Trig functions for angles beyond 90° — ASTC rule (All, Sin, Tan, Cos): Worked examples

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are each tagged with the cell they cover, and together they hit all of them.

Cell	What makes it tricky	Example
C1 QII, sine	positive sign hides behind an obtuse angle	Ex 1
C2 QIII, cosine	both coordinates negative	Ex 2
C3 QIV, tangent	sign flips from the "friendly" value	Ex 3
C4 Negative angle	measured clockwise instead of anticlockwise	Ex 4
C5 Beyond one turn	must strip off full $360°$ rotations first	Ex 5
C6 Boundary / degenerate	terminal ray on an axis → a ratio is $0$ or undefined	Ex 6
C7 Limiting behaviour	approaching a vertical asymptote of $tan$	Ex 7
C8 Real-world word problem	a compass bearing dressed as an angle	Ex 8
C9 Exam twist	$90°$ shift forces a co-function swap, not plain ASTC	Ex 9

Figure — Trig functions for angles beyond 90° — ASTC rule (All, Sin, Tan, Cos)

Look at the four coloured wedges: each has a little acute angle drawn back to the horizontal axis — that acute angle is $θ^{'}$ , and it is what supplies the size of every answer below.

Ex 1 — Cell C1 (QII, sine)

Worked example Evaluate

$sin 135°$ Forecast: obtuse angle, sine... will it be positive or negative? Guess before reading.

Find the quadrant. $135°$ lies between $90°$ and $180°$ → QII. Why this step? Sign depends entirely on quadrant; nothing else can be decided first.
Find the reference angle. In QII, $θ^{'} = 180° - θ = 180° - 135° = 45°$ . Why this step? The reference angle is the gap from the nearest piece of the $x$ -axis, which in QII is the negative $x$ -axis at $180°$ . This gives the magnitude.
Attach the sign. QII is the S in ASTC → sine is positive. Why this step? In QII the point $P = (x, y)$ has $y > 0$ , and $sin θ = y$ .
Combine. $sin 135° = + sin 45° = \frac{2}{2}$ .

Verify: $45°$ is the diagonal of the unit square, whose $y$ -height is $\frac{1}{2} = \frac{2}{2} \approx 0.707$ . Positive, as a point in the upper half-plane must be. ✓

Ex 4 — Cell C4 (negative angle)

Ex 5 — Cell C5 (beyond one full turn)

Ex 6 — Cell C6 (boundary / degenerate: ray on an axis)

Worked example Evaluate

$sin 180°$ , $cos 90°$ , and $tan 90°$ Forecast: the ray lands exactly on an axis. No quadrant, no reference angle — read the coordinates directly.

Here ASTC has nothing to do, because the terminal ray is between quadrants, sitting on an axis. We just read $P = (x, y)$ off the Unit Circle.

$sin 180°$ : at $180°$ the ray points along the negative $x$ -axis, so $P = (- 1, 0)$ . Then $sin 180° = y = 0$ . Why? Sine is the height $y$ ; on the horizontal axis the height is zero.
$cos 90°$ : at $90°$ the ray points straight up, $P = (0, 1)$ . Then $cos 90° = x = 0$ . Why? Cosine is the horizontal coordinate $x$ ; straight up has no horizontal reach.
$tan 90°$ : $tan 90° = \frac{y}{x} = \frac{1}{0}$ → undefined. Why? Division by zero. The terminal ray is vertical, so it has zero "run" — a slope of infinite steepness has no finite value.

Verify: on the circle $x^{2} + y^{2} = 1$ : point $(- 1, 0)$ gives $sin = 0$ ; point $(0, 1)$ gives $cos = 0$ ; and $tan$ at a vertical ray must be undefined since $cos 90° = 0$ sits in the denominator. ✓

Ex 7 — Cell C7 (limiting behaviour of $tan$ )

Worked example What does

$tan θ$ do as $θ \to 90°$ ? Forecast: just below $90°$ , is $tan$ a big positive or big negative number? And just above?

Approach from below, $θ = 89°, 89.9°, \dots$ : these are in QI, where $x > 0$ is tiny and $y \approx 1$ . So $tan = y / x = \frac{near 1}{tiny +} \to + \infty$ . Why? Dividing a number near $1$ by a shrinking positive number grows without bound, positively.
Approach from above, $θ = 91°, 90.1°, \dots$ : these are in QII, where $x < 0$ is tiny and $y \approx 1$ . So $tan = \frac{near 1}{tiny -} \to - \infty$ . Why? Now the denominator is a shrinking negative number, flipping the sign.
Conclusion: $θ = 90°$ is a vertical asymptote of $tan$ ; the value leaps from $+ \infty$ to $- \infty$ across it. See Graphs of Sin Cos Tan.

