Unit circle definition of trig functions — all 6 trig functions for any angle: Worked examples

Before we start, three small tools we'll lean on constantly — each explained here so this page reads on its own.

The scenario matrix

Here is the complete list of case-classes this topic can throw at you. Every cell gets covered by at least one worked example below.

#	Case class	What's tricky	Covered by
A	Quadrant I, "nice" angle	baseline, no signs to worry about	Ex 1
B	Quadrant II (x<0, y>0)	cos negative, sin positive	Ex 2
C	Quadrant III (x<0, y<0)	both negative	Ex 3
D	Quadrant IV (x>0, y<0)	sin negative, cos positive	Ex 4
E	Axis / degenerate ( $x = 0$ or $y = 0$ )	some functions undefined — ALL four axis points	Ex 5
F	Negative angle (clockwise)	even/odd behaviour	Ex 6
G	Huge angle ( $> 360°$ )	periodicity, subtract full loops (in deg and rad)	Ex 7
H	Limiting behaviour ( $θ \to 90°$ )	"blows up" vs "undefined"	Ex 8
I	Real-world word problem	translate motion → coordinates (deg and rad)	Ex 9
J	Exam twist (given one function + quadrant)	reconstruct the rest via the circle	Ex 10

The master figure below fixes our conventions: the four quadrants with their coordinate signs, the CCW = positive direction, the start point $(1, 0)$ , and the exact landing arrows for the four "nice" angles $45°, 150°, 210°, 300°$ used in Ex 1–4. Refer back to it whenever a step says "which quadrant" — the coloured wedges tell you the signs at a glance.

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

Figure 1 (alt-text / caption). A violet unit circle centred at the origin with the horizontal axis labelled $x = cos θ$ and the vertical axis labelled $y = sin θ$ . Each quadrant is annotated with its coordinate signs: Q I (upper-right) $x > 0, y > 0$ ; Q II (upper-left) $x < 0, y > 0$ ; Q III (lower-left) $x < 0, y < 0$ ; Q IV (lower-right) $x > 0, y < 0$ . A navy square marks the start point $(1, 0)$ , and a curved navy arrow near the top shows CCW = positive. Four coloured radius arrows point to the landing dots for $45°$ (orange, Q I), $150°$ (magenta, Q II), $210°$ (violet, Q III) and $300°$ (orange, Q IV) — the angles worked in Ex 1–4.

Case A — Quadrant I baseline

Worked example Ex 1 (cell A). Find all six trig functions at

$θ = 45°$ . Forecast: guess first — will any of the six be negative? Will any be undefined?

Step 1. Locate the landing point. Rotating $45°$ CCW from $(1, 0)$ lands halfway up the first quarter (the orange arrow in the master figure), at equal height and width. Why this step? Coordinates are the trig values, so we must know where we land before anything else.

Step 2. Get the coordinates. On the unit circle a $45°$ point satisfies $x = y$ and $x^{2} + y^{2} = 1$ , so $2 x^{2} = 1$ , giving $x = y = \frac{1}{2} = \frac{2}{2}$ . Why this step? We derive it from the circle equation instead of memorising — the picture guarantees $x = y$ by symmetry along the $45°$ line.

Step 3. Read off the core two: $cos 45° = x = \frac{2}{2}$ , $sin 45° = y = \frac{2}{2}$ . Why this step? The parent's definition is literally $cos θ = x$ and $sin θ = y$ — the cosine is the horizontal coordinate and the sine is the vertical coordinate of the landing point, so "reading off" is the whole rule, not a shortcut.

Step 4. Build the other four by division: $tan 45° = \frac{y}{x} = 1, cot 45° = \frac{x}{y} = 1, sec 45° = \frac{1}{x} = 2, csc 45° = \frac{1}{y} = 2 .$ Why this step? The parent note defines these as ratios/reciprocals of $sin$ and $cos$ ; we never memorise separate tables.

