SAT Trigonometry

SAT trigonometry is a small, predictable slice of the Math section. Expect about 2–3 trigonometry questions per test. They sit inside Geometry and Trigonometry, which is roughly 15% of Math — about 5–7 questions across both modules, shared with area, volume, circles, and lines. The full topic map is in the SAT Math study guide.

Those items focus on right-triangle trig (SOH-CAH-TOA), special right triangles, and radian measure. You find a missing side or a ratio using sine, cosine, and tangent, and you apply the Pythagorean theorem when the third side is not drawn. The same skills show up when a right triangle sits on the coordinate plane; the distance formula there is the Pythagorean theorem, covered in coordinate geometry.

The SAT tests the co-function identity sin x = cos(90 − x) and the two special triangles: 30-60-90 (sides 1-√3-2) and 45-45-90 (sides 1-1-√2). The Digital SAT reference sheet at the start of each Math module prints those side ratios, a² + b² = c², and the note that a full circle is 360° or 2π radians. Knowing the ratios cold is still faster than hunting the sheet. You are not asked to recite them with no support, and you will not need the law of sines, the law of cosines, secant, cosecant, or cotangent.

You do need the unit-circle values that the special triangles already give you: sin and cos of 0°, 30°, 45°, 60°, and 90°. That is the unit circle as SAT trigonometry uses it, not a full precalculus chapter.

SOH-CAH-TOA names the three ratios from a specific acute angle.

sin = opposite / hypotenuse cos = adjacent / hypotenuse tan = opposite / adjacent

Opposite and adjacent are named from the angle you were given. Switch the angle and those two sides swap. The hypotenuse never swaps: it is the side across from the right angle, and it is the longest side.

Two-second setup: mark the reference angle, write H on the hypotenuse, O on the side across from that angle, and A on the remaining side. Then pick the ratio that uses the two sides you know, or the side you know and the side you need. Always mark the reference angle on the diagram before labeling the sides. That single step prevents the most common trig error.

If the problem gives an angle and a side and asks for another side, write the ratio that includes both of those sides before you touch the calculator. sin(θ) = O/H is an equation. Multiply or divide to isolate the unknown. Do not invert the fraction because the calculator is open.

Which ratio to choose: if the hypotenuse is the given side or the side you need, the ratio is sine or cosine, never tangent. Tangent uses only the two legs. If both legs are known and the hypotenuse is not, tangent answers the question without a square root.

In right triangle ABC, angle C is 90°, AC = 3, and BC = 4. Find sin A, cos A, and tan A.

The hypotenuse is AB. 3² + 4² = 9 + 16 = 25, so AB = √25 = 5. Take the square root. 25 is c², not c. a² + b² = c² gives c², and the hypotenuse is √(a² + b²).

Angle A is at A, so the opposite side is BC = 4 and the adjacent side is AC = 3.

sin A = opposite/hypotenuse = 4/5 cos A = adjacent/hypotenuse = 3/5 tan A = opposite/adjacent = 4/3

Check the other acute angle. Angle B is opposite AC, so sin B = 3/5. That equals cos A. Two angles that add to 90° have swapped sine and cosine values, which is sin x = cos(90 − x) on this triangle: A and B add to 90°, so sin B = cos A.

A right triangle has an acute angle θ, opposite side 8, and hypotenuse 17. Find cos θ and tan θ.

The adjacent side is the missing one: 8² + a² = 17², so 64 + a² = 289, a² = 225, and a = 15. Length is positive. Forgetting the square root would leave 225, which is not a side.

cos θ = adjacent/hypotenuse = 15/17 tan θ = opposite/adjacent = 8/15 sin θ = 8/17, which you could have written before finding 15.

This is the 8-15-17 triple. When you see 8 and 17 on a right triangle, test 15 before running a long calculator square root. Then still check 64 + 225 = 289.

The same check works for 5-12-13. If sin θ = 5/13 and θ is acute, the opposite side is 5 and the hypotenuse is 13, so the adjacent side is 12 because 25 + 144 = 169. Then cos θ = 12/13 and tan θ = 5/12.

Special right triangles are the SAT trigonometry shortcut. Memorize the ratios. They appear on nearly every SAT, and recalling them is faster than the calculator or a hunt through the reference sheet.

30-60-90 sides are in the ratio 1 : √3 : 2, or in general x : x√3 : 2x. The shortest side (x) is opposite 30°. The side x√3 is opposite 60°. The hypotenuse is 2x, opposite the right angle. The longer leg is opposite the larger acute angle, so √3 is opposite 60°, not opposite 30°.

45-45-90 sides are in the ratio 1 : 1 : √2, or x : x : x√2. The two legs are equal because the two acute angles are equal. The hypotenuse is the leg times √2.

Build the famous sine and cosine values from these ratios instead of memorizing a detached list. In a 30-60-90 triangle with x = 1, the side opposite 30° is 1 and the hypotenuse is 2, so sin 30° = 1/2 and cos 60° = 1/2. The side opposite 60° is √3, so sin 60° = √3/2 and cos 30° = √3/2. In a 45-45-90 triangle with leg 1, sin 45° = cos 45° = 1/√2 = √2/2, and tan 45° = 1.

