This is just a briefer restatement of the same rule, which I find easier to remember. I don't think there is a particular reason why it works. (If there were, there would be some nice generalization, which there does not seem to be.)
$\theta \hskip2cm 0^\circ \hskip 1cm 30^\circ \hskip 1cm 45^\circ \hskip1cm 60^\circ \hskip1cm 90^\circ $
$\sin(\theta) \hskip1cm {\sqrt{0} \over 2}\hskip1cm {\sqrt{1}\over 2}\hskip1cm {\sqrt{2}\over 2}\hskip1cm {\sqrt{3}\over 2}\hskip1cm {\sqrt{4}\over 2}$
$\cos(\theta) \hskip1cm {\sqrt{4} \over 2}\hskip1cm {\sqrt{3}\over 2}\hskip1cm {\sqrt{2}\over 2}\hskip1cm {\sqrt{1}\over 2}\hskip1cm {\sqrt{0}\over 2}$
Edit: You might say that these values correspond to looking at triangles where the Pythagorean theorem becomes one of the following:
$$4=0+4=1+3=2+2=3+1=4+0$$
and that these triangles happen to have very nice angles.
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.
Best Answer
Michael Hardy has the correct answer for B. There is a specific side to find, but if the angle is not between the known sides there could be two solutions. For C, it is saying that it is false that there can be two answers. You have argued that there is only one solution to the equation, which agrees with the book. In D you are right-sometimes there are two answers, but not all the time. It is also the case if your given angle is greater than 90 degrees.