Unit Circle Help

[Aviator_13]

Hey guys i just wanted some clarification on how we change angles during questions. Whats the deal with knocking back some angles by 2pi and them not changing and how come you can add 90 degrees to some angles and change it from sin to cos. Just a little bit confused. Is the only way to remember all these things by just memorising them or should some logical steps go through your mind when your conducting these steps? I dont even know if any of that made sense but i hope you kinda get what i mean

[RuiAce]

It's up to you to know all of these trigonometric identities. To be fair there are a lot, but you (were supposed to have) covered all of them in Year 11 MX1, so you're expected to know them.
\[ \text{The following identities are called complementary identities}\\ \begin{align*} \sin (90^\circ - x) &= \cos x\\ \cos (90^\circ - x) &= \sin x \\ \tan (90^\circ - x) &= \cot x\\ \cot (90^\circ - x) &= \tan x \end{align*}\\ \text{and you should be able to figure it out for cosec and sec.} \]
Note that this is the whole point of the prefix co-.

Note also that for the other identities on your mind, cosec/sec/cot can be done by considering the reciprocal of sin/cos/tan.
\[ \text{All of the other identities are essentially your ASTC identities.}\\ \text{For example, since only 'sine is positive' in the second quadrant,}\\ \begin{align*}\sin (180^\circ - x) &= \sin x\\ \cos (180^\circ -x ) &= -\cos x\\ \tan (180^\circ - x) &= -\tan x \end{align*} \]
You should then be able to figure out third quadrant identities with \(180^\circ -x \), and fourth quadrant identities with \( 360^\circ - x\).

And finally, for the \(360^\circ\) case, the easiest way of thinking about it is really because the trigonometric functions are (at most) periodic in \(360^\circ\) (i.e. \(2\pi\)). But alternatively, you can imagine that the "fifth quadrant" is the exact same as the "first quadrant", and hence
\begin{align*}\sin (360^\circ + x) &= \sin x\\ \cos (360^\circ + x) &= \cos x\\ \tan (360^\circ +x) &= \tan x \end{align*}
(I use at most because technically \(\tan x\) is periodic in \(180^\circ\).)

And the idea is, well if you go around the unit circle again, the 9th quadrant will be the same as the 5th quadrant, which will be the same as the 1st quadrant. This gives us \( \sin (720^\circ + x) = \sin x\) and etc. Same goes for any quadrants. This also goes backwards, so we end up with \( \sin (-360^\circ + x) = \sin x\) as well (I think of this as the "negative 3rd quadrant" being the same as the 1st quadrant).

I will also leave converting these to radians as your exercise.