Trigonometry

📐 Right Triangle Trigonometry

Part 1 of 7 — SOH-CAH-TOA, Choosing the Reference Angle & Inverse Trig

Right-triangle trig is the foundation of every other trig question on the ACT Math section (45 questions in 50 minutes, 4 answer choices each). If you can label a triangle correctly and pick the right ratio, you can answer most of these questions in under a minute.

The Three Ratios

For an acute angle θ in a right triangle:

Step 1 Is Always: Choose the Reference Angle

"Opposite" and "adjacent" have no meaning until you know which angle you are standing at. The same side is opposite one acute angle and adjacent to the other.

1. Mark the angle the question is about (the given angle, or the angle you are asked to find).
2. Hypotenuse: the side across from the right angle. It is always the longest side, and it never changes.
3. Opposite: the leg that does not touch your angle.
4. Adjacent: the leg that touches your angle (and is not the hypotenuse).

Example of the switch: In right triangle ABC with the right angle at C, legs AC = 8 and BC = 15, and hypotenuse AB = 17:

Notice that $\sin A = \cos B$ and $\cos A = \sin B$. The two acute angles are complementary (they add to 90°), so the sine of one always equals the cosine of the other, and their tangents are reciprocals. The ACT tests this directly: if $\sin A = 0.28$, then $\cos B = 0.28$ with no calculation.

Step 2: Pick the Ratio That Uses the Two Sides in Play

Look at the side you know and the side you want. Exactly one ratio connects them.

Step 3: Multiply or Divide?

Write the ratio as an equation, then solve.

• Unknown on top → multiply. $\sin 35^\circ = \dfrac{x}{20}$ gives $x = 20\sin 35^\circ$.
• Unknown on the bottom → divide. $\cos 61^\circ = \dfrac{8}{h}$ gives $h = \dfrac{8}{\cos 61^\circ}$.

Many ACT questions stop here and ask "Which expression gives the length…?" You do not need a calculator for those, only the correct setup. Sanity check: a leg must come out shorter than the hypotenuse. Dividing a leg by a sine or cosine (both less than 1 for acute angles) makes it longer, which is right only when you are finding the hypotenuse.

Finding an Angle: Inverse Trig

When you know two sides and want the angle, use an inverse function:

$\theta = \sin^{-1}\left(\frac{\text{opp}}{\text{hyp}}\right), \quad \theta = \cos^{-1}\left(\frac{\text{adj}}{\text{hyp}}\right), \quad \theta = \tan^{-1}\left(\frac{\text{opp}}{\text{adj}}\right)$

• $\sin^{-1}$ means "the angle whose sine is…". It is not $\frac{1}{\sin}$ (that is cosecant, covered in Part 4).
• On a calculator, check you are in degree mode when the answers are in degrees.
• Example: $\sin\theta = 0.5$ means $\theta = \sin^{-1}(0.5) = 30^\circ$.

Exact Values from Special Right Triangles

Building the Whole Triangle from One Ratio

If you are told $\tan X = \frac{3}{4}$, the legs are $3k$ and $4k$ for some $k$, so the hypotenuse is $5k$. You can then write any ratio from either angle. Know the common Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples).

The Pythagorean Identity

Divide $a^2 + b^2 = c^2$ by $c^2$ and you get

$\sin^2\theta + \cos^2\theta = 1$

So if $\cos\theta = \frac{5}{13}$ for an acute angle, then $\sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$ and $\sin\theta = \frac{12}{13}$. (Square first, then subtract; $1 - \frac{5}{13}$ is a common wrong move.)

Area with Trig

The two legs of a right triangle are a base and its height, so area $= \frac{1}{2}(\text{leg})(\text{leg})$. If you know the hypotenuse and an angle, find both legs with sine and cosine first.

Worked Examples

Question: In right triangle PQR, the right angle is at R, hypotenuse PQ = 18, and angle P measures 28°. Find QR and PR.

Solution:

1. Stand at angle P. QR does not touch P, so it is opposite. PR touches P, so it is adjacent.
2. Opposite with hypotenuse → sine: $\sin 28^\circ = \frac{QR}{18}$, so $QR = 18\sin 28^\circ \approx 18(0.4695) \approx 8.45$.
3. Adjacent with hypotenuse → cosine: $PR = 18\cos 28^\circ \approx 18(0.8829) \approx 15.89$.
4. Check: both legs are shorter than 18, and $8.45^2 + 15.89^2 \approx 324 = 18^2$. ✓

Question: A wheelchair ramp rises 3 feet over a horizontal distance of 10 feet. What angle does the ramp make with the ground?

Solution:

1. At the ground angle, the 3-foot rise is opposite and the 10-foot run is adjacent. No hypotenuse is involved → tangent.
2. $\tan\theta = \frac{3}{10} = 0.3$, so $\theta = \tan^{-1}(0.3) \approx 16.7^\circ$.
3. If the question asks for an expression, the answer is simply $\tan^{-1}\left(\frac{3}{10}\right)$. ✓

Question: In right triangle XYZ, the right angle is at Y and $\tan X = \frac{5}{12}$. Find $\sin X$ and $\cos Z$.

