7\pi /6)x$ and $y=\answer [given]{\tan (11\pi /6)}x$. You can confirm this by looking at the graph below: $$\graph {r=1+2\sin (\theta ),y=\tan (7\pi /6)x,y=\tan (11\pi /6)x}$$