Exercises — Unit circle definition of trig functions — all 6 trig functions for any angle

WHAT the point means: $x=0$, $y=-1$ — you are at the very bottom of the circle (the "6 o'clock" spot). WHICH angle: starting at $(1,0)$ (3 o'clock) and going CCW, the bottom is three-quarters of the way around: $\theta =270°$. Read off coordinates directly (that IS the definition): $\cos 270°=x=0,\sin 270°=y=-1.$ No formula needed — the coordinates are the answer.

$200°$ is past $180°$ but before $270°$, so it lands in the bottom-left region = Quadrant III. In the bottom-left, you are to the left of centre ($x<0$) and below centre ($y<0$). $\cos 200°=x<0,\sin 200°=y<0.$

Landing point: a half-turn CCW from $(1,0)$ lands at the far left: $P=(-1,0)$. Core two (read coordinates): $\cos 180°=x=-1$, $\sin 180°=y=0$. The other four are built from these — check for zero denominators: $\tan 180°=\frac{y}{x}=\frac{0}{-1}=0.$ $\cot 180°=\frac{x}{y}=\frac{-1}{0}=\text{undefined}(\text{divide by }y=0).$ $\sec 180°=\frac{1}{x}=\frac{1}{-1}=-1.$ $\csc 180°=\frac{1}{y}=\frac{1}{0}=\text{undefined}(\text{divide by }y=0).$ WHY two are undefined: $y=0$ sits in the bottom of $\cot$ and $\csc$; you cannot divide by zero, so those functions have no value here — not "a big number," simply undefined.

WHY $45°$ is special: the landing point sits on the line $y=x$ (equal right and up), and it must obey $x^2+y^2=1$. Set $y=x$: $x^2+x^2=1\Rightarrow 2x^2=1\Rightarrow x=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ (positive, since Q1 is right-and-up). $\cos 45°=\frac{\sqrt{2}}{2},\sin 45°=\frac{\sqrt{2}}{2}.$ $\tan 45°=\frac{y}{x}=\frac{\sqrt{2}/2}{\sqrt{2}/2}=1.$ $\tan 45°=1$ makes sense: equal rise and run means slope $1$.

Step 1 — quadrant (choose the sign): $210°=180°+30°$, so it's in Quadrant III → both $x<0$ and $y<0$. Step 2 — reference angle (get the magnitude): the reference angle is the acute angle to the nearest x-axis: $210°-180°=30°$. The size of $x$ and $y$ here equals their size at $30°$. Step 3 — known $30°$ magnitudes: $\cos 30°=\frac{\sqrt{3}}{2}$, $\sin 30°=\frac{1}{2}$. Step 4 — attach Q3 signs (both negative): $\cos 210°=-\frac{\sqrt{3}}{2},\sin 210°=-\frac{1}{2}.$ Check on the circle: ${\left(-\frac{\sqrt{3}}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2=\frac{3}{4}+\frac{1}{4}=1$ ✓.

Quadrant: $135°=90°+45°$ → Quadrant II → $x<0$, $y>0$. Reference angle: distance to the nearest x-axis is $180°-135°=45°$. Magnitudes from $45°$: $\cos =\sin =\frac{\sqrt{2}}{2}$. Attach Q2 signs: $\cos 135°=-\frac{\sqrt{2}}{2}$ (negative $x$), $\sin 135°=+\frac{\sqrt{2}}{2}$ (positive $y$). Now build the requested functions: $\tan 135°=\frac{\sin }{\cos }=\frac{+\sqrt{2}/2}{-\sqrt{2}/2}=-1.$ $\sec 135°=\frac{1}{\cos }=\frac{1}{-\sqrt{2}/2}=-\frac{2}{\sqrt{2}}=-\sqrt{2}.$ WHY $\tan$ is negative here: in Q2 the "up" is positive but "right" is negative, so their ratio $y/x$ is negative — matches "All Students..." where only Sine (and its reciprocal) stay positive in Q2.

