Nested trig functions (incl. inverse trig functions) Edit: Although this problem has received a kind answer, I would still appreciate more comprehensive explanation. I am still rather confused.
This problem has confused me a bit:

The standard method is one I know well: Draw a right triangle, set the adjacent and hypotenuse to:$(2x+1)$ and $\sqrt {17}$, respectively; then use the Pythagorean theorem to solve for the third side. Then it is easy to find the sine by taking the opposite over the hypotenuse.
Alas, I feel uncomfortable about this method -- it feels like it's glossing over something. Why can we assume that $\arccos(2x+1/\sqrt {17})$ describes an angle in a RIGHT triangle? The range of $\arccos$ is, after all, zero to $\pi$ -- which obviously includes some values that are too large for a right triangle. 
So what gives? Why does the method mentioned above work? Shouldn't we be using unit circle trigonometry rather than the basic right triangle definition? 
And, while we're at it, I may as well bring up another similar problem that confused me: 
$$\sec(\arctan(\frac{x}{x-1})$$
The teacher said the answer was $\frac{\sqrt{2x^2+1}}{\lvert x-1\rvert}$
Why the absolute value on the denominator? I still can't figure that out. Of course, in a triangle all sides must have positive length, but that doesn't mean we can't have negative values for trig functions!
 A: When you work with inverse trig functions, it's important to know their ranges. In your first example, we know that $\cos^{-1}$ range is $[0, \pi]$ so the sine value will be non-negative. We do not know whether $2x+1$ is negative or positive but it does not really matter because when we apply Pythagorean theorem, we square $2x+1$, right?. We need to be careful, however, when we take a square root because sine can be negative. Luckily, we know that it's not the case here based on the range of $\arccos$ function.In your second sample, the range for $\arctan$ is $[-\pi/2, \pi/2]$ so the value of cosine must be non-negative and we need absolute value.Your observation is correct that the right triangle cannot accommodate negative values for trig functions so I always recommend using unit circle whenever possible. 
A: You need not draw triangles. The relation
$$
\lvert\sin\alpha\rvert=\sqrt{1-\cos^2\alpha}
$$
holds for every angle $\alpha$ because of $\sin^2\alpha+\cos^2\alpha=1$.
Since $\arccos t$ by definition returns an angle in $[0,\pi]$, you know that
$$
\sin(\arccos t)=\sqrt{1-\cos^2\arccos t}=\sqrt{1-t^2}
$$
for every $t\in[-1,1]$.
What about $\sec\arctan\frac{x}{x-1}$? You know that $\arctan t$ returns a value in $(-\pi/2,\pi/2)$ where the secant is positive. Thus you can use
$$
\lvert\sec\alpha\rvert=\sqrt{1+\tan^2\alpha}
$$
and remove the absolute value:
$$
\sec\arctan\frac{x}{x-1}=\sqrt{1+\tan^2\arctan\frac{x}{x-1}}
=\sqrt{1+\frac{x^2}{(x-1)^2}}=\frac{\sqrt{2x^2-2x+1}}{|x-1|}
$$
A: $$\begin{aligned}\sin\arccos\dfrac{2x+1}{\sqrt{17}}&=\sqrt{1-\left(\dfrac{2x+1}{\sqrt{17}}\right)^2}\\&=\sqrt{\dfrac{17-4x^2-1-4x}{17}}\\&=\dfrac{2}{\sqrt{17}}\sqrt{4-x^2-x}\end{aligned}$$

$$\begin{aligned}\sec\arctan\dfrac{x}{x-1}&=\sqrt{1+\tan^2\arctan\dfrac{x}{x-1}}\\&=\sqrt{1+\left(\dfrac{x}{x-1}\right)^2}\\&=\sqrt{\dfrac{(x-1)^2+x^2}{(x-1)^2}}\\&=\dfrac{\sqrt{2x^2-2x+1}}{\mid x-1\mid}\end{aligned}$$
Clearly range of $\arctan \theta$ is $\left[-\pi/2, \pi/2\right]$. Note that $\arctan\tan\theta \equiv \arctan \sin\theta/\cos\theta$, so $\cos \theta$ must be positive hence the absolute value.
