# What is Sin(arcsin(4/5)+arctan(12/5))?

Aug 5, 2015

$\arcsin \left(\frac{4}{5}\right)$ is some $\alpha$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\sin \alpha = \frac{4}{5}$

$\arctan \left(\frac{12}{5}\right)$ is some $\beta$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\tan \beta = \frac{12}{5}$

We are asked to find $\sin \left(\alpha + \beta\right)$. (Do you have some suspicions about why I rewrote the problem this way?)

We know that

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ .

We know $\sin \alpha = \frac{4}{5}$ which is positive and with the restriction we already mentioned, we conclude that $0 \le \alpha \le \frac{\pi}{2}$. We want $\cos \alpha$.

At this point in our study of trigonometry we have already developed at least one method to find $\cos \alpha$ (We may have three or four or more methods, but we only need one right now.)

Use your chosen method to get $\cos \alpha = \frac{3}{5}$

So we have:
$\sin \alpha = \frac{4}{5} \text{ }$ and $\text{ } \cos \alpha = \frac{3}{5}$

Similarly, given $\tan \beta = \frac{12}{5}$ and $0 \le \beta \le \frac{\pi}{2}$, find $\sin \beta$ and $\cos \beta$

$\sin \beta = \frac{12}{13} \text{ }$ and $\text{ } \cos \beta = \frac{5}{13}$

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$= \frac{4}{5} \frac{5}{13} + \frac{3}{5} \frac{12}{13}$

$= \frac{56}{65}$.

If you want to write this without the names $\alpha$ and $\beta$ 9Or something like that, you could write:

$\sin \left(\arcsin \left(\frac{4}{5}\right) + \arctan \left(\frac{12}{5}\right)\right)$

$= \sin \left(\arcsin \left(\frac{4}{5}\right)\right) \cos \left(\arctan \left(\frac{12}{5}\right)\right) + \cos \left(\arcsin \left(\frac{4}{5}\right)\right) \sin \left(\arctan \left(\frac{12}{5}\right)\right)$

But I, personally do not think that is more clear. (I do think it is good for students to see it this way, however.)