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How to Use the One Trig Identity on the SAT

Every right triangle hides three ratios — sine, cosine, and tangent. Learn SOH-CAH-TOA once, connect it to the unit circle, and any SAT triangle question becomes plug-and-play.

Math · Right triangles & trigonometry2:10Published July 5, 2026

On YouTube: SAT Trig: One Identity, Every Problem

What this lesson covers

SOH-CAH-TOA: sin = opp/hyp, cos = adj/hyp, tan = opp/adj
Worked on a 3-4-5: sin 3/5, cos 4/5, tan 3/4
The unit circle: the point lands at (cos, sin) = (0.8, 0.6)
Practice: a 5-12-13 triangle
The three traps (opp/adj flip, wrong ratio, which side is the hypotenuse)

Questions worked in the video

0:29A 3-4-5 right triangle — find sin, cos, and tan of angle θ.
1:09A 5-12-13 right triangle — find sin, cos, and tan of the marked angle.

Worked examples

The questions the video works, written out: the setup, each step, the answer and the trap.

0:29Example 1

In a right triangle with legs 3 and 4 and hypotenuse 5, angle $\theta$ is opposite the side of length 3. Find $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .

Label the sides from the angle's point of view. The hypotenuse is always the longest side, across from the right angle: 5. Opposite $\theta$ is 3. The remaining leg, the one touching $\theta$ , is adjacent: 4.
SOH: $\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{3}{5}$ .
CAH: $\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{4}{5}$ .
TOA: $\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{3}{4}$ .
Check with the one identity worth memorizing: $\displaystyle \sin^2\theta + \cos^2\theta = \frac{9}{25} + \frac{16}{25} = 1$ . Also $\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}$ .

Answer: $\displaystyle \sin\theta = \frac{3}{5}$ , $\displaystyle \cos\theta = \frac{4}{5}$ , $\displaystyle \tan\theta = \frac{3}{4}$

The three ratios are the three ways to pair the sides once you know which is opposite, which is adjacent and which is the hypotenuse. The trap is that opposite and adjacent swap when you switch to the other acute angle: for the angle across from the 4, sine is $\displaystyle \frac{4}{5}$ and cosine is $\displaystyle \frac{3}{5}$ . That swap is the identity $\sin\theta = \cos(90^\circ - \theta)$ , which the SAT tests directly.

0:49Example 2

Place the same angle $\theta$ at the center of a unit circle, measured from the positive x-axis. What are the coordinates of the point where the angle's ray meets the circle?

On a unit circle the radius is 1, and that radius is the hypotenuse of a right triangle whose legs are the point's x and y coordinates.
Scale the 3-4-5 triangle so its hypotenuse is 1: divide every side by 5. The legs become $\displaystyle \frac{4}{5} = 0.8$ along x and $\displaystyle \frac{3}{5} = 0.6$ along y.
So the point is (0.8, 0.6), which is $(\cos\theta, \sin\theta)$ : on the unit circle, cosine is the x-coordinate and sine is the y-coordinate.
Check: the point must sit on the circle, and $0.8^2 + 0.6^2 = 0.64 + 0.36 = 1$ .

Answer: (0.8, 0.6), that is $(\cos\theta, \sin\theta)$

Dividing by a hypotenuse of 1 changes nothing, so on the unit circle the ratios become the coordinates themselves. The trap is writing the point as (sin, cos) and getting (0.6, 0.8); cosine goes with x because the adjacent side runs along the x-axis.

1:09Example 3

A right triangle has legs 5 and 12 and hypotenuse 13. For the angle opposite the side of length 5, find the sine, cosine and tangent.

Confirm the hypotenuse: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ , so 13 is the side across from the right angle. Opposite the marked angle is 5, adjacent is 12.
$\displaystyle \sin = \frac{5}{13}$ , $\displaystyle \cos = \frac{12}{13}$ , $\displaystyle \tan = \frac{5}{12}$ .
Check: $\displaystyle \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1$ .

Answer: $\displaystyle \sin = \frac{5}{13}$ , $\displaystyle \cos = \frac{12}{13}$ , $\displaystyle \tan = \frac{5}{12}$

The sides do all the work; no calculator and no angle measure needed. The trap answers are $\displaystyle \frac{12}{13}$ for sine, from taking the adjacent side instead of the opposite, and $\displaystyle \frac{12}{5}$ for tangent, from flipping the ratio. Always find the hypotenuse first, then read opposite and adjacent relative to the one angle the question names.

Chapters

Lesson transcript

The narration of the video, word for word, under its chapter headings.

0:00Every triangle = 3 ratios

Welcome to SAT Climb. Every right triangle hides three ratios: sine, cosine, and tangent. Learn them once, and any triangle question becomes plug and play.

0:16The setup: a 3-4-5

Here's a 3-4-5 right triangle. Pick the angle θ. What are its sine, cosine, and tangent? All three come from the sides.

0:29SOH-CAH-TOA

Sine = opposite / hypotenuse = 3/5. Cosine = adjacent / hypotenuse = 4/5. Tangent = opposite / adjacent = 3/4. Same triangle, three ratios.

0:49On the unit circle

Put that angle at the center of a unit circle. The point lands at (cos, sin) = (0.8, 0.6). On the unit circle, cos is the x and sin is the y.

1:09Your turn

Your turn. A 5-12-13 triangle. Find the sine and cosine of the marked angle. The sides are all you need.

1:35Three traps

Three traps. Opposite vs adjacent flip with the angle you pick. Grabbing the wrong ratio. And forgetting the hypotenuse is the longest side, across the right angle.

1:56Recap

Right-triangle trig: solved. Three ratios, every triangle. Start free at satclimb.com.

Read the written version: the Right triangles & trigonometry strategy guide, then try 25 hard Right triangles & trigonometry questions with full explanations. Both are free, no account needed.

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