<s>[INST] if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$ if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$
Here is my thought process: 
the identity says: 
$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$
We know $\sin(x)=1/3$ and $\sec(y)=\frac{1}{\cos(y)} = \frac{4}{5}$, so
$\sin(x+y)=\frac{1}{3}\times\frac{4}{5}+\cos(x)\sin(y)$
But I'm stuck here because I don't know how to find $\cos(x)$ or $\sin(y)$
 [/INST] You have started off well. To get the values of $\cos x$ and $\sin y$, you should use the identity
$$\sin^2\theta + \cos^2\theta \equiv 1$$
I will show you how to find $\cos x$. 
You can use the same method to find $\sin y$ yourself.
For example, if $\sin x = \frac{1}{3}$ then we have
\begin{eqnarray*}
\sin^2 x + \cos^2x &\equiv& 1 \\ \\
\left(\frac{1}{3}\right)^{\! 2} + \cos^2x &=& 1 \\ \\
\frac{1}{9} + \cos^2x &=& 1 \\ \\
\cos^2x &=& \frac{8}{9} \\ \\
\cos x&=& \pm\frac{2}{3}\sqrt{2}
\end{eqnarray*}
Next you need to decide on the $\pm$. Looking at the graph of $y=\cos x$ we see that $\cos x > 0$ if $0^{\circ} \le x < 90^{\circ}$ or $270^{\circ} < x \le 360^{\circ}$. So, for example: if we know that $x$ is acute then
$$\cos x = +\frac{2}{3}\sqrt{2}$$
If we know that $x$ is obtuse then
$$\cos x = -\frac{2}{3}\sqrt{2}$$
If $x$ is reflex and less that $270^{\circ}$ then 
$$\cos x = +\frac{2}{3}\sqrt{2}$$
If $x$ is reflex and more than $270^{\circ}$ then
$$\cos x = -\frac{2}{3}\sqrt{2}$$
The same is true modulo $360^{\circ}$.
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