Verify: $x^{2} + y^{2} = \frac{1}{2} + \frac{1}{2} = 1$ ✓ (on the circle). Why bother verifying? The circle equation is an independent check we did not use to get the answer, so if it fails we've made a sign or arithmetic slip somewhere — it catches errors before they spread. All positive, none undefined — matches "All" in Quadrant I. ✓

Case B — Quadrant II

Worked example Ex 2 (cell B). Find

$sin$ , $cos$ , $tan$ at $θ = 150°$ . Forecast: which of these three end up negative?

Step 1. Quadrant: $150°$ is between $90°$ and $180°$ → Quadrant II (upper-left; magenta wedge in the master figure). So $x < 0$ , $y > 0$ . Why this step? The signs of the coordinates decide the signs of the answers — this must come first.

Step 2. Reference angle: distance to the horizontal axis. Here the nearest x-axis is the negative one at $180°$ , so $θ_{ref} = 180° - 150° = 30°$ . Why this step? The reference angle gives the magnitudes from the familiar $30°$ values (its triangle is a mirror-copy of the $30°$ triangle).

Step 3. Magnitudes at $30°$ : $cos 30° = \frac{3}{2}$ , $sin 30° = \frac{1}{2}$ . Why do we know these? From the special $30°$ - $60°$ - $90°$ right triangle (an equilateral triangle cut in half): its legs are in ratio $1 : 3 : 2$ , which is exactly SOH-CAH-TOA giving $sin 30° = \frac{1}{2}$ , $cos 30° = \frac{3}{2}$ . See Right-triangle trigonometry (SOH-CAH-TOA).

Step 4. Apply Q2 signs: $cos 150° = - \frac{3}{2}$ (x negative), $sin 150° = + \frac{1}{2}$ (y positive). Why this step? Side lengths are positive but coordinates carry signs — this is exactly the classic mistake the parent warns about.

Step 5. $tan 150° = \frac{sin}{cos} = \frac{1/2}{- 3 /2} = - \frac{1}{3} = - \frac{3}{3}$ . Why this step? $tan = y / x$ by definition; a positive $y$ over a negative $x$ must be negative.

Verify: $(- \frac{3}{2})^{2} + (\frac{1}{2})^{2} = \frac{3}{4} + \frac{1}{4} = 1$ ✓ — confirms the point sits on the circle, so the signs and magnitudes are mutually consistent. Q2 = "Sine positive only", so $tan$ negative — consistent. ✓

Case C — Quadrant III

Worked example Ex 3 (cell C). Find

$cos$ , $sin$ , $tan$ at $θ = 210°$ . Forecast: Quadrant III has both coordinates negative — so which function comes out positive?

Step 1. Quadrant: $210° = 180° + 30°$ → Quadrant III (lower-left; the violet arrow in the master figure). $x < 0, y < 0$ . Why this step? An angle past $180°$ but before $270°$ lands in the bottom-left of the circle, where both coordinates are negative — that decides every sign that follows.

Step 2. Reference angle: measured from the negative x-axis at $180°$ , so $θ_{ref} = 210° - 180° = 30°$ . Why this step? Same acute helper; only the sign rule changes between quadrants.

Step 3. Magnitudes: $cos 30° = \frac{3}{2}$ , $sin 30° = \frac{1}{2}$ . Why do we know these? Same $30°$ - $60°$ - $90°$ half-equilateral triangle as in Ex 2 — the reference angle lets us reuse those fixed values.

Step 4. Apply Q3 signs (both negative): $cos 210° = - \frac{3}{2}$ , $sin 210° = - \frac{1}{2}$ . Why this step? In Q3 the landing point is below and to the left of the origin, so both the x-coordinate (cos) and the y-coordinate (sin) carry a minus sign.

Step 5. $tan 210° = \frac{- 1/2}{- 3 /2} = \frac{1}{3} = \frac{3}{3}$ — positive! Two negatives divide to a positive. Why this step? This is the "T" in All-Students-Take-Calculus: only tan/cot survive positive in Q3, because $y / x$ with both negative cancels the signs.

