Trigonometry

Trigonometry on the SAT: Every Topic You Need to Know

Last updated: March 2026 · Intermediate
Before you start

You should be comfortable with:

The Digital SAT includes 1 to 4 trigonometry questions out of roughly 44 math questions total, and they tend to land at medium-to-hard difficulty. That is a small slice of the test — but for students who review the right material, these are often free points. The key reason: trig on the SAT is predictable. The same half-dozen concepts appear over and over, and once you recognize the patterns, the problems take 30 to 60 seconds each.

One important detail: no trig formulas are provided on the SAT reference sheet. You get area and volume formulas, but not SOH CAH TOA, not the Pythagorean identity, and not the cofunction relationship. You need to have them memorized before test day.

What Trig Topics Are Tested on the SAT

The SAT draws from a short list of trig concepts. Here is what you need to know, roughly in order of how often each one appears:

  • SOHCAHTOA — Setting up and solving basic trig ratios (sin\sin, cos\cos, tan\tan) in a right triangle. This is the most common type of trig question.
  • Special right triangles — The 30-60-9030\text{-}60\text{-}90 and 45-45-9045\text{-}45\text{-}90 triangles, often combined with trig ratios.
  • Cofunction identitysin(x)=cos(90°x)\sin(x) = \cos(90° - x). The SAT loves this relationship. It appears in some form on nearly every test.
  • Radian-degree conversion — Quick conversions, especially the special angles: π6\frac{\pi}{6}, π4\frac{\pi}{4}, π3\frac{\pi}{3}, and π2\frac{\pi}{2}.
  • Unit circle values — Knowing the exact values of sin\sin and cos\cos for special angles (0°, 30°, 45°, 60°, 90°).
  • Pythagorean identitysin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 and its rearrangements.

What Is NOT on the SAT

You do not need the law of sines, the law of cosines, graphing trig functions, sum/difference formulas, double-angle formulas, or proofs of trig identities. If you have limited study time, skip those topics for SAT prep and focus on the six items above.

How Trig Questions Appear

SAT trig problems follow a few recognizable templates. Knowing the format helps you identify what the question is really asking.

Format 1 — Right triangle with a trig ratio: “In right triangle ABC, angle C is 90°. What is sin(A)\sin(A)?” You label the sides, set up the SOHCAHTOA ratio, and solve. You may need the Pythagorean theorem to find a missing side first.

Format 2 — Cofunction setup: “If sin(x°)=0.6\sin(x°) = 0.6, what is cos(90°x°)\cos(90° - x°)?” The answer is just 0.60.6 — the cofunction identity tells you they are equal. The SAT disguises this in various ways, sometimes using complementary angle phrasing instead of the 90°x90° - x form.

Format 3 — Radian conversion with special triangles: “In a right triangle, one angle measures π6\frac{\pi}{6} radians. The side opposite this angle is 7. What is the hypotenuse?” You convert π6\frac{\pi}{6} to 30°30°, recognize the 30-60-9030\text{-}60\text{-}90 triangle, and use the side ratios.

In every case, the underlying math is straightforward. The difficulty comes from recognizing which concept applies and translating the SAT’s wording into a setup you know how to solve.

SAT-Style Practice Problems

Work through each problem before opening the solution. These mirror the difficulty and style of real SAT questions.

Problem 1: In right triangle ABCABC, angle C=90°C = 90°, AC=5AC = 5, and BC=12BC = 12. What is sin(A)\sin(A)?

First, find the hypotenuse using the Pythagorean theorem:

AB=AC2+BC2=25+144=169=13AB = \sqrt{AC^2 + BC^2} = \sqrt{25 + 144} = \sqrt{169} = 13

Angle AA is at vertex AA. The side opposite angle AA is BC=12BC = 12, and the hypotenuse is AB=13AB = 13.

sin(A)=oppositehypotenuse=BCAB=1213\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{12}{13}

Answer: 1213\dfrac{12}{13}

Problem 2: If cos(x°)=0.68\cos(x°) = 0.68, what is sin(90°x°)\sin(90° - x°)?

Apply the cofunction identity: sin(90°x°)=cos(x°)\sin(90° - x°) = \cos(x°).

sin(90°x°)=cos(x°)=0.68\sin(90° - x°) = \cos(x°) = 0.68

Answer: 0.680.68

Problem 3: In a 30-60-9030\text{-}60\text{-}90 triangle, the shortest side is 7. What is the length of the hypotenuse?

In a 30-60-9030\text{-}60\text{-}90 triangle, the sides are in the ratio 1:3:21 : \sqrt{3} : 2. The shortest side is opposite the 30°30° angle, and the hypotenuse is always twice the shortest side.

hypotenuse=2×7=14\text{hypotenuse} = 2 \times 7 = 14

Answer: 1414

Problem 4: Convert 5π6\dfrac{5\pi}{6} radians to degrees.

Multiply by 180°π\frac{180°}{\pi}:

5π6×180°π=5×180°6=900°6=150°\frac{5\pi}{6} \times \frac{180°}{\pi} = \frac{5 \times 180°}{6} = \frac{900°}{6} = 150°

Answer: 150°150°

Problem 5: If sin(θ)=35\sin(\theta) = \dfrac{3}{5} and θ\theta is in Quadrant I, what is cos(θ)\cos(\theta)?