Associate 30° with the smaller value (1/2) and 60° with the larger value (√3/2). If you swap them, rebuild the triangle with short side 1 rather than guessing.

The side opposite 30° in a 30-60-90 triangle is 5. Find the other two sides, then sin 30°, cos 30°, and tan 30°.

The short side is x, so x = 5. The hypotenuse is 2x = 10. The side opposite 60° is x√3 = 5√3.

Check the definition, not only the memory ratio.

sin 30° = opposite/hypotenuse = 5/10 = 1/2 cos 30° = adjacent/hypotenuse = (5√3)/10 = √3/2 tan 30° = opposite/adjacent = 5/(5√3) = 1/√3 = √3/3

The other acute angle is 60°.

sin 60° = (5√3)/10 = √3/2 cos 60° = 5/10 = 1/2 tan 60° = (5√3)/5 = √3

sin 30° = cos 60° = 1/2 and sin 60° = cos 30° = √3/2. That is the co-function identity on the two angles that add to 90°.

A 45-45-90 triangle has each leg equal to 6. The hypotenuse is 6√2.

sin 45° = 6/(6√2) = 1/√2 = √2/2 cos 45° = √2/2 tan 45° = 6/6 = 1

A common swap puts √3 opposite 30°. The longer leg is opposite the larger acute angle, so √3 is opposite 60°.

A full circle is 360° and also 2π radians, so π radians = 180°. That single fact converts every angle the SAT writes in radians.

Radians to degrees: multiply by 180/π. Degrees to radians: multiply by π/180.

The pairs worth knowing because they match the special triangles:

30° = π/6 45° = π/4 60° = π/3 90° = π/2 180° = π 360° = 2π

Arc length depends on which unit the angle is in. In degrees, arc length = (θ/360) × 2πr. In radians, arc length = rθ, with θ in radians. Sector area follows the same fraction of the circle: (θ/360) of πr² in degrees, or (θ/(2π)) of πr² in radians. Circle formulas and the θ/360 setup are in the circles, area, and volume guide. Use that page for area and volume, and use the conversion here before you substitute.

Do not feed a radian measure into the degree formula θ/360. π/3 is not “π/3 degrees.” If the problem says radians, multiply by 180/π first, or use rθ and divide by 2π.

A central angle measures π/3 radians on a circle of radius 12. Convert the angle to degrees and find the arc length.

Degrees: (π/3) × (180/π) = 180/3 = 60°. The π cancels. π/3 radians is 60°, the special angle above.

Arc length in radians: rθ = 12 × (π/3) = 4π.

The fraction method agrees. The angle is (π/3) / (2π) = 1/6 of the circle. Circumference is 2π(12) = 24π. One sixth of 24π is 4π. Same answer. Using θ/360 on the radian number π/3 would be the mistake; 60/360 = 1/6 is the degree form of the same fraction.

Other direction: convert 150° to radians. 150 × (π/180) = 150π/180 = 5π/6.

Grid-in style: what is the degree measure of an angle of π/5 radians? (π/5) × (180/π) = 180/5 = 36. The answer is 36.

One more arc, in degrees, so the two formulas stay distinct. Radius 9, central angle 120°. The degree fraction is 120/360 = 1/3. Arc length = (1/3) × 2π(9) = 6π. In radians, 120° = 2π/3, and rθ = 9 × (2π/3) = 6π.

The unit circle is the circle of radius 1 centered at the origin. A ray from the origin at angle θ, measured from the positive x-axis, hits that circle at the point (cos θ, sin θ). The x-coordinate is cosine. The y-coordinate is sine. Tangent is sin/cos, which is y/x, and it is undefined when x = 0 (at 90°).

In quadrant I the SAT values are exactly the special-triangle ratios with hypotenuse 1:

0° (0 radians): (1, 0), so cos 0° = 1 and sin 0° = 0 30° (π/6): (√3/2, 1/2), so cos 30° = √3/2 and sin 30° = 1/2 45° (π/4): (√2/2, √2/2), so sin 45° = cos 45° = √2/2 60° (π/3): (1/2, √3/2), so cos 60° = 1/2 and sin 60° = √3/2 90° (π/2): (0, 1), so cos 90° = 0 and sin 90° = 1

You can rebuild any of these from a 30-60-90 or 45-45-90 triangle drawn inside the circle. Drop a perpendicular to the x-axis. The horizontal leg is adjacent (cosine), the vertical leg is opposite (sine), and the hypotenuse is the radius, 1.

A right triangle on the coordinate plane is the same picture scaled up. If the hypotenuse is 2 instead of 1, multiply the unit-circle coordinates by 2. The 30° point (√3/2, 1/2) becomes (√3, 1), which is the 1 : √3 : 2 triangle again. Angle questions in the xy-plane are still SOH-CAH-TOA after you label the legs.

Point P is on the unit circle in quadrant I. The angle from the positive x-axis to OP is 60°. What are the coordinates of P?

