Digital SAT Math · Geometry & Trigonometry

Trigonometry beyond basic right triangles

Right-triangle SOH-CAH-TOA is not enough for every trig item on the Digital SAT. A short slice of the test puts an angle in standard position on the coordinate plane, asks for sin⁡\sin, cos⁡\cos, or tan⁡\tan at a special or quadrantal measure, or asks you to convert degrees ↔ radians. The arithmetic is tiny; the charge is for where the angle sits and which sign that quadrant forces.

On the test

DomainGeometry and Trigonometry (score report)
What it looks likeAn angle in degrees or radians (often with π\pi); a unit circle or ray from the origin; a request for sin⁡θ\sin\theta, cos⁡θ\cos\theta, tan⁡θ\tan\theta, a coordinate, or a converted measure
Often asked“What is cos⁡ ⁣(5π6)\cos\!\left(\dfrac{5\pi}{6}\right)?”, “Convert 225∘225^\circ to radians”, “The terminal point is (x,y)(x, y); what is yy?”
FormatMultiple choice and student-produced response
CalculatorAllowed; mode matters — Desmos/your calculator in degree mode will mangle a π\pi-radian input

Recognition cues: π\pi in an angle measure; the words unit circle, standard position, terminal side, radians; a diagram with a circle of radius 1 centered at the origin; choices that differ only by a sign.

What this page owns

This page owns conversion, unit-circle values at quadrantal and special angles, and sign by quadrant. Anything that is only SOH-CAH-TOA on a drawn right triangle belongs on right-triangle trigonometry. Arc length, sector area, and the circle equation (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2 belong on circles (and coordinate geometry for the equation form). Here the circle is only a unit circle used to read (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta).

Pattern recognition

Five shapes cover essentially every item:

  1. Convert — degrees → radians or radians → degrees with π rad=180∘\pi\,\mathrm{rad}=180^\circ.
  2. Quadrantal — 00, π/2\pi/2, π\pi, 3π/23\pi/2 (and coterminal copies); answers are −1-1, 00, or 11.
  3. Special angle, Q1 value — reference 30∘/45∘/60∘30^\circ/45^\circ/60^\circ (or π/6,π/4,π/3\pi/6,\pi/4,\pi/3) with the usual half-radical values.
  4. Sign by quadrant — same magnitude as the reference angle; sign from ASTC (or from the unit-circle coordinates).
  5. Coterminal reduce first — angle outside [0,2π)[0,2\pi) reduced by ±2π\pm 2\pi before reading the value.

Method

  1. Convert with one identity only. π\pi radians =180∘= 180^\circ. Degrees → radians: multiply by π/180\pi/180. Radians → degrees: multiply by 180/π180/\pi. Cancel π\pi when it sits in both numerator and denominator; reduce the fraction.
  2. Place the angle. Sketch a ray from the origin, counterclockwise from the positive xx-axis (negative angles go clockwise). Name the quadrant or the axis.
  3. Reference angle from the xx-axis. In QI the reference is the angle itself. QII: π−θ\pi-\theta (or 180∘−θ180^\circ-\theta). QIII: θ−π\theta-\pi (or θ−180∘\theta-180^\circ). QIV: 2π−θ2\pi-\theta (or 360∘−θ360^\circ-\theta). The reference is always between 00 and π/2\pi/2.
  4. Recall the Q1 value, then sign. Memorize (or rebuild from 3030-6060-9090 and 4545-4545-9090):
Referencesin⁡\sincos⁡\costan⁡\tan
00001100
π/6\pi/6 (30∘30^\circ)1/21/23/2\sqrt{3}/21/31/\sqrt{3}
π/4\pi/4 (45∘45^\circ)2/2\sqrt{2}/22/2\sqrt{2}/211
π/3\pi/3 (60∘60^\circ)3/2\sqrt{3}/21/21/23\sqrt{3}
π/2\pi/2 (90∘90^\circ)1100undefined
  1. Sign from the quadrant (ASTC). All positive in I; Sine (and csc) positive in II; Tangent (and cot) positive in III; Cosine (and sec) positive in IV. On the unit circle: x=cos⁡θx=\cos\theta, y=sin⁡θy=\sin\theta, so the sign of cosine is the sign of xx and the sign of sine is the sign of yy.
Quadrantsin⁡\sincos⁡\costan⁡\tan
I++++++
II++−-−-
III−-−-++
IV−-++−-

Worked example 1 — convert, then stop

Stem. Convert 150∘150^\circ to radians.