Solution:

1. Opposite X is YZ = 5k; adjacent to X is XY = 12k. Hypotenuse XZ = 13k (a 5-12-13 triple).
2. $\sin X = \frac{5}{13}$.
3. Now stand at Z: the adjacent leg is YZ = 5k, so $\cos Z = \frac{5}{13}$, the same as $\sin X$, exactly as the complementary-angle rule predicts. ✓

Quick Check: Label, Then Choose the Ratio 🎯

Which Ratio Connects These Sides? 🔍

For each situation, choose the trig ratio you would set up (from the angle you are given).

Compute It 🧮

1. In a 30°-60°-90° triangle, the hypotenuse is 16. How long is the leg opposite the 30° angle?

2. In a right triangle, $\sin\theta = 0.6$ and the hypotenuse is 25. How long is the leg opposite θ?

3. What acute angle, in degrees, has a tangent of 1?

ACT-Style Practice

ACT right-triangle questions usually come in one of three shapes. Try each before opening the answer.

The wire is the hypotenuse; the height on the pole is opposite the 64° ground angle. Height $= 30\sin 64^\circ \approx 27.0$ feet.

Adjacent to A is 8k and the hypotenuse is 17k, so the leg opposite A is 15k (8-15-17). From B, opposite is 8k and adjacent is 15k, so $\tan B = \frac{8}{15}$.

ACT Tip: Draw and label the triangle even when one is printed. Writing "O", "A", "H" next to the sides takes five seconds and prevents the most common mistake on this topic: using the ratio from the wrong angle.

ACT-Style Questions 📋

Key Takeaways

• Choose the reference angle first. Opposite = the leg that does not touch the angle; adjacent = the leg that does; the hypotenuse is across from the right angle.
• SOH-CAH-TOA: pick the ratio that uses the side you know and the side you want.
• Unknown on top → multiply; unknown on the bottom → divide. A leg is always shorter than the hypotenuse.
• Inverse trig finds angles: $\theta = \tan^{-1}\left(\frac{\text{opp}}{\text{adj}}\right)$, and so on. $\sin^{-1}$ is not $\frac{1}{\sin}$.
• Complementary angles: in a right triangle, $\sin A = \cos B$.
• One ratio gives the whole triangle: use triples (3-4-5, 5-12-13, 8-15-17, 7-24-25) or $\sin^2\theta + \cos^2\theta = 1$.
• Special triangles: 45°-45°-90° is $x : x : x\sqrt{2}$; 30°-60°-90° is $x : x\sqrt{3} : 2x$.

🗼 Trig Ratios & Applications

Part 2 of 7 — Elevation & Depression, Two-Triangle Problems, Law of Sines & Law of Cosines

Part 1 gave you the ratios. This part is about setting up word problems, where the triangle is not drawn for you, and about the two laws that handle triangles with no right angle.

Angles of Elevation and Depression

Both angles are measured from a horizontal line, never from a vertical one.

Why the angle of depression equals the angle of elevation

Picture a person on a cliff looking down at a boat. Draw the horizontal line through the person's eye and the horizontal ground line through the boat. Those two horizontal lines are parallel, and the line of sight crosses both of them as a transversal. The angle of depression (at the top) and the angle of elevation (at the boat) are alternate interior angles, so they are equal.

Practical consequence: move the angle of depression down to the bottom of the picture. At the ground end, the vertical height is opposite the angle and the horizontal distance is adjacent, which gives you a standard SOH-CAH-TOA setup.

Setting up any word problem

1. Sketch the situation: a vertical line for the height, a horizontal line for the ground, a slanted line for the line of sight (or ladder, wire, ramp).
2. Mark the right angle where the vertical meets the horizontal.
3. Place the angle at the ground end (move a depression angle down if needed).
4. Label the known and unknown sides as opposite, adjacent, or hypotenuse from that angle.
5. Choose the ratio and decide whether to multiply or divide.

Watch for eye height. If a person whose eyes are 5 feet above the ground sights the top of a tree, the triangle starts at eye level. Add the 5 feet back at the end.

Two-Triangle Problems

Some ACT problems use two right triangles that share a side.

• Stacked objects (a flagpole on a building, a statue on a pedestal): both tops are sighted from the same spot, so both triangles share the same horizontal distance $d$. The upper object's height is the difference $d\tan(\text{bigger angle}) - d\tan(\text{smaller angle})$. You cannot subtract the angles first: $d\tan 8^\circ$ is not the same as $d\tan 48^\circ - d\tan 40^\circ$.
• Moving observer (the angle of elevation changes after walking closer): write one equation for each position using the same unknown height, then solve the system. Sometimes the triangle formed by the two sight lines is isosceles, which is a shortcut.

Triangles Without a Right Angle

SOH-CAH-TOA only works in right triangles. For any other triangle, use one of these two laws. The convention: side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, side $c$ is opposite angle $C$.

Law of Sines

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Law of Cosines

$c^2 = a^2 + b^2 - 2ab\cos C$

The ACT usually states these formulas in the question when you need them. Your job is to know which one applies and to plug in the right sides and angles.

Which Law Applies?

Rule of thumb: the Law of Sines needs a matched pair, an angle together with the side opposite it. If you have no matched pair, you need the Law of Cosines.