Verify: $\frac{3}{4} + \frac{1}{4} = 1$ ✓ (point on the circle). $tan > 0$ matches Q3 — a quick sign sanity-check that would flag any slip. ✓

Case D — Quadrant IV

Worked example Ex 4 (cell D). Find

$sin$ , $cos$ , $sec$ at $θ = 300°$ . Forecast: in the lower-right quarter, is $sec$ (the $1/ cos$ function) positive or negative?

Step 1. Quadrant: $300°$ is between $270°$ and $360°$ → Quadrant IV (lower-right; the second orange arrow in the master figure). $x > 0, y < 0$ . Why this step? Past $270°$ but before a full loop you land bottom-right: to the right of the origin ( $x > 0$ ) but below it ( $y < 0$ ) — this fixes the signs.

Step 2. Reference angle: nearest horizontal axis is the positive x-axis at $360°$ , so $θ_{ref} = 360° - 300° = 60°$ . Why this step? In Q4 the acute gap to the x-axis is measured back from the full loop; that gap is what matches a known angle's magnitudes.

Step 3. Magnitudes at $60°$ : $cos 60° = \frac{1}{2}$ , $sin 60° = \frac{3}{2}$ . Why do we know these? The same $30°$ - $60°$ - $90°$ triangle from Ex 2, just read from the $60°$ corner: the roles of the two legs swap, giving $cos 60° = \frac{1}{2}$ , $sin 60° = \frac{3}{2}$ .

Step 4. Apply Q4 signs: $cos 300° = + \frac{1}{2}$ , $sin 300° = - \frac{3}{2}$ . Why this step? Q4's landing point has a positive x (so cos is $+$ ) and a negative y (so sin is $-$ ).

Step 5. $sec 300° = \frac{1}{cos 300°} = \frac{1}{1/2} = 2$ — positive (reciprocal of a positive). Why this step? $sec$ inherits the sign of $cos$ ; Q4 has cosine positive ("C"osine survives in Q4).

Verify: $(\frac{1}{2})^{2} + (- \frac{3}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1$ ✓ (on the circle). $sec > 0$ matches Q4 — consistent. ✓

Case E — Axis / degenerate inputs (ALL four axis points)

Worked example Ex 5 (cell E). Find all six trig functions at each axis angle:

$θ = 0°, 90°, 180°, 270°$ (and note $360°$ ). Forecast: at each axis point one coordinate is $0$ . Predict before computing which two functions break at each angle.

Step 1. Landing points, one per axis (each a quarter-turn apart): $0° \to (1, 0), 90° \to (0, 1), 180° \to (- 1, 0), 270° \to (0, - 1) .$ Why this step? Every quarter-turn lands you exactly on an axis, so one coordinate is $0$ and the other is $\pm 1$ — these are the "danger points" where a denominator can vanish.

Step 2. Read the core two straight off ( $cos = x$ , $sin = y$ ), then build the four ratios, flagging any division by zero:

$θ$	$x = cos$	$y = sin$	$tan = \frac{y}{x}$	$cot = \frac{x}{y}$	$sec = \frac{1}{x}$	$csc = \frac{1}{y}$
$0°$	$1$	$0$	$0$	undef	$1$	undef
$90°$	$0$	$1$	undef	$0$	undef	$1$
$180°$	$- 1$	$0$	$0$	undef	$- 1$	undef
$270°$	$0$	$- 1$	undef	$0$	undef	$- 1$

Why this step? A function is undefined exactly when its denominator coordinate is $0$ : when $x = 0$ (at $90°, 270°$ ) the $x$ -in-the-bottom functions $tan, sec$ break; when $y = 0$ (at $0°, 180°$ ) the $y$ -in-the-bottom functions $cot, csc$ break. It is never a number there — just no value.