On the unit circle, P = (cos 60°, sin 60°) = (1/2, √3/2).

Check it with a triangle. Hypotenuse OP = 1. The side adjacent to 60° is the x-leg: cos 60° = 1/2, so x = 1/2. The side opposite 60° is the y-leg: sin 60° = √3/2, so y = √3/2. Same point.

Scale it the way a coordinate-plane question often does. Point Q is on the ray OP and OQ = 2. Then Q = 2 × P = (1, √3). Drop a perpendicular from Q to the x-axis at (1, 0). The legs are 1 and √3 and the hypotenuse is 2, a 30-60-90 triangle. The leg adjacent to the origin is 1 and the hypotenuse is 2, so cos θ = 1/2 and the angle at the origin is 60°. The opposite leg is √3. Q is (1, √3), not (√3, 1). Swapping the coordinates swaps 30° and 60°.

(√3, 1) would be the 30° point scaled by 2, because cos 30° = √3/2 and sin 30° = 1/2.

The identity SAT trigonometry actually writes into a question is the co-function relationship:

sin x = cos(90 − x) cos x = sin(90 − x)

In degree notation the same identity is sin x = cos(90° − x). The parentheses and the degree sign do not change the meaning. Two acute angles that add to 90° swap sine and cosine. If a right triangle has acute angles 28° and 62°, then sin 28° = cos 62°, because 28 + 62 = 90. You do not need a decimal to answer “sin 28° equals which of the following?” when one choice is cos 62°.

The second relationship worth knowing is the Pythagorean identity, and it is the Pythagorean theorem on the unit circle, not a new formula. Because the point (cos θ, sin θ) is on the circle of radius 1:

cos²θ + sin²θ = 1

If an acute angle has sin θ = 5/13, then cos²θ = 1 − 25/169 = 144/169, so cos θ = 12/13 (positive in quadrant I). That is the 5-12-13 triangle again. The SAT is much more likely to ask for the co-function match, or for the missing ratio in a right triangle, than to print sin²θ + cos²θ = 1 as a line to simplify. Use the identity as a check. The ACT goes further into identities; that comparison is in ACT geometry and trigonometry.

Which of the following equals sin 35°?

A) cos 35° B) cos 55° C) sin 55° D) tan 55°

Use sin x = cos(90 − x). Here x = 35, so 90 − 35 = 55, and sin 35° = cos 55°. The answer is B.

Why the others fail: cos 35° is the adjacent ratio for the same angle, not the complement. sin 55° = cos 35°, which swaps the wrong way. tan 55° is opposite/adjacent and is not equal to a sine of the complement.

A second form, with no choices. In a right triangle the acute angles are x° and (90 − x)°. If cos x° = 7/25, what is sin(90 − x)°?

sin(90 − x) = cos x, so sin(90 − x)° = 7/25. No new triangle is required. If the question had asked for cos(90 − x) or for sin x, then you would build the triangle: opposite of x is the missing side, 25² − 7² = 625 − 49 = 576, and √576 = 24. Then sin x = 24/25 and cos(90 − x) = 24/25.

Stopping at 576 is the square-root mistake from the 3-4-5 example. 576 is a square, not the side.

Labeling opposite and adjacent from the wrong angle. Mark the reference angle before you write O or A.

Stopping at c². a² + b² = c² gives the square of the hypotenuse. The hypotenuse in the 3-4-5 example is 5, not 25. The side next to the 8-and-17 triangle is 15, not 225. The side next to 7 and 25 is 24, not 576.

Mixing up sin 30° = 1/2 and sin 60° = √3/2. Associate 30° with the smaller value (1/2) and 60° with the larger value (√3/2). Putting √3 opposite 30° is the same swap in triangle form.

Using tan when you needed sin, because both involve the opposite side. Tan does not use the hypotenuse. If the hypotenuse is the given or requested side, tan is the wrong ratio.

Treating a radian angle as degrees in an arc problem. Convert with 180/π, or use rθ. π/3 radians is 60°, not π/3 degrees.

Reading the unit circle backwards. The x-coordinate is cosine and the y-coordinate is sine. The point (√3/2, 1/2) is 30°, not 60°.

Draw and label the right triangle for every trig problem, even when you think you can visualize it. Write H, O, and A from the stated angle before any formula.

Drill until they are automatic: a 3-4-5 ratio problem, an 8-15-17 side find, one 30-60-90 with the short side given, one 45-45-90, one co-function item that never asks for a decimal, one radian-to-degree conversion, and one unit-circle point in quadrant I. Then mix them. A real module does not announce the skill.

ScoreBooster's Geometry topic includes trig problems with step-by-step solution breakdowns that show exactly where students typically make labeling errors. Those sessions mix trig with circles and coordinate questions, which is how the test switches diagrams. After a miss, ask which side was opposite the angle you actually marked. Most trig misses are labeling, not arithmetic.

For arc length, area, and volume beside these ratios, use the circles and area guide. When the triangle is plotted on axes, use coordinate geometry for slope, midpoint, and distance, then come back here for sine and cosine.