Step 1 — apply the factor. Degrees to radians multiplies by π/180\pi/180:

150∘⋅π180=150π180=5π6150^\circ\cdot\frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}

Step 2 — reduce. 150150 and 180180 share a factor of 3030: 5π/65\pi/6.

Check. 5π/65\pi/6 is a bit less than π\pi, so a bit less than 180∘180^\circ — matches 150∘150^\circ. Inverted factor 150⋅180/π150\cdot 180/\pi is huge and not a special angle. Answer: 5π6\dfrac{5\pi}{6}.

Trap watch. Multiplying by 180/π180/\pi instead of π/180\pi/180 is the inverted conversion. Writing π/6\pi/6 drops the 55. Writing 5π/35\pi/3 doubles the angle into QIV.

Worked example 2 — quadrant, reference, sign

Stem. What is the value of cos⁡ ⁣(5π6)\cos\!\left(\dfrac{5\pi}{6}\right)?

Step 1 — place. π/2<5π/6<π\pi/2 < 5\pi/6 < \pi, so quadrant II.

Step 2 — reference from the xx-axis.

π−5π6=π6\pi - \frac{5\pi}{6} = \frac{\pi}{6}

Step 3 — Q1 value. cos⁡(π/6)=3/2\cos(\pi/6) = \sqrt{3}/2.

Step 4 — sign in QII. Cosine is negative in QII (ASTC: only sine is positive).

cos⁡ ⁣(5π6)=−32\cos\!\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}

Check. On the unit circle the terminal point in QII near the negative xx-axis has xx negative and about −3/2-\sqrt{3}/2, y=+1/2y = +1/2. Answer: −32-\dfrac{\sqrt{3}}{2}.

Trap watch. Dropping the sign gives +3/2+\sqrt{3}/2. Measuring the reference from the yy-axis as 5π/6−π/2=π/35\pi/6-\pi/2=\pi/3 swaps to ±1/2\pm 1/2. Reporting sin⁡(5π/6)=1/2\sin(5\pi/6)=1/2 answers the wrong function.

Worked example 3 — unit-circle coordinate

Stem. An angle θ\theta in standard position has terminal point PP on the unit circle in quadrant III, and the reference angle is π/3\pi/3. What is the yy-coordinate of PP?

Step 1 — yy means sine. On the unit circle, y=sin⁡θy=\sin\theta.

Step 2 — Q1 value for reference π/3\pi/3. sin⁡(π/3)=3/2\sin(\pi/3)=\sqrt{3}/2.

Step 3 — sign in QIII. Sine is negative in QIII.

y=−32y = -\frac{\sqrt{3}}{2}

Check. QIII means both xx and yy negative, so cosine would be −1/2-1/2 for the same reference. The stem asked for yy, not xx. Answer: −32-\dfrac{\sqrt{3}}{2}.

Trap watch. Reporting +3/2+\sqrt{3}/2 drops the QIII sign. Reporting −1/2-1/2 uses the cosine value (or confuses 30∘30^\circ with 60∘60^\circ). Reporting the reference angle itself as a coordinate is not a coordinate.

Practice

Answer before opening the explanation. Three items hand you a unit circle figure — the value you need is the coordinate or the angle the diagram forces, not a number sitting in the sentence. Two are student-produced response. Every wrong choice below is one named slip: inverted conversion, wrong-axis reference, sign drop, or the wrong special value.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

Which of the following is the radian measure of a 90° angle?

Show the answer Choice A

Why it is right

Degrees convert to radians by multiplying by π/180. So 90 · (π/180) = 90π/180 = π/2. A right angle is one-quarter of a full turn; a full turn is 2π radians, so one quarter is (2π)/4 = π/2, which matches.

Why each other choice fails

Choice B
Halves 90° incorrectly into a 45° angle, whose radian measure is π/4. The conversion is 90 · π/180 = π/2, not π/4.
Choice C
Uses a full rotation (360° = 2π radians) instead of a quarter rotation. 90° is one-fourth of 360°, so the radian measure is one-fourth of 2π.
Choice D
Multiplies 90 by π and forgets to divide by 180. The conversion factor is π/180, not π alone.