Using the Law of Cosines well

• The angle in the formula is always the angle opposite the side on the left. To find angle $C$, put the side opposite $C$ by itself on the left.
• If $C = 90^\circ$, then $\cos C = 0$ and the formula becomes the Pythagorean theorem. The $-2ab\cos C$ term is the correction for a non-right angle.
• Obtuse check: if $c^2 > a^2 + b^2$, then $\cos C$ is negative and angle $C$ is obtuse. If $c^2 < a^2 + b^2$, angle $C$ is acute.
• The largest angle is always opposite the longest side.

Bonus formula: area with two sides and the included angle

$\text{Area} = \frac{1}{2}ab\sin C$

This is the familiar $\frac{1}{2}bh$, because the height to side $a$ is $b\sin C$.

Worked Examples

Question: From the top of a 120-foot cliff, the angle of depression to a boat is 18°. How far is the boat from the base of the cliff?

Solution:

1. Move the 18° angle down to the boat (alternate interior angles).
2. At the boat, the 120-foot cliff is opposite and the distance $d$ is adjacent → tangent.
3. $\tan 18^\circ = \frac{120}{d}$, so $d = \frac{120}{\tan 18^\circ} \approx \frac{120}{0.3249} \approx 369$ feet. ✓

Question: In triangle ABC, angle A = 50°, angle B = 65°, and side a = 10. Find side b.

Solution:

1. You have a matched pair (A with a) → Law of Sines.
2. $\frac{10}{\sin 50^\circ} = \frac{b}{\sin 65^\circ}$, so $b = \frac{10\sin 65^\circ}{\sin 50^\circ} \approx \frac{10(0.9063)}{0.7660} \approx 11.8$.
3. Check: the larger angle (65°) is opposite the longer side (11.8 > 10). ✓

SAS: Two sides are 6 and 10 with an included angle of 120°. Find the third side.

$c^2 = 36 + 100 - 2(6)(10)\cos 120^\circ = 136 - 120\left(-\tfrac{1}{2}\right) = 196, \quad c = 14$

Because the angle is obtuse, the cosine is negative and the third side comes out longer than it would in a right triangle.

SSS: A triangle has sides 3, 5, and 7. Find its largest angle (opposite 7).

$49 = 9 + 25 - 2(3)(5)\cos C \implies 15 = -30\cos C \implies \cos C = -\tfrac{1}{2} \implies C = 120^\circ$ ✓

Quick Check: Set Up the Situation 🎯

Pick the Right Tool 🔍

Choose the method you would use first for each triangle.

ACT-Style Practice

Both triangles share the 50-foot adjacent side. Top of flagpole: $50\tan 48^\circ$. Roof: $50\tan 40^\circ$. Flagpole $= 50\tan 48^\circ - 50\tan 40^\circ \approx 55.5 - 42.0 \approx 13.6$ feet. Subtracting the angles first ($50\tan 8^\circ \approx 7.0$) gives a wrong answer, because the 8° gap is not part of a right triangle with the 50-foot leg.

Compare $13^2 = 169$ with $8^2 + 9^2 = 145$. Since $169 > 145$, $\cos C < 0$, so the largest angle is obtuse.

ACT Tip: When the stem says "The law of sines states…" or "The law of cosines states…", the formula is a hint about which one to use. Your work is matching each side with its opposite angle.

ACT-Style Questions 📋

Key Takeaways

• Elevation and depression are measured from the horizontal. The angle of depression from the top equals the angle of elevation from the bottom (alternate interior angles of parallel horizontals).
• Move the angle to the ground end, where the height is opposite and the horizontal distance is adjacent.
• Add eye height when the triangle starts above the ground.
• Stacked objects: subtract the two heights, $d\tan(\text{big}) - d\tan(\text{small})$, never the angles.
• Law of Sines $\frac{a}{\sin A} = \frac{b}{\sin B}$ needs a matched angle-side pair (AAS, ASA, SSA).
• Law of Cosines $c^2 = a^2 + b^2 - 2ab\cos C$ handles SAS and SSS; $c^2 > a^2 + b^2$ means angle C is obtuse.
• The ACT usually gives you the law; you supply the setup.

⭕ The Unit Circle

Part 3 of 7 — Radians, Unit-Circle Values, Quadrant Signs (ASTC) & Reference Angles

Right triangles only handle angles between 0° and 90°. The unit circle extends sine, cosine, and tangent to every angle, including obtuse angles, angles past 180°, and negative angles.

Radians ↔ Degrees

A radian measures an angle by the arc it cuts off on a circle: on a circle of radius 1, an angle of 1 radian cuts off an arc of length 1. A full circle is $2\pi$ radians, so

$180^\circ = \pi \text{ radians}$

Shortcut for radians with π: replace π with 180° and simplify. $\frac{5\pi}{3} \to \frac{5(180^\circ)}{3} = 300^\circ$.

A radian measure without π is still a real angle: 1 radian ≈ 57.3°, so 2 radians ≈ 114.6°.

Arc length: on a circle of radius $r$, a central angle of $\theta$ radians cuts off an arc of length $s = r\theta$.

Angles in Standard Position

An angle is in standard position when its vertex is at the origin and its initial side lies on the positive x-axis. Positive angles rotate counterclockwise; negative angles rotate clockwise. Angles that share a terminal side, such as 30°, 390°, and −330°, are coterminal (they differ by multiples of 360° or $2\pi$) and have the same trig values.