Step 3. Note $360°$ : a full loop returns to $(1, 0)$ , so it is identical to $0°$ in every entry. Why this step? It closes the loop — there is no "new" fifth axis case; the fourth quarter-turn brings you home.

Verify: every row satisfies $x^{2} + y^{2} = 1$ : $1^{2} + 0^{2}$ , $0^{2} + 1^{2}$ , $(- 1)^{2} + 0^{2}$ , $0^{2} + (- 1)^{2}$ all $= 1$ ✓. Why verify here especially? The undefined entries are where careless students write " $0$ " or " $\infty$ "; the circle check confirms the coordinates, and the denominator rule then unambiguously decides "undefined". ✓

Case F — Negative (clockwise) angle

Worked example Ex 6 (cell F). Evaluate

$cos (- 120°)$ and $sin (- 120°)$ . Forecast: a negative angle means walking clockwise. Which coordinate stays the same as $+ 120°$ , and which flips?

Step 1. Even/odd rules from the circle: walking $- θ$ (clockwise) lands at the mirror image across the x-axis of the $+ θ$ point, which keeps $x$ and negates $y$ . So $cos (- θ) = cos θ$ (even) and $sin (- θ) = - sin θ$ (odd). "Even" means a function unchanged by flipping the input's sign; "odd" means it flips output sign too. More: Even and odd functions. Why this step? Clockwise- $120°$ lands at the mirror image (across the x-axis) of CCW- $120°$ ; the picture is the proof — same horizontal reach, opposite height.

Step 2. Compute the positive version. $120°$ is Q2, reference angle $180° - 120° = 60°$ , so $cos 120° = - \frac{1}{2}$ , $sin 120° = + \frac{3}{2}$ . Why this step? We only ever need the positive-angle value; the parity rule then converts it.

Step 3. Apply the parities: $cos (- 120°) = cos 120° = - \frac{1}{2}$ ; $sin (- 120°) = - sin 120° = - \frac{3}{2}$ . Why this step? Cosine is even so it copies; sine is odd so it flips sign.

Verify: $(- \frac{1}{2})^{2} + (- \frac{3}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1$ ✓. And $- 120°$ lands in the lower-left (Q3-like) region, where both coords are negative — matches our signs, a good sanity check. ✓

Case G — Huge angle (past a full loop), degrees and radians

Worked example Ex 7 (cell G). Simplify

$tan (1125°)$ , then redo it in radians. Forecast: $1125°$ is way past a full circle. How many full loops, and where do you actually stop?

Step 1. Strip full loops of $360°$ . Since $1125 = 3 \times 360 + 45$ , we have $1125° = 3 \cdot 360° + 45°$ . Why this step? Periodicity: each $360°$ returns you to the identical point, so it changes nothing. See Periodicity and $2\pi$.

Step 2. Reduce: $tan (1125°) = tan (45°)$ . Why this step? Only the leftover $45°$ decides the coordinates.

Step 3. From Ex 1, $tan 45° = 1$ . Why this step? We already found the $45°$ landing point $(x, y)$ with $x = y$ , so $tan = y / x = 1$ .

Step 4. Radian version. Convert once at the start: $1125° \times \frac{π}{180} = \frac{25 π}{4}$ . Strip full loops of $2 π = \frac{8 π}{4}$ : $\frac{25 π}{4} - 3 \cdot \frac{8 π}{4} = \frac{25 π - 24 π}{4} = \frac{π}{4}$ , and $\frac{π}{4} = 45°$ . Why this step? Same idea, different unit — a full loop is $2 π$ rad instead of $360°$ ; the leftover $\frac{π}{4}$ is exactly $45°$ , so the answer is unchanged.