Question 2 Standard

What is the degree measure of an angle of 3π/4 radians?

Show the answer Choice B

Why it is right

Radians convert to degrees by multiplying by 180/π. So (3π/4) · (180/π) = 3 · 180 / 4 = 540/4 = 135. The π cancels, leaving a pure degree measure. 135° sits in quadrant II, consistent with 3π/4 being three-quarters of the way from 0 to π.

Why each other choice fails

Choice A
Uses 100 in place of 180 in the conversion, computing (3/4)·100 = 75. The identity is π rad = 180°, not 100°.
Choice C
Converts 4π/3 instead of 3π/4: (4π/3)·(180/π) = 240. The numerator and denominator of the coefficient were swapped.
Choice D
Treats the angle as three-quarters of a full turn: (3/4)·360 = 270. A full turn in radians is 2π, not π, so three-quarters of a turn is 3π/2, not 3π/4.

Question 3 Standard

The figure shows an angle of 3π/2 radians in standard position on the unit circle. What is the value of cos(3π/2)?

3π/2
Angle 3π/2 in standard position on the unit circle.
Show the answer Choice C

Why it is right

The terminal side of 3π/2 lies on the negative y-axis, so the unit-circle point is (0, −1). Cosine is the x-coordinate of that point, which is 0. Equivalently, 3π/2 is a quadrantal angle with cos = 0 and sin = −1.

Why each other choice fails

Choice A
Reports the cosine of π (or the sine of 3π/2 with the wrong function). At 3π/2 the point is (0, −1), so cosine is 0 and sine is −1.
Choice B
Reports the cosine of 0 (the positive x-axis). The figure's ray points down the negative y-axis, not along the positive x-axis.
Choice D
Invents a special-angle half value for a quadrantal angle. Quadrantal cosines are only −1, 0, or 1 — never 1/2.

Question 4 Standard

What is the value of sin(2π/3)?

Show the answer Choice D

Why it is right

2π/3 is between π/2 and π, so it lies in quadrant II. The reference angle from the positive x-axis is π − 2π/3 = π/3. Then sin(π/3) = √3/2, and sine is positive in quadrant II, so sin(2π/3) = √3/2.

Why each other choice fails

Choice A
Uses the correct magnitude √3/2 but attaches a negative sign. Sine is positive in quadrant II; the negative would be correct for sine in QIII or QIV.
Choice B
Reports cos(2π/3) with the sign flipped, or uses the 30° reference value instead of 60°. The reference for 2π/3 is π/3, whose sine is √3/2, not 1/2.
Choice C
Combines the cosine magnitude of the reference angle with a negative sign — cos(π/3) = 1/2 and cos is negative in QII, which is cos(2π/3), not sin(2π/3).

Question 5 Standard Student-produced response

An angle measures 7π/6 radians. What is the degree measure of this angle?

Show the answer 210

Why it is right

Multiply by 180/π to convert radians to degrees: (7π/6) · (180/π) = 7 · 180 / 6 = 1260 / 6 = 210. So 7π/6 radians equals 210°. That is 30° past 180°, which matches a reference angle of π/6 in quadrant III.

Answers students type instead

30
Reports only the reference angle in degrees (π/6 → 30°) and forgets to rebuild the original angle in its quadrant.
150
Converts 5π/6 instead of 7π/6, or subtracts 30° from 180° and stops as if the angle were in QII. 7π/6 is past π, so the degree measure is 180 + 30 = 210, not 150.
420
Multiplies 7/6 by 360 instead of 180, treating the coefficient as a fraction of a full turn rather than a fraction of π.

Question 6 Harder

The figure shows an angle of 2π/3 radians in standard position on the unit circle. What is the x-coordinate of the point where the terminal side meets the unit circle?

2π/3
Angle 2π/3 in standard position on the unit circle.
Show the answer Choice A

Why it is right

On the unit circle the x-coordinate is cos(θ). For θ = 2π/3 (quadrant II), the reference angle is π − 2π/3 = π/3, and cos(π/3) = 1/2. Cosine is negative in quadrant II, so cos(2π/3) = −1/2. That is the x-coordinate of the terminal point.