The Big Idea: (cos θ, sin θ)

The unit circle has radius 1 and center at the origin. Where the terminal side of θ meets the circle, the point is

$(x, y) = (\cos\theta, \sin\theta), \qquad \tan\theta = \frac{y}{x}$

x is cosine, y is sine. That is the whole definition. Since every point on the unit circle has coordinates between −1 and 1, sine and cosine are always between −1 and 1. Tangent has no such limit.

Quadrant Signs: ASTC

The signs of x and y in each quadrant decide the signs of the trig values.

Memory aid: "All Students Take Calculus," starting in Quadrant I and moving counterclockwise. Tangent is positive in Quadrant III because it is a negative divided by a negative.

Reference Angles

The reference angle is the acute angle between the terminal side and the x-axis (never the y-axis).

The Three-Step Method for Any Angle

1. Quadrant: find where the terminal side lands.
2. Reference angle: find the acute angle to the x-axis, and take its trig value.
3. Sign: attach + or − using ASTC.

Example: $\sin 240^\circ$. Quadrant III; reference angle 60°; sine is negative in Quadrant III. So $\sin 240^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$.

This is also why angles that are mirror images share values up to sign: $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$, because the 150° point is the 30° point reflected across the y-axis (same y, opposite x).

Points Not on the Unit Circle

If the terminal side passes through any point $(x, y)$, let $r = \sqrt{x^2 + y^2}$ (the distance from the origin). Then

$\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x}$

The signs take care of themselves because x and y carry their own signs; $r$ is always positive.

Worked Examples

• $240^\circ \cdot \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3}$
• $\frac{5\pi}{6} \cdot \frac{180}{\pi} = 150^\circ$
• $2 \text{ radians} \cdot \frac{180}{\pi} \approx 114.6^\circ$ (no π, so the answer is not a "nice" angle) ✓

• $\sin 315^\circ$: Quadrant IV, reference 45°, sine negative → $-\frac{\sqrt{2}}{2}$.
• $\cos\frac{2\pi}{3}$: that is 120°, Quadrant II, reference 60°, cosine negative → $-\frac{1}{2}$.
• $\tan 210^\circ$: Quadrant III, reference 30°, tangent positive → $\frac{\sqrt{3}}{3}$. ✓

Question: $\sin\theta = -\frac{5}{13}$ and θ is in Quadrant III. Find $\cos\theta$ and $\tan\theta$.

Solution:

1. Reference triangle: opposite 5, hypotenuse 13, so the other leg is 12.
2. In Quadrant III, x is negative: $\cos\theta = -\frac{12}{13}$.
3. Tangent is positive in Quadrant III: $\tan\theta = \frac{-5/13}{-12/13} = \frac{5}{12}$. ✓

Question: The terminal side of θ passes through $(-6, 8)$. Find sin θ, cos θ, and tan θ.

Solution: $r = \sqrt{36 + 64} = 10$. So $\sin\theta = \frac{8}{10} = \frac{4}{5}$, $\cos\theta = -\frac{6}{10} = -\frac{3}{5}$, $\tan\theta = \frac{8}{-6} = -\frac{4}{3}$. The point is in Quadrant II, and only sine is positive, as ASTC predicts. ✓

Quick Check: Radians, Signs & Reference Angles 🎯

Convert & Find 🧮

1. Convert $\frac{5\pi}{12}$ radians to degrees.

2. Convert $\frac{3\pi}{2}$ radians to degrees.

3. What is the reference angle, in degrees, for 160°?

ACT-Style Practice

The reference triangle is 8-15-17. In Quadrant IV, y is negative, so $\sin\theta = -\frac{15}{17}$.

Both are in Quadrant II, where sine is positive. Their reference angles are 80° and 10°, and $\sin 80^\circ > \sin 10^\circ$, so sin 100° is greater. Reference angles turn a strange-looking comparison into a familiar one.

$s = r\theta = 6 \cdot \frac{2\pi}{3} = 4\pi$. (The angle must be in radians for this formula.)

ACT Tip: If an answer choice has the right size but the wrong sign, the question is testing ASTC. Decide the sign from the quadrant before you look at the choices.

ACT-Style Questions 📋

Key Takeaways

• 180° = π radians. Degrees → radians: multiply by $\frac{\pi}{180}$. Radians → degrees: multiply by $\frac{180}{\pi}$ (or replace π with 180°).
• On the unit circle, a point is (cos θ, sin θ) and $\tan\theta = \frac{y}{x}$.
• ASTC: All positive in QI, Sine in QII, Tangent in QIII, Cosine in QIV.
• Reference angle = acute angle to the x-axis: $180^\circ - \theta$, $\theta - 180^\circ$, or $360^\circ - \theta$.
• Any angle: quadrant → reference-angle value → sign.
• Off the unit circle: $r = \sqrt{x^2 + y^2}$, then $\sin\theta = \frac{y}{r}$, $\cos\theta = \frac{x}{r}$.
• Arc length $s = r\theta$ with θ in radians.