Verify: $1125 - 3 \cdot 360 = 45$ ✓, and in radians $\frac{25 π}{4} - 3 \cdot 2 π = \frac{π}{4}$ ✓, with $tan \frac{π}{4} = 1$ ✓. Both unit systems agree — reassuring that the conversion was done correctly. ✓

Case H — Limiting behaviour (blows up vs undefined)

The figure below plots $tan θ$ as $θ$ climbs toward $90°$ : the curve rockets upward toward a vertical dashed line at $90°$ but never touches a value there — that gap is the "undefined".

Figure 2 (alt-text / caption). A magenta curve of $y = tan θ$ over roughly $0°$ to $89.9°$ against peach background, axes in navy labelled "angle theta (degrees)" and "tan(theta)". The curve is gentle near $0°$ then steepens dramatically, shooting up past the marked points at $89°$ (orange dot, height about $57$ ) and $89.9°$ (violet dot, height about $573$ ). A vertical dashed navy line at $θ = 90°$ marks the asymptote, annotated "undefined at exactly 90 deg" — the curve approaches it but has no point on it.

Worked example Ex 8 (cell H). Compare

$tan (89°)$ , $tan (89.9°)$ , and $tan (90°)$ . Forecast: as the angle creeps toward $90°$ , is $tan$ approaching a specific big number, or is it exploding — and what happens exactly at $90°$ ?

Step 1. Recall $tan θ = \frac{y}{x}$ , with $x = cos θ$ shrinking to $0$ as $θ \to 90°$ (you're nearing the top of the circle, where horizontal reach $\to 0$ ). Why this step? The behaviour is entirely about the denominator $x$ collapsing.

Step 2. Numeric approach: $tan 89° \approx 57.29, tan 89.9° \approx 572.96.$ The value grows without bound — a "vertical asymptote" you'll meet again in Graphs of trig functions. Why this step? Watching the actual numbers explode makes "approaches infinity" concrete rather than abstract.

Step 3. At exactly $90°$ : the landing point is $(0, 1)$ , so $tan 90° = \frac{1}{0}$ = undefined. Not "a huge number" — no value at all. Why this step? This is the parent's third mistake spelled out: a limit trending to infinity is not the same as a value.

Verify: $tan 89° \approx 57.29$ and $tan 89.9° \approx 572.96$ (checked numerically), confirming the blow-up while $θ = 90°$ itself has $cos = 0$ → undefined. The growing numbers and the undefined endpoint together prove they are different situations. ✓

Case I — Real-world word problem

Worked example Ex 9 (cell I). A wheel of radius

$1 m$ has a marker starting at the 3-o'clock spot $(1, 0)$ . It rotates CCW through $θ = 225°$ (equivalently $\frac{5 π}{4}$ rad). How far right of centre and how far above centre is the marker? Forecast: $225°$ is bottom-left — so both "right" and "above" should come out negative (i.e. left and below).

Step 1. Translate the physics. "Distance right of centre" is the $x$ -coordinate $= cos θ$ ; "distance above centre" is the $y$ -coordinate $= sin θ$ , both in metres because the radius is $1 m$ . Why this step? On a radius- $1$ wheel the coordinates are the trig values, in the wheel's length units.

Step 2. Confirm the angle in radians too: $225° \times \frac{π}{180} = \frac{5 π}{4}$ rad. Why this step? Physics and engineering usually state rotations in radians; $\frac{5 π}{4}$ and $225°$ are the same turn, so either can be used.

Step 3. Quadrant + reference: $225° = 180° + 45°$ → Q3, $θ_{ref} = 45°$ , both coordinates negative. Why this step? Q3 places the marker below-left of centre, forcing both distances negative.

Step 4. Magnitudes at $45°$ are $\frac{2}{2}$ each, so $x = cos 225° = - \frac{2}{2} \approx - 0.707 m, y = sin 225° = - \frac{2}{2} \approx - 0.707 m .$ Why this step? Negative $x$ = $0.707 m$ to the left; negative $y$ = $0.707 m$ below centre.