Why each other choice fails

Choice B
Drops the quadrant II sign. The magnitude 1/2 is correct for the reference angle π/3, but cosine must be negative when the terminal side is in QII.
Choice C
Reports sin(2π/3) with a wrong (negative) sign, or measures the reference from the y-axis and then signs as cosine. The y-coordinate is +√3/2; the x-coordinate is −1/2.
Choice D
Reports the positive sine of the angle — the y-coordinate rather than the x-coordinate. On the unit circle, x = cos θ and y = sin θ.

Question 7 Harder

What is the value of cos(5π/4)?

Show the answer Choice B

Why it is right

5π/4 lies between π and 3π/2, so it is in quadrant III. The reference angle measured to the nearest x-axis is 5π/4 − π = π/4. The first-quadrant value is cos(π/4) = √2/2. Cosine is negative in quadrant III (ASTC: only tangent is positive there), so cos(5π/4) = −√2/2. On the unit circle both coordinates of the terminal point are negative at this angle.

Why each other choice fails

Choice A
Keeps the Q1 magnitude and drops the sign. In quadrant III both sine and cosine are negative, so the cosine cannot be positive.
Choice C
Uses a 30°/60° special value (√3/2) instead of the 45° value required by reference π/4. The reference for 5π/4 is π/4, not π/6.
Choice D
Combines the wrong special magnitude with a positive sign — two errors that do not cancel into the key.

Question 8 Harder

What is the value of tan(5π/3)?

Show the answer Choice C

Why it is right

5π/3 is in quadrant IV (between 3π/2 and 2π). The reference angle is 2π − 5π/3 = π/3. Then tan(π/3) = √3, and tangent is negative in quadrant IV, so tan(5π/3) = −√3. As a check: sin(5π/3) = −√3/2 and cos(5π/3) = 1/2, so the quotient is (−√3/2)/(1/2) = −√3.

Why each other choice fails

Choice A
Drops the quadrant IV sign. Tangent is positive in QI and QIII only; in QIV it is negative.
Choice B
Uses tan(π/6) = 1/√3 instead of tan(π/3) = √3 — the 30° value for a 60° reference — and also drops the needed negative sign.
Choice D
Uses the 30° tangent magnitude with the correct QIV sign. The reference for 5π/3 is π/3, whose tangent is √3, not 1/√3.

Question 9 Harder Student-produced response

What is the value of sin(7π/6)?

Show the answer -1/2

Why it is right

7π/6 is between π and 3π/2, so it is in quadrant III. The reference angle is 7π/6 − π = π/6. Then sin(π/6) = 1/2, and sine is negative in quadrant III, so sin(7π/6) = −1/2. On the unit circle the terminal point is (−√3/2, −1/2); the y-coordinate is the sine.

Answers students type instead

1/2
Drops the quadrant III sign and reports the positive Q1 sine of the reference angle π/6.
√3/2
Uses sin(π/3) instead of sin(π/6) — wrong-axis or 30°/60° swap — and usually without the required negative sign.
-√3/2
Reports cos(7π/6) instead of sin(7π/6). Cosine is the x-coordinate −√3/2; sine is the y-coordinate −1/2.

Question 10 Hardest

The figure shows an angle of 11π/6 radians in standard position on the unit circle. What is the y-coordinate of the point where the terminal side meets the unit circle?

11π/6
Angle 11π/6 in standard position on the unit circle.
Show the answer Choice D

Why it is right

On the unit circle the y-coordinate is sin(θ). For θ = 11π/6 (quadrant IV), the reference angle is 2π − 11π/6 = π/6. Then sin(π/6) = 1/2, and sine is negative in quadrant IV, so sin(11π/6) = −1/2. That is the y-coordinate of the terminal point.

Why each other choice fails

Choice A
Drops the quadrant IV sign. The magnitude matches the reference π/6, but the terminal side is below the x-axis, so y must be negative.
Choice B
Uses the cosine magnitude of the reference (or sin of π/3) with a positive sign — both the wrong function/value and the wrong sign for y in QIV.
Choice C
Reports cos(11π/6) with a wrong sign, or swaps sine and cosine after taking reference π/3. The x-coordinate is +√3/2; the y-coordinate is −1/2.

Question 11 Hardest

Angle θ satisfies cos θ = −√3/2 and sin θ < 0, with 0 ≤ θ < 2π. What is θ?