🔁 Trig Identities

Part 4 of 7 — Reciprocal Functions, Quotient & Pythagorean Identities, Double-Angle Formulas

An identity is an equation that is true for every angle where both sides are defined. On the ACT, identities show up in two ways: "Which expression is equivalent to…?" and "If $\cos\theta = \ldots$, what is $\sec\theta$ (or $\sin 2\theta$)?" A short list of identities covers almost all of them.

The Reciprocal Functions

Pairing trap: secant goes with cosine and cosecant goes with sine, the opposite of what the letters suggest. One way to remember it: each pair has exactly one "co-" (sine/cosecant, cosine/secant, tangent/cotangent).

Notation trap: $\sin^{-1}x$ is the inverse sine (an angle). $(\sin x)^{-1} = \frac{1}{\sin x} = \csc x$ is the reciprocal. They are different.

Reciprocals keep the sign: if $\cos\theta = -\frac{2}{7}$, then $\sec\theta = -\frac{7}{2}$.

Quotient Identities

$\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$

These come straight from the unit circle: $\tan\theta = \frac{y}{x}$, and $y = \sin\theta$, $x = \cos\theta$.

Pythagorean Identities

$\sin^2\theta + \cos^2\theta = 1$

Divide every term by $\cos^2\theta$ or by $\sin^2\theta$ to get the other two:

$1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta$

Be ready to spot the rearranged forms:

Remember that $\sin^2\theta$ means $(\sin\theta)^2$. The identity works for any angle, so $\sin^2 17^\circ + \cos^2 17^\circ = 1$ with no calculator.

Cofunction Identities

Complementary angles swap sine and cosine:

$\sin(90^\circ - \theta) = \cos\theta, \quad \cos(90^\circ - \theta) = \sin\theta, \quad \tan(90^\circ - \theta) = \cot\theta$

So if $\sin\theta = \cos 25^\circ$ for an acute angle θ, then $\theta = 65^\circ$.

Double-Angle Identities

When the ACT needs these, it often prints them in the question, but you should recognize them on sight:

$\sin 2\theta = 2\sin\theta\cos\theta$

$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$

The big trap: $\sin 2\theta \neq 2\sin\theta$. Doubling the angle does not double the value. (Test it: $\sin 60^\circ \approx 0.87$, but $2\sin 30^\circ = 1$.) Likewise $\sin(A + B) \neq \sin A + \sin B$.

To use $\sin 2\theta$ when you are given only $\sin\theta$, find $\cos\theta$ first with the Pythagorean identity (and the correct sign for the quadrant).

Strategy 1: Simplifying an Expression

1. Rewrite everything in sines and cosines (sec, csc, tan, cot all have sin/cos forms).
2. Simplify the fractions: multiply by reciprocals, cancel common factors.
3. Look for a Pythagorean pattern like $1 - \cos^2\theta$.

Example: $\csc\theta \cdot \tan\theta = \frac{1}{\sin\theta} \cdot \frac{\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} = \sec\theta$.

Backup plan: plug in an angle such as 30° or 60° into the original expression and into each answer choice. The equivalent one gives the same number. (Avoid 45°, where sine and cosine are equal and can hide a wrong choice.)

Strategy 2: Finding One Value from Another

Given one ratio, sketch a right triangle (or use $\sin^2\theta + \cos^2\theta = 1$), then attach the sign from the quadrant.

In terms of a variable: if $\sin x = k$ for an acute angle $x$, draw opposite $k$ and hypotenuse 1. The adjacent leg is $\sqrt{1 - k^2}$, so

$\cos x = \sqrt{1 - k^2}, \qquad \tan x = \frac{k}{\sqrt{1 - k^2}}$

Worked Examples

Question: Simplify $\dfrac{1 - \cos^2\theta}{\sin\theta\cos\theta}$.

Solution:

1. Replace $1 - \cos^2\theta$ with $\sin^2\theta$: $\dfrac{\sin^2\theta}{\sin\theta\cos\theta}$.
2. Cancel one $\sin\theta$: $\dfrac{\sin\theta}{\cos\theta} = \tan\theta$. ✓

Question: Simplify $\sec\theta\cot\theta$.

Solution: $\dfrac{1}{\cos\theta} \cdot \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} = \csc\theta$. ✓

Question: θ is acute and $\cos\theta = \frac{3}{5}$. Find $\sin 2\theta$ and $\cos 2\theta$.

Solution:

1. 3-4-5 triangle: $\sin\theta = \frac{4}{5}$.
2. $\sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$.
3. $\cos 2\theta = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}$. (Negative is fine: θ ≈ 53°, so 2θ ≈ 106° is in Quadrant II.) ✓

Question: $\cos x = k$ for an acute angle $x$. Express $\tan x$ in terms of $k$.

Solution: Adjacent $k$, hypotenuse 1, opposite $\sqrt{1 - k^2}$. So $\tan x = \dfrac{\sqrt{1 - k^2}}{k}$. ✓

Quick Check: Reciprocals & Basic Identities 🎯

Rewrite in Sines and Cosines 🔍

ACT-Style Practice

$\frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} = \sin\theta \cdot \frac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta$.

$\cos\theta = \sqrt{1 - 0.64} = 0.6$, so $\sin 2\theta = 2(0.8)(0.6) = 0.96$. Doubling 0.8 to get 1.6 is impossible, since a sine is never greater than 1.