Attack each piece with the right tool. Term 1 — odd symmetry: $\sin$ is odd (clockwise flips $y$): $\sin (-30°)=-\sin 30°=-\frac{1}{2}$. Term 2 — periodicity: $390°=360°+30°$, one full loop back to the same point, so $\cos 390°=\cos 30°=\frac{\sqrt{3}}{2}$. Term 3 — direct coordinate: at $180°$ the point is $(-1,0)$, so $\tan 180°=\frac{0}{-1}=0$. Add them: $-\frac{1}{2}+\frac{\sqrt{3}}{2}+0=\frac{\sqrt{3}-1}{2}.$ Numerically $\approx \frac{1.732-1}{2}\approx 0.366$.

WHY use the Pythagorean identity: we know $y=\sin \theta$ and need $x=\cos \theta$; the circle equation $x^2+y^2=1$ links them directly — that's exactly the tool that turns one coordinate into the other. Step 1: ${\cos }^2\theta =1-{\sin }^2\theta =1-{\left(\frac{3}{5}\right)}^2=1-\frac{9}{25}=\frac{16}{25}$. Step 2 — take the root, then choose sign by quadrant: $\cos \theta =\pm \frac{4}{5}$. In Q2 the x-coordinate is negative, so $\cos \theta =-\frac{4}{5}.$ Step 3 — build the rest: $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3/5}{-4/5}=-\frac{3}{4},\sec \theta =\frac{1}{\cos \theta }=-\frac{5}{4}.$ Sanity check: Q2 → only sine positive; indeed $\sin >0$ while $\cos ,\tan ,\sec$ all came out negative ✓.

Start from the circle itself: the point $(x,y)$ satisfies $x^2+y^2=1.$ WHY divide by $x^2$: we want $\tan =y/x$ and $\sec =1/x$ to appear, and both have $x$ in the denominator — so dividing the whole equation by $x^2$ manufactures exactly those ratios. $\frac{x^2}{x^2}+\frac{y^2}{x^2}=\frac{1}{x^2}⟹1+{\left(\frac{y}{x}\right)}^2={\left(\frac{1}{x}\right)}^2.$ Rename with definitions $\tan \theta =y/x$, $\sec \theta =1/x$: $1+{\tan }^2\theta ={\sec }^2\theta .■$ Excluded angles: we divided by $x^2$, which is illegal when $x=0$. That happens at $\theta =90°,270°,\dots$ (top and bottom of the circle) — exactly where $\tan$ and $\sec$ are already undefined. Consistent: the identity is only claimed where both sides exist.

Geometric WHAT: adding $180°$ rotates the landing point by a half-turn about the origin. A half-turn sends the point $(x,y)$ to the point directly opposite through the centre, which is $(-x,-y)$. See it: in the figure below, the amber point $(x,y)$ and its half-turn image $(-x,-y)$ are on a straight line through the centre, equal distance either side. Read the new coordinates: $\cos (\theta +180°)=(\text{new }x)=-x=-\cos \theta ,$ $\sin (\theta +180°)=(\text{new }y)=-y=-\sin \theta .$ WHY "every $\theta$": the half-turn map $(x,y)\mapsto (-x,-y)$ is true for all points on the circle — no quadrant, sign, or size restriction. It even works at the axis points, e.g. $\theta =0$: $(1,0)\mapsto (-1,0)$ gives $\cos 180°=-1$ ✓.

Way 1 — even symmetry then reference angle: $\cos$ is even, so $\cos (-150°)=\cos 150°$. Now $150°$ is Q2 (reference angle $30°$, cosine negative): $\cos 150°=-\frac{\sqrt{3}}{2}$. Way 2 — shift by $180°$: write $-150°=30°-180°$, so $\cos (-150°)=\cos (30°-180°)$. By L5.2 (a $180°$ shift flips sign), $\cos (30°-180°)=-\cos 30°=-\frac{\sqrt{3}}{2}$. Agreement: both give $-\frac{\sqrt{3}}{2}$ ✓. Two independent tools, same answer — that's the consistency the unit circle guarantees.

Read a landing point → give all 6 ::: L1–L2 Reduce any angle via quadrant + reference angle ::: L3 Use periodicity, odd/even, and ${\sin }^2+{\cos }^2=1$ together ::: L4 Prove a Pythagorean-type identity from $x^2+y^2=1$ ::: L5 Prove a shift identity from a coordinate map, all cases ::: L5