# Deriving an Area Formula in Terms of $\theta$

Suppose we create a rain gutter using an aluminum sheet that is 12 inches wide. After marking off a length of 4 inches from each edge, the length is bent up at an angle $\theta$, see the following figure.

I wish to express the area of the opening in terms of $\theta$, so after some trigonometry, I was able to get

$$A(\theta) = 16\cos \theta \sin \theta \hspace{20 pt} \text{where} \, \, 0 \leq \theta < \frac{\pi}{2} \tag{1}$$

But when $\theta = \frac{\pi}{2}$, $A(\theta) = 0$. Observing that the area I want when $\theta = \frac{\pi}{2}$ is a triangle, it is clear that the area can be found by doubling the area of the two right triangles

$$2 \cdot \Big(\frac{1}{2}\Big) \cdot (4) \cdot (4) = 16$$

But since $\sin(\frac{\pi}{2}) = 1$, we can express the area in terms of $\theta$ as

$$16\sin(\theta) \tag{2}$$

when $\theta = \frac{\pi}{2}$.

So the formula that I have is a piecewise function

$$A(\theta) = \begin{cases} 16\sin(\theta)\cos(\theta) & 0 \leq \theta < \frac{\pi}{2} \\\\ 16\sin(\theta) & \theta = \frac{\pi}{2} \end{cases}$$

But the book has $A(\theta) = 16\sin(\theta)(\cos(\theta) + 1)$, so I'm not sure what I'm doing wrong. I could be wrong about $(2)$, but even then I do not know how to proceed to get the correct equation.

• Are you forgetting to include the area of the rectangular region? This part would correct your formula. – abiessu Mar 26 '16 at 20:51

The area of the two triangles is, as you have noticed, $2\cdot\frac 12\cdot 4\sin\theta \cdot 4\cos\theta$. This appropriately goes to $0$ as $\theta$ goes to $\pi\over 2$. The area of the rectangle in the middle is $4\cdot 4\sin\theta$, so the area of the whole region is then
$$16\sin\theta\cos\theta+16\sin\theta=16\sin\theta(\cos\theta+1)=8\sin2\theta+16\sin\theta$$