Try θ = 60°: $\frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$, and $\cos 60^\circ = \frac{1}{2}$. They match. Algebraically, $\sin\theta \cdot \frac{\cos\theta}{\sin\theta} = \cos\theta$.

ACT Tip: When every answer choice is a single trig function, the expression almost always simplifies by "convert to sin and cos, then cancel." When the choices contain squares, look for a Pythagorean identity.

ACT-Style Questions 📋

Key Takeaways

• Reciprocals: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$. Secant goes with cosine.
• Quotients: $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$.
• Pythagorean: $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$.
• Cofunction: $\sin(90^\circ - \theta) = \cos\theta$.
• Double angle: $\sin 2\theta = 2\sin\theta\cos\theta$; $\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$. Never $2\sin\theta$.
• Simplifying: convert to sin and cos, cancel, look for Pythagorean patterns; plug in 30° or 60° to check.
• From one ratio: build the triangle, then use the quadrant for the sign.

📈 Graphing Trig Functions

Part 5 of 7 — Amplitude, Period, Phase Shift, Midline, Maximum & Minimum

ACT graph questions give you an equation and ask for a feature (amplitude, period, maximum), or give you a graph and ask for the equation. Everything comes from four numbers: A, B, C, and D.

The Parent Graphs

Both $y = \sin x$ and $y = \cos x$ repeat every $2\pi$ and stay between −1 and 1. Memorize their five key points over one cycle; each key point is a quarter-period apart.

Sine starts on the midline going up. Cosine starts at the maximum. The cosine graph is the sine graph shifted $\frac{\pi}{2}$ to the left.

The General Form

$y = A\sin(Bx + C) + D \qquad \text{or} \qquad y = A\cos(Bx + C) + D$

Amplitude

The amplitude is $\lvert A\rvert$, always positive. A negative A flips the graph over its midline: $y = -\cos x$ starts at a minimum instead of a maximum, and $y = -\sin x$ starts on the midline going down. The amplitude of $y = -5\cos(2x) + 3$ is 5, not −5.

Period

B squeezes or stretches the graph horizontally. Bigger B → shorter period → more cycles.

• $y = \sin(3x)$: period $\frac{2\pi}{3}$ (three cycles fit in $2\pi$).
• $y = \cos\left(\frac{x}{3}\right)$: period $\frac{2\pi}{1/3} = 6\pi$.
• $y = \sin\left(\frac{\pi}{4}x\right)$: period $\frac{2\pi}{\pi/4} = 8$. When B contains π, the period is often a whole number.

To double the period, halve B. To halve the period, double B. Changing A or D never changes the period. On $0 \le x \le 2\pi$, the graph of $y = \sin(Bx)$ completes exactly B cycles when B is a positive whole number.

Working backward: if a cycle takes $P$ units, then $B = \frac{2\pi}{P}$.

Phase shift

Factor B out of the parentheses to see the shift clearly:

$y = \sin\left(2x - \frac{\pi}{2}\right) = \sin\left(2\left(x - \frac{\pi}{4}\right)\right)$

The graph shifts $\frac{\pi}{4}$ to the right. Using the formula: $C = -\frac{\pi}{2}$, so $-\frac{C}{B} = \frac{\pi/2}{2} = \frac{\pi}{4}$. Two common errors: forgetting to divide by B (answering $\frac{\pi}{2}$), and reading the minus sign as "left." In $\sin(x - h)$ the shift is $h$ units right; in $\sin(x + h)$ it is $h$ units left.

Midline, maximum, minimum

D moves the whole graph up or down. The midline is $y = D$, the maximum is $D + \lvert A\rvert$, and the minimum is $D - \lvert A\rvert$. For $y = 3\sin x + 7$: midline 7, maximum 10, minimum 4.

Reading an Equation from a Graph

Locating a Specific Maximum or Minimum

The sine function reaches its maximum when its input equals $\frac{\pi}{2}$, and its minimum when the input equals $\frac{3\pi}{2}$. Cosine reaches its maximum when its input is 0 (or $2\pi$) and its minimum at $\pi$. So to find where $y = A\sin(Bx + C) + D$ (with A > 0) peaks, solve $Bx + C = \frac{\pi}{2}$. The height there is $A + D$.

A Quick Word on Tangent

$y = \tan x$ has period π, not $2\pi$, and vertical asymptotes where $\cos x = 0$ (at $x = \frac{\pi}{2} + k\pi$). It has no amplitude because it has no maximum or minimum. For $y = \tan(Bx)$, the period is $\frac{\pi}{\lvert B\rvert}$.

Worked Examples

Question: For $y = -2\cos(3x) + 1$, find the amplitude, period, midline, maximum, minimum, and the starting point at x = 0.

Solution:

• Amplitude $\lvert -2\rvert = 2$; period $\frac{2\pi}{3}$; midline $y = 1$.
• Maximum $1 + 2 = 3$; minimum $1 - 2 = -1$.
• At x = 0, $\cos 0 = 1$, so $y = -2(1) + 1 = -1$: the graph starts at its minimum because A is negative. ✓

Question: For $y = 4\sin(2x - \pi) + 3$, where is the first maximum with x > 0?