Verify: $(- 0.707)^{2} + (- 0.707)^{2} \approx 0.5 + 0.5 = 1 = r^{2}$ ✓ — the marker really is on the wheel's rim, $1 m$ from centre. Why check this? It confirms our two distances are physically consistent with a $1 m$ radius; a wrong magnitude would fail here. Units: metres throughout. ✓

Case J — Exam twist (reconstruct from one clue)

Worked example Ex 10 (cell J). Given

$sin θ = \frac{3}{5}$ and $θ$ is in Quadrant II, find $cos θ$ and $tan θ$ — without a calculator. Forecast: you know $y$ ; can you recover $x$ using only the circle equation? What sign will $x$ have?

Step 1. Use the Pythagorean identity $sin^{2} θ + cos^{2} θ = 1$ (recall $sin^{2} θ$ means $(sin θ)^{2}$ ). This is the circle equation $x^{2} + y^{2} = 1$ with $x = cos θ$ , $y = sin θ$ substituted in. See Pythagorean identities. Why this step? The circle links $x$ and $y$ ; knowing one leg on a radius- $1$ triangle pins down the other's size.

Step 2. Solve for the magnitude: $cos^{2} θ = 1 - (\frac{3}{5})^{2} = 1 - \frac{9}{25} = \frac{16}{25}$ , so $∣ cos θ ∣ = \frac{4}{5}$ . Why this step? Squaring loses the sign, so this only gives the size $\frac{4}{5}$ — the sign is still open.

Step 3. Choose the sign from the quadrant. Q2 has $x < 0$ , so $cos θ = - \frac{4}{5}$ . Why this step? The identity gives only size; the quadrant supplies the missing sign — the exam is testing exactly this.

Step 4. $tan θ = \frac{sin θ}{cos θ} = \frac{3/5}{- 4/5} = - \frac{3}{4}$ . Why this step? $tan = y / x$ by definition; positive $y$ over negative $x$ is negative.

Verify: $(\frac{3}{5})^{2} + (- \frac{4}{5})^{2} = \frac{9}{25} + \frac{16}{25} = 1$ ✓. Why verify? Our reconstructed $cos$ was forced to land the point back on the unit circle; passing this check confirms the magnitude and that $\frac{3}{5}, \frac{4}{5}$ form a valid coordinate pair. Q2 → $tan < 0$ , consistent with the "Sine only" positivity rule. ✓

Recall Quick self-test (reveal after guessing)

Sign of $cos$ in Quadrant III? ::: Negative ( $x < 0$ ). $tan (- 45°) = ?$ ::: $- 1$ (tan is odd, $tan 45° = 1$ ). Which two functions are undefined at $θ = 270°$ ? ::: $tan$ and $sec$ (they divide by $x = 0$ ). Which two are undefined at $θ = 0°$ ? ::: $cot$ and $csc$ (they divide by $y = 0$ ). $sin (1125°)$ reduces to $sin$ of what angle? ::: $sin 45° = \frac{2}{2}$ (subtract $3 \times 360°$ , or $3 \times 2 π$ ). If $sin θ = \frac{3}{5}$ in Q2, then $cos θ = ?$ ::: $- \frac{4}{5}$ . Convert $225°$ to radians. ::: $\frac{5 π}{4}$ (multiply by $\frac{π}{180}$ ).

Connections

Unit circle definition of trig functions — all 6 trig functions for any angle — the parent this page drills.
Reference angles — the acute helper used in Ex 2–4, 6, 9.
Pythagorean identities — the reconstruction tool in Ex 10.
Right-triangle trigonometry (SOH-CAH-TOA) — the $0°$ – $90°$ special case and source of the $30°$ / $60°$ / $45°$ values.
Even and odd functions — the parity logic of Ex 6.
Periodicity and $2\pi$ — the loop-stripping of Ex 7.
Graphs of trig functions — the asymptote behaviour of Ex 8.
Radian measure — the degree↔radian conversion used in Ex 7 and Ex 9.

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