1 I'm a little confused about how you choose to use either sine or cosine or tangent over the others. Are they interchangeable given the same information you have about a right triangle? What are the circumstances that should dictate the use of one over the other? Or is it preference?
I assume that as in part 2 you have a right angled triangle. In this case you can work out the angle given two of the sides. Depending on which sides you have, you should choose sin, cos or tan, as shown in the diagram below.
$$\begin{align}
\sin d &= \frac{\text{opposite side}}{\text{hypoteneuse side}} \\
\cos d &= \frac{\text{adjacent side}}{\text{hypoteneuse side}} \\
\tan d &= \frac{\text{opposite side}}{\text{adjacent side}} \\
\end{align}$$
There are relationships between the different trig functions, e.g. $\sin d = \sqrt{1 - \cos^2 d}$. However it's a lot easier to find them directly given a standard right-angled triangle problem.
For reference, the identities that you can use are:
$$\begin{align}
\sin^2 x+\cos^2 x&=1 \\
\tan^2 x+1&=\sec^2 x, \;\;\;\;\;\,
\text{where } \sec x = 1/\cos x\\
1+\cot^2 x&=\mathrm{cosec}^2 x, \;\;\;
\text{where } \cot x = 1/\tan x, \mathrm{cosec}\,x=1/\sin x
\end{align}$$
2 If you know the sides you can work out the angles.
Yes
If you know the angle you can work out the length of the sides. Is this a correct assumption?
No, see the right-hand side of the diagram. You can have similar triangles where the angles are all the same but the side lengths are scaled up or down by some factor, like Russian dolls.
If you think about it, this makes sense, because all the sin, cos, or tan gives you is the ratio between sides. E.g. for sin, how many times bigger is the opposite side than the hypoteneuse.
So, is it correct so assume that if you know one of the angles besides the 90 degree angle and 1 length of one side you can determine the sine, cosine and tangent of that triangle?
Strictly speaking, we talk about the sine, cosine and tangent of angles not triangles. A typical problem setting is that you are given one of the angles and one of the sides in a right-angled triangle. You can then work out the length of another side. E.g. if you have an angle $d$ and the opposite side you can rearrange the sine formula to find the hypoteneuse side.
$$\begin{align}
\sin d &= \frac{\text{opposite side}}{\text{hypoteneuse side}} \\
\text{hypoteneuse side}\times\sin d &= \text{opposite side} \\
\text{hypoteneuse side} &= \frac{\text{opposite side}}{\sin d}
\end{align}$$
Also note that if you have one angle and the 90 degree angle you can work out the third angle because the angles add to 180 degrees. And you can then work out all the side -- but you need at least one side to fix the scale factor as discussed above.
And if you know two sides you can always determine the angle? Is that a correct assumption?
Yes that's correct. See part 1.
Each of the angles makes an acute angle to one of the coordinate axes, even though the full angle runs into some quadrant of the unit circle other than the first quadrant. You will want to work out what that acute angle (between 0º and 90º) is for each of the given angles, find the trig values for them from a calculator, then assign the appropriate sign (positive or negative) to each of those trig values.
Here is a graph of the angles:
Best Answer
First of all, I upvoted your query re very positive approach, re
"but I'm looking for a physical derivation for this".
To understand this, in the realm of trigonometry,
where sine and cosine are functions of angles,
you need to consider the sine and cosine functions
against the backdrop of the unit circle.
Imagine a unit circle centered at the origin, that hits the $x$ and $y$
axes at points (1,0), (0,1), (-1,0), and (0, -1).
Consider any point in the unit circle that is in the first (upper right) quadrant. The point will have coordinates $(x,y).$
Let $\theta$ denote the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y). Since the radius of the circle is 1, $\cos \theta = x$ and
$\sin \theta = y.$
Now imagine traveling around the arc of the unit circle until you reach the point (0,1).
This point may be construed to represent $90^{\circ}$, just as one complete revolution around the circle can be construed to represent $360^{\circ}.$
It is easy to see that $\cos(90^{\circ}) = 0$ and $\sin(90^{\circ}) = 1.$
Now imagine traveling around the arc to any point on the unit circle that is in the 2nd (upper left) quadrant.
Here, the point $(x,y)$ in the 2nd quadrant will have $x < 0$ and $y > 0.$
Again, just as before, consider $\theta$ to be the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y).
Here, by convention, $\cos \theta$ (again) $ = x$
and $\sin \theta$ (again) $ = y.$
Thus, it is easy to see that when $(x,y)$ is in the 2nd quadrant, and
$\theta$ is the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y)
that $\cos \theta$ will by convention be $< 0$ and
$\sin \theta$ will by convention be $ > 0.$
A clear advantage of these conventions is that they facilitate the formulas shown at https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities.
Another advantage of these conventions, which may not seem important in the realm of trigonometry, is that they facilitate the cosine and sine functions being continuous functions. This is a pandora's box that may not be worth exploring in the realm of trigonometry, but is still worth a very casual mention.
See https://www.mathopenref.com/triggraphsine.html.