Verify (numerically): $tan 89° \approx 57.29$ (large positive), $tan 91° \approx - 57.29$ (large negative). The magnitudes match because $89°$ and $91°$ share reference angle $1°$ ; only the sign flips as we cross the axis. ✓

Ex 8 — Cell C8 (real-world word problem: a bearing)

Worked example A drone flies on a compass

bearing of 210°. Its speed is $40$ m/s. Find the east–west and north–south components of its velocity. Forecast: bearings are measured clockwise from North, not the maths way. Which quadrant of ordinary maths angle is this, and will both components be negative?

Translate bearing → standard maths angle. Bearings count clockwise from North (up); maths angles count anticlockwise from East (right). Conversion: $θ_{maths} = 90° - bearing = 90° - 210° = - 120°$ , coterminal with $240°$ . Why this step? We must put the direction into the Unit Circle frame before ASTC applies.
Quadrant & reference. $240°$ → QIII, $θ^{'} = 240° - 180° = 60°$ .
Apply ASTC. QIII → both cosine and sine negative.
- East component $= 40 cos 240° = 40 \times (- cos 60°) = 40 \times (- \frac{1}{2}) = - 20$ m/s (i.e. $20$ m/s West).
- North component $= 40 sin 240° = 40 \times (- sin 60°) = 40 \times (- \frac{3}{2}) = - 203 \approx - 34.64$ m/s (i.e. $34.64$ m/s South). Why? A bearing of $210°$ points into the south-west, so both the eastward and northward components should indeed be negative — the sign check agrees with common sense.
Combine into a sanity magnitude. $(- 20)^{2} + (- 203)^{2} = 400 + 1200 = 1600 = 40$ m/s.

Verify: the recombined speed is $40$ m/s, exactly the given speed — components must always rebuild the original magnitude. ✓ Both negative ⇒ south-west, matching a $210°$ bearing. ✓

Ex 9 — Cell C9 (exam twist: the $90°$ shift co-function trap)

Worked example Simplify

$sin (90° + θ)$ for $θ = 40°$ , then state the general identity. Forecast: most students blast ASTC and keep it as $sin$ . That's the trap. What actually happens?

Spot the danger. The shift is $90°$ (an odd multiple of $90°$ ), not $180°$ or $360°$ . Odd $90°$ shifts force a co-function swap: $sin \leftrightarrow cos$ . See Trigonometric Identities. Why? Rotating the terminal ray by $90°$ swaps the roles of the horizontal and vertical coordinates, so the function itself changes, not just its sign.
Find the sign with ASTC. For small $θ = 40°$ , the angle $90° + 40° = 130°$ lands in QII, where sine is positive. Why? We still need ASTC — but only to fix the sign of the new function.
Write the identity. $sin (90° + θ) = + cos θ$ .
Numeric check. $sin 130° = cos 40°$ .

Verify: $sin 130° \approx 0.766$ and $cos 40° \approx 0.766$ — equal, confirming the co-function swap (and that plain " $sin (90° + θ) = sin θ$ " would be wrong: $sin 40° \approx 0.643 \neq = 0.766$ ). ✓

Active recall

Recall Quick self-test on the matrix

How do you convert $- 120°$ before applying ASTC? ::: Add $360°$ to get the coterminal $240°$ . Reference angle of $780°$ ? ::: Reduce to $60°$ (subtract $360°$ twice), which is already acute → $θ^{'} = 60°$ . Why is $tan 90°$ undefined? ::: $cos 90° = 0$ sits in the denominator $y / x$ ; division by zero. As $θ \to 90 °^{-}$ , $tan θ \to$ ? ::: $+ \infty$ (QI, tiny positive $x$ ). As $θ \to 90 °^{+}$ , $tan θ \to$ ? ::: $- \infty$ (QII, tiny negative $x$ ). Does $sin (90° + θ)$ stay as sine? ::: No — an odd $90°$ shift swaps to $+ cos θ$ . Bearing $210°$ points into which compass region? ::: South-west (both E and N components negative).

Connections

Unit Circle — where boundary values ( $sin 180° = 0$ , $cos 90° = 0$ ) are read straight off.
Reference Angles — the magnitude supplier used in every example.
Trigonometric Identities — the co-function swap in Ex 9.
Radians and Degrees — periodicity used in Ex 4 and Ex 5.
Graphs of Sin Cos Tan — the $tan$ asymptote of Ex 7 seen as a graph.

Worked examples — Trig functions for angles beyond 90° — ASTC rule (All, Sin, Tan, Cos)