Solution:

1. Phase shift: $-\frac{C}{B} = \frac{\pi}{2}$ to the right.
2. Sine peaks when its input is $\frac{\pi}{2}$: $2x - \pi = \frac{\pi}{2} \implies 2x = \frac{3\pi}{2} \implies x = \frac{3\pi}{4}$.
3. Height $= A + D = 4 + 3 = 7$. First maximum: $\left(\frac{3\pi}{4}, 7\right)$.
4. Check: $y = 4\sin(2x)$ peaks at $\frac{\pi}{4}$, and $\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$. ✓

Question: A graph of the form $y = A\cos(Bx) + D$ has a maximum at $(0, 6)$, and the next minimum is at $\left(\frac{\pi}{2}, -2\right)$. Find the equation.

Solution:

1. Amplitude $= \frac{6 - (-2)}{2} = 4$; midline $D = \frac{6 + (-2)}{2} = 2$.
2. Max to next min is half a cycle, so the period is $2 \cdot \frac{\pi}{2} = \pi$, and $B = \frac{2\pi}{\pi} = 2$.
3. A maximum at x = 0 means positive cosine: $y = 4\cos(2x) + 2$. ✓

Quick Check: Features from the Equation 🎯

Read the Features 🧮

For $y = 6\sin\left(\frac{\pi}{4}x\right) - 1$:

1. What is the amplitude?

2. What is the period?

3. What is the maximum value?

ACT-Style Practice

The period is $\frac{2\pi}{2} = \pi$. To halve it to $\frac{\pi}{2}$, double B: replace the 2 with a 4. Multiplying the whole function by a number changes the amplitude, and adding a number shifts the graph up; neither touches the period.

The period is $\frac{2\pi}{5}$, so $\frac{2\pi}{B} = \frac{2\pi}{5}$ and $B = 5$.

Tangent's basic period is π, so the period is $\frac{\pi}{3}$, not $\frac{2\pi}{3}$.

ACT Tip: For "which equation matches the graph" questions, eliminate choices in this order: midline (D), amplitude (A), starting behavior (sign of A, sine vs cosine), then period (B). Two or three checks usually leave one choice.

ACT-Style Questions 📋

Key Takeaways

• Sine starts on the midline rising; cosine starts at a maximum. A negative A flips either one.
• Amplitude $= \lvert A\rvert$ (never negative). Period $= \frac{2\pi}{\lvert B\rvert}$. Midline $y = D$.
• Max $= D + \lvert A\rvert$, min $= D - \lvert A\rvert$.
• Phase shift $= -\frac{C}{B}$: factor B out; $(x - h)$ moves right, $(x + h)$ moves left.
• From a graph: amplitude $= \frac{\max - \min}{2}$, midline $= \frac{\max + \min}{2}$, max to next min is half a period, then $B = \frac{2\pi}{\text{period}}$.
• To find a peak, set the inside equal to $\frac{\pi}{2}$ (sine) or 0 (cosine).
• Double the period → halve B. Tangent has period $\frac{\pi}{\lvert B\rvert}$ and no amplitude.

🎡 Modeling with Sinusoids

Part 6 of 7 — Ferris Wheels, Tides, Temperatures & Other Periodic Situations

Anything that repeats on a regular cycle (a rider on a Ferris wheel, the water level at a dock, the average temperature through a year, hours of daylight, a weight bouncing on a spring) can be modeled with a sine or cosine function. ACT modeling questions ask you to interpret a given model or build one from a description. Both use the graph features from Part 5, now with units.

What Each Parameter Means in Context

For $y = A\sin(Bt) + D$ or $y = A\cos(Bt) + D$:

Units: amplitude, midline, maximum, and minimum are in the output units (feet, meters, degrees). The period is in the input units (minutes, hours, months). A choice that reports a period in feet is wrong on its face.

Translating a description into numbers

• Diameter given? Amplitude = half the diameter. A 50-foot wheel has amplitude 25, not 50.
• Lowest point given instead of the center? Center = lowest point + radius. A 60-foot wheel whose bottom is 4 feet off the ground has its center at 4 + 30 = 34 feet.
• High and low values given? $A = \frac{\text{high} - \text{low}}{2}$, $D = \frac{\text{high} + \text{low}}{2}$.
• Time from a high to the next low? That is half a period.
• Then compute $B = \frac{2\pi}{\text{period}}$. A 4-minute revolution gives $B = \frac{2\pi}{4} = \frac{\pi}{2}$; a 12-hour tide cycle gives $B = \frac{\pi}{6}$.

Choosing Sine or Cosine from the Starting Point

Where the object is at $t = 0$ decides the form, with $A > 0$:

Ferris wheel boarding: riders board at the bottom, so a model with t = 0 at boarding is

$h(t) = -r\cos\left(\frac{2\pi}{T}t\right) + c$

where $r$ is the radius, $T$ is the time per revolution, and $c$ is the height of the center.

Interpreting a Given Model

Most ACT questions about a printed model ask one of these:

Evaluating: for $d(t) = 3\cos\left(\frac{\pi}{6}t\right) + 8$ at $t = 4$, the inside is $\frac{4\pi}{6} = \frac{2\pi}{3}$, and $\cos\frac{2\pi}{3} = -\frac{1}{2}$ (Quadrant II), so $d(4) = 3\left(-\frac{1}{2}\right) + 8 = 6.5$. This is where Part 3's unit circle pays off.

