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 $\sin(y)$
 A: Draw a right triangle for each angle, label the known sides, then find the missing side. The unknown ratio should be obvious. 

Note that the Pythagorean theorem and the Pythagorean identity are equivalent to one another.
$$\text{opposite}^2+\text{adjacent}^2 = \text{hypotenuse}^2$$
$$\frac{\text{opposite}^2+\text{adjacent}^2}{\text{hypotenuse}^2} = \frac{\text{hypotenuse}^2}{\text{hypotenuse}^2}$$
$$\bigg(\frac{\text{opposite}}{\text{hypotenuse}}\bigg)^2 + \bigg(\frac{\text{adjacent}}{\text{hypotenuse}}\bigg)^2=\bigg( \frac{\text{hypotenuse}}{\text{hypotenuse}}\bigg)^2$$
$$\sin^2\theta+\cos^2\theta=1^2$$

Lastly, try to derive the sum of angles formula by referring to the last diagram. Hint: use the law of sines.

A: 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 $360^{\circ}$.
