Why does my solutions manual take away or add 180 when finding $\theta$? These are the provided notes:



These are the provided questions:



I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)
 A: Consider the following diagram:

The range of the arccosine function is $[0, \pi]$.  Since you are working in degrees, this corresponds to $[0^\circ, 180^\circ]$.  Thus, you can calculate the measure of the obtuse angle $\theta$ whose cosine is $-3/5$ directly since 
$$\cos\theta = -\frac{3}{5} \Longrightarrow \theta = \arccos\left(-\frac{3}{5}\right)$$ 
However, this will not work for sine or tangent since the range of the arcsine function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (or, in degrees, $[-90^\circ, 90^\circ]$) and the range of the arctangent function is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (in degrees, $(-90^\circ, 90^\circ))$.
For an obtuse angle $\theta$ such that $\sin\theta = \frac{1}{4}$, we draw a right triangle in the second quadrant with opposite side of length $|y| = |1| = 1$, hypotenuse of length $r = 4$, and adjacent side of length $|x| = |-\sqrt{15}| = \sqrt{15}$.  The reference triangle we draw in the first-quadrant, has adjacent side of length $\sqrt{15}$, opposite side of length $1$, and hypotenuse of length $4$. The hypotenuse of the reference triangle forms an angle of $180^\circ - \theta$ with the positive $x$-axis, as shown in the diagram.  Since the sine function satisfies the property 
$$\sin(180^\circ - \theta) = \sin\theta$$
we obtain 
$$\sin(180^\circ - \theta) = \frac{1}{4}$$
Moreover, since $\theta$ is an obtuse angle, $180^\circ -\theta$ is an acute angle, so it falls within the range of the arcsine function. Thus,
$$180^\circ - \theta = \arcsin\left(\frac{1}{4}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arcsin\left(\frac{1}{4}\right)$$
It would be inconvenient to draw a right triangle for an obtuse angle $\theta$ such that $\sin\theta = 0.7890$.  However, we can still use the property that $\sin(180^\circ - \theta) = \sin\theta$ to conclude that 
$$\sin(180^\circ - \theta) = 0.7890$$
Since $180^\circ - \theta$ is an acute angle, we obtain
$$180^\circ - \theta = \arcsin(0.7890)$$
Now, solve for $\theta$.
For an obtuse angle $\theta$ such that $\tan\theta = -\frac{6}{11}$, we draw a second-quadrant angle with adjacent side of length $|x| = |-6| = 6$ and opposite side of length $|y| = |11| = 11$.  The hypotenuse has length $\sqrt{157}$.  The reference triangle in the first-quadrant has adjacent side of length $6$, opposite side of length $11$, and hypotenuse of length $\sqrt{157}$.  The angle the hypotenuse of the reference triangle forms with the positive $x$-axis is $180^\circ - \theta$.  Since $\theta$ is obtuse, $180^\circ - \theta$ is acute.  Since the tangent function has the property that 
$$\tan(180^\circ - \theta) = -\tan\theta$$
we obtain
$$\tan(180^\circ - \theta) = -\left(-\frac{6}{11}\right) = \frac{6}{11}$$
Since $180^\circ - \theta$ is an acute angle, it falls within the range of the arctangent function.  Thus,
$$180^\circ - \theta = \arctan\left(\frac{6}{11}\right)$$
Solving for $\theta$ yields
$$\theta = 180^\circ - \arctan\left(\frac{6}{11}\right)$$
For the obtuse angle $\theta$ such that $\tan\theta = -3.8522$, we use the property that $\tan(180^\circ - \theta) = -\tan\theta$ to conclude that 
$$\tan(180^\circ - \theta) = -(-3.8522) = 3.8522$$
Since $180^\circ - \theta$ is an acute angle,
$$180^\circ - \theta = \arctan(3.8522)$$
Now, solve for $\theta$.
A: The point is, I think here, that the sine, cosine and tangent functions are all many-to-one, that is there are an infinite number of different values of $\theta$ that will give the same value of $\sin\theta$ (and the same for $\cos\theta$ and $\tan\theta$). If you consider the graph of, for example, $y=\sin\theta$, you can see it goes up and down (oscillates) indefinitely, so that if say $\sin\theta=1/2$, the horizontal line $y=1/2$ will intersect the sine curve at many (infinite) places, at $30^o, 150^o, 390^o$ etc. 
For answering questions, I would recommend doing a sketch of the relevant trig function and the horizontal line corresponding to the value of $\sin\theta$ or $\cos\theta$ etc you are given, and then looking at the intersection points and using the symmetry of the curve to determine the values you don't know. 
A: If we are just looking at acute and obtuse angles
$\sin (180 - \theta ) = \sin \theta $ tells you that sin is positive for both acute and obtuse angles. Try it, type in to your calculator any angle from 0 to 180 and you get a positive numbers.
The cos and tan expressions tell you that for obtuse angles that is bigger than 90 and less than 180 you get negative answers. Again try it on your calculator.
So looking at the questions provided
a) -0.2566 is negative for cos so it is an obtuse angle. So if your calculator gives you an acute angled answer you subtract is from 180 to get an obtuse angle.
b) same applies swapping tan for cas in a) as when tan is negative it is an obtuse angle.
C) sin is positive for both acute and obtuse angles so you get 2 answers. One is acute and than subtracting the acute angle from 180 gives you the second answer which is obtuse.
A: The question asks what the angles are if they are obtuse, i.e. between 90 and 180 degrees. The notes provide the conversions, e.g. if you get sin(30) = 0.5 then you convert it to an obtuse angle through sin(30) = sin(180 - 30) from the first note.
A: Try Googling the following terms.


*

*unit circle

*standard position

*terminal size

*initial side

*reference angle


Draw a diagram, and label the point where the terminal side intersects the unit circle as $A(\sin\theta, \cos\theta)$. Plot also point $B(\sin(180-\theta), \cos(180-\theta))$.
Look at diagram and compare points $A$ and $B$.