Show the answer Choice A

Why it is right

The cosine magnitude √3/2 is the Q1 cosine of π/6, so the candidates with cos = −√3/2 are the QII and QIII angles with reference π/6: 5π/6 (QII) and 7π/6 (QIII). The extra condition sin θ < 0 forces quadrant III, so θ = 7π/6. Check: cos(7π/6) = −√3/2 and sin(7π/6) = −1/2 < 0.

Why each other choice fails

Choice B
Picks the quadrant II angle with the same cosine. At 5π/6, cosine is −√3/2 but sine is +1/2, which violates sin θ < 0.
Choice C
Picks a quadrant IV angle with reference π/6. There cos is +√3/2 (positive), not −√3/2, even though sine is negative.
Choice D
Uses reference π/3 instead of π/6: cos(4π/3) = −1/2, not −√3/2. The magnitude √3/2 belongs to a 30° reference, not 60°.

Question 12 Hardest

What is the value of cos(17π/6)?

Show the answer Choice B

Why it is right

First reduce by a full turn: 17π/6 − 2π = 17π/6 − 12π/6 = 5π/6. Now evaluate cos(5π/6). That angle is in quadrant II with reference π − 5π/6 = π/6, so cos(π/6) = √3/2 and cosine is negative in QII: cos(5π/6) = −√3/2. Therefore cos(17π/6) = −√3/2.

Why each other choice fails

Choice A
Reduces correctly to 5π/6 but drops the quadrant II sign, reporting the positive cosine of the reference angle.
Choice C
Uses reference π/3 after reduction (or confuses 5π/6 with 2π/3): cos(2π/3) = −1/2. The reduced angle 5π/6 has reference π/6, not π/3.
Choice D
Either drops the sign after using the wrong reference, or evaluates cos of a QI angle such as π/3 without reducing 17π/6 at all.

Common mistakes

  1. Inverted conversion factor — multiplying degrees by 180/π180/\pi instead of π/180\pi/180, or the reverse for radians → degrees. The leftover unit (or a huge non-special number) is the tell.
  2. Reference angle from the wrong axis — measuring to the yy-axis instead of the xx-axis, which swaps 30∘30^\circ and 60∘60^\circ values (12\frac{1}{2} ↔ 32\frac{\sqrt{3}}{2}).
  3. Sign drop in QII–IV — keeping the Q1 magnitude and forgetting ASTC. Cosine in QII and QIII is negative; sine in QIII and QIV is negative.
  4. Calculator in the wrong mode — feeding π/3\pi/3 to a calculator still in degrees (or 6060 in radians). Exact special values on paper beat a mode error.
  5. Stopping at the reference angle — reporting π/6\pi/6 or 30∘30^\circ when the stem asked for sin⁡\sin or cos⁡\cos of the original angle.
  6. Confusing sine with cosine — on the unit circle, x=cos⁡θx=\cos\theta and y=sin⁡θy=\sin\theta; swapping them is a free wrong choice for the test writer.
  7. Skipping coterminal reduction — reading cos⁡(17π/6)\cos(17\pi/6) as if 17π/617\pi/6 were already between 00 and 2π2\pi, instead of reducing by 2π2\pi first.
  8. Treating quadrantal angles as specials — inventing a half-radical for cos⁡π\cos\pi or sin⁡(3π/2)\sin(3\pi/2) instead of 0,±10,\pm 1.

FAQ

Do I have to memorize the whole unit circle? Memorize Q1 specials and the quadrantal points; build every other quadrant with reference angle + sign. That is faster and less fragile than memorizing sixteen separate ordered pairs.

Is π\pi always radians? On the Digital SAT, an angle written with π\pi and no degree symbol is in radians. A degree symbol means degrees. Do not mix them inside one conversion step without the π/180\pi/180 factor.

What is a reference angle? The acute angle between the terminal side and the nearest xx-axis. It is never measured to the yy-axis, and it is never negative.

Can Desmos replace the unit circle? It can check a decimal approximation after you set the mode correctly. It will not hand you 2/2\sqrt{2}/2 in exact form, and it will happily compute the wrong mode. Use it as a check, not as step 1.

How do I enter a negative SPR answer? Type the minus sign. Fractions such as -1/2 are accepted; do not leave the sign off and hope.