Use the Pythagorean identity: sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1.

cos2θ=1sin2θ=1(35)2=1925=1625\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}

Since θ\theta is in Quadrant I, cosine is positive:

cosθ=45\cos\theta = \frac{4}{5}

Answer: 45\dfrac{4}{5}

Problem 6: In right triangle PQRPQR with the right angle at QQ, PR=10PR = 10 and PQ=6PQ = 6. What is tan(R)\tan(R)?

First, find the missing side QRQR:

QR=PR2PQ2=10036=64=8QR = \sqrt{PR^2 - PQ^2} = \sqrt{100 - 36} = \sqrt{64} = 8

For angle RR: the side opposite RR is PQ=6PQ = 6, and the side adjacent to RR is QR=8QR = 8.

tan(R)=oppositeadjacent=PQQR=68=34\tan(R) = \frac{\text{opposite}}{\text{adjacent}} = \frac{PQ}{QR} = \frac{6}{8} = \frac{3}{4}

Answer: 34\dfrac{3}{4}

Problem 7: If sin(x°)=cos(2x°+15°)\sin(x°) = \cos(2x° + 15°) and both angles are acute, find xx.

Use the cofunction identity: sin(x°)=cos(90°x°)\sin(x°) = \cos(90° - x°).

Since sin(x°)=cos(2x°+15°)\sin(x°) = \cos(2x° + 15°), the arguments of cosine must be equal:

90x=2x+1590 - x = 2x + 15

75=3x75 = 3x

x=25x = 25

Check: sin(25°)=cos(90°25°)=cos(65°)\sin(25°) = \cos(90° - 25°) = \cos(65°), and 2(25)+15=652(25) + 15 = 65. It checks out.

Answer: x=25x = 25

Problem 8: A right triangle has legs aa and bb and hypotenuse cc. If sin(A)=ac=0.8\sin(A) = \dfrac{a}{c} = 0.8, what is cos(A)\cos(A)?

Use the Pythagorean identity:

cos2(A)=1sin2(A)=1(0.8)2=10.64=0.36\cos^2(A) = 1 - \sin^2(A) = 1 - (0.8)^2 = 1 - 0.64 = 0.36

cos(A)=0.36=0.6\cos(A) = \sqrt{0.36} = 0.6

This is a 3-4-53\text{-}4\text{-}5 triangle scaled up: sin(A)=0.8=45\sin(A) = 0.8 = \frac{4}{5} and cos(A)=0.6=35\cos(A) = 0.6 = \frac{3}{5}.

Answer: 0.60.6

SAT Trig Strategy Tips

These habits will help you work faster and avoid common mistakes on test day.

  • Draw the triangle and label all sides before you write any ratio. Sketching takes five seconds and prevents sign and ratio errors.
  • Check whether angles are in degrees or radians. The SAT uses both. If you see a degree symbol (like 35°35°), it is degrees. If there is no degree symbol (like π6\frac{\pi}{6} or just a number), it is radians.
  • Know the cofunction identity cold. The relationship sin(x)=cos(90°x)\sin(x) = \cos(90° - x) appears in some form on almost every test. When you see sin\sin and cos\cos in the same question with angles that add to 90°90°, the cofunction identity is the key.
  • If you see sin2+cos2\sin^2 + \cos^2 in any form, it equals 1. The SAT sometimes disguises this as 1sin2θ1 - \sin^2\theta or 1cos2θ1 - \cos^2\theta. Recognize the pattern and substitute immediately.
  • Memorize the common Pythagorean triples: 3-4-53\text{-}4\text{-}5, 5-12-135\text{-}12\text{-}13, and 8-15-178\text{-}15\text{-}17. These show up constantly in SAT triangles. Spotting a triple lets you skip the Pythagorean theorem entirely.
  • Do not overthink it. SAT trig is designed to be solvable in under a minute. If your approach requires more than three or four steps, you are probably missing a shortcut.

Key Takeaways

  • The SAT tests a small, predictable set of trig topics: SOHCAHTOA, special triangles, the cofunction identity, radian conversion, unit circle values, and the Pythagorean identity
  • No trig formulas are on the reference sheet — you must memorize them before test day
  • The cofunction identity (sin(x)=cos(90°x)\sin(x) = \cos(90° - x)) is the single most common trig concept on the SAT
  • Most SAT trig problems are right-triangle problems in disguise — label the triangle and the answer usually follows in one or two steps
  • These questions are free points for prepared students. A focused review of the six core concepts is all you need.

Build your foundation with these pages:

  • SOH CAH TOA — the trig ratios that underlie most SAT trig questions
  • Special Angles — exact values for 30°, 45°, 60° and their radian equivalents
  • Unit Circle — how sine and cosine relate to coordinates on the circle
  • Trig Identities — the Pythagorean identity and cofunction relationships

Return to Trigonometry for the full topic list.

Last updated: March 28, 2026