Quarter-cycle landmarks: in one period, a sinusoid moves through max, midline, min, midline, max in equal quarter-period steps. On a Ferris wheel with an 8-minute revolution, a rider who boards at the bottom is level with the center after 2 minutes, at the top after 4 minutes, level with the center again after 6, and back at the bottom after 8.

Models with a Horizontal Shift

Sometimes the cycle does not start at t = 0. A model like

$T(m) = 20\sin\left(\frac{\pi}{6}(m - 4)\right) + 55$

for monthly temperature has a 12-month period and a shift of 4 months: the temperature crosses its average of 55 going up at m = 4, peaks a quarter-period (3 months) later at m = 7 with 75, crosses the average going down at m = 10, and bottoms out at 35 at m = 13, which is m = 1 of the next year.

Worked Examples

Question: A Ferris wheel is 60 feet in diameter, its lowest point is 4 feet above the ground, and it makes one revolution every 8 minutes. A rider boards at the lowest point at t = 0. Write h(t), the rider's height in feet after t minutes, and find the height at t = 2 and t = 4.

Solution:

1. Radius $r = 30$; center $c = 4 + 30 = 34$; $B = \frac{2\pi}{8} = \frac{\pi}{4}$.
2. Starting at the minimum → negative cosine: $h(t) = -30\cos\left(\frac{\pi}{4}t\right) + 34$.
3. $h(2) = -30\cos\frac{\pi}{2} + 34 = 0 + 34 = 34$ feet (a quarter turn: level with the center).
4. $h(4) = -30\cos\pi + 34 = 30 + 34 = 64$ feet (the top). ✓

Question: At a dock, high tide of 11 feet occurs at midnight and the next low tide, 3 feet, occurs at 6 a.m. Write the depth d(t), t hours after midnight, and find the depth at 2 a.m.

Solution:

1. $A = \frac{11 - 3}{2} = 4$, $D = \frac{11 + 3}{2} = 7$.
2. High to low is half a cycle: period $= 12$ hours, so $B = \frac{2\pi}{12} = \frac{\pi}{6}$.
3. Maximum at t = 0 → cosine: $d(t) = 4\cos\left(\frac{\pi}{6}t\right) + 7$.
4. $d(2) = 4\cos\frac{\pi}{3} + 7 = 4\left(\frac{1}{2}\right) + 7 = 9$ feet. The next high tide is at noon. ✓

Question: A city's average monthly temperature, in °F, is modeled by $T(m) = 20\sin\left(\frac{\pi}{6}(m - 4)\right) + 55$, where m = 1 is January. In which month is it hottest, and what is that temperature?

Solution: Sine peaks when its input is $\frac{\pi}{2}$: $\frac{\pi}{6}(m - 4) = \frac{\pi}{2} \implies m - 4 = 3 \implies m = 7$ (July). The maximum is $55 + 20 = 75$°F. ✓

Quick Check: Interpret the Model 🎯

Questions 1 and 2 use $h(t) = 18\sin\left(\frac{\pi}{15}t\right) + 22$, the height in feet of a point on a waterwheel t seconds after it is first observed.

Match the Starting Point to the Model 🔍

Assume A > 0 and B > 0. Choose the form that matches each situation at t = 0.

ACT-Style Practice

Amplitude 25 (the radius), midline 30, $B = \frac{2\pi}{4} = \frac{\pi}{2}$, midline rising → sine: $h(t) = 25\sin\left(\frac{\pi}{2}t\right) + 30$.

Minimum $= 10 - 4 = 6$ meters. Cosine is at its minimum when its input is π: $\frac{\pi}{6}t = \pi \implies t = 6$, so 6 a.m.

ACT Tip: Before computing anything, write down three numbers from the story: the middle value, the distance from the middle to the top, and the time for one cycle. Those are D, A, and the period, and they answer most modeling questions.

ACT-Style Questions 📋

Key Takeaways

• Amplitude = half of (max − min) = the radius of a wheel; midline = average = the center height; period = time for one cycle.
• Diameter → halve it. Lowest point given → center = lowest point + radius.
• High to next low is half a period. Then $B = \frac{2\pi}{\text{period}}$.
• Starting point decides the form: max → $A\cos$, min → $-A\cos$, midline rising → $A\sin$, midline falling → $-A\sin$.
• Boarding a Ferris wheel at the bottom: $h(t) = -r\cos\left(\frac{2\pi}{T}t\right) + c$.
• Evaluate by substituting t and using unit-circle values with the correct quadrant sign.
• Units check: heights and depths are outputs; periods are times.

🧭 Integrated ACT Trig Review

Part 7 of 7 — Choosing the Right Tool, Avoiding the Classic Traps & a Mixed Problem Set

Parts 1–6 each taught one tool. On the real test the questions arrive mixed, often with two tools in one problem, and the challenge is recognizing which tool a question needs. This part gives you a decision map, a list of the traps the answer choices are built from, and two mixed practice sets.

Decision Map: What Is the Question Really Asking?