Precalc Study Lab

5.1 Angles & Radian Measure

Degrees to Radians

\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}

Multiply the degree measure by π/180

Radians to Degrees

\theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi}

Multiply the radian measure by 180/π

Arc Length

s = r\theta

Where s is arc length, r is radius, and θ is the angle in radians

Area of a Sector

A = \frac{1}{2}r^2\theta

Only works when θ is in radians

Coterminal Angles

\theta \pm 360°n \quad \text{or} \quad \theta \pm 2\pi n \quad (n = 1, 2, 3, \ldots)

Add or subtract full rotations to find angles that land in the same position

Key Points

To convert: degrees → radians multiply by $\frac{\pi}{180}$ ; radians → degrees multiply by $\frac{180}{\pi}$
Arc length $s = r\theta$ and sector area $A = \frac{1}{2}r^2\theta$ require $\theta$ in radians
Coterminal angles differ by full rotations: $\theta \pm 360°$ or $\theta \pm 2\pi$
$180° = \pi$ radians — this single fact drives every conversion

Watch Out For

Forgetting to convert degrees to radians before using $s = r\theta$ or $A = \frac{1}{2}r^2\theta$ — these formulas only work with radians
Converting the wrong direction: multiplying by $\frac{\pi}{180}$ goes degrees → radians, not the other way around

5.2 Unit Circle — Sine & Cosine

Unit circle definitions

\cos\theta = x, \quad \sin\theta = y

where $(x, y)$ is the point on the unit circle at angle $\theta$

Pythagorean identity

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

follows directly from $x^2 + y^2 = 1$

Key first-quadrant values

\sin 0 = 0,\; \sin\frac{\pi}{6} = \frac{1}{2},\; \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2},\; \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2},\; \sin\frac{\pi}{2} = 1

Reference angle

\text{ref}(\theta) = \text{acute angle between terminal side and } x\text{-axis}

use the reference angle to find trig values in any quadrant

Even/odd identities

\cos(-\theta) = \cos\theta, \quad \sin(-\theta) = -\sin\theta

cosine is even, sine is odd

Key Points

On the unit circle: $\cos\theta = x$ -coordinate, $\sin\theta = y$ -coordinate
Sine values for key angles follow the pattern $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ ; cosine is the reverse order
All Students Take Calculus → QI all positive, QII sine positive, QIII tangent positive, QIV cosine positive
Reference angle is always measured to the $x$ -axis: QII subtract from $\pi$ , QIII subtract $\pi$ , QIV subtract from $2\pi$

Watch Out For

Measuring the reference angle to the $y$ -axis instead of the $x$ -axis — it's always the acute angle to the nearest part of the $x$ -axis
Getting the ASTC signs wrong: cosine is the $x$ -coordinate (negative on the left side), sine is the $y$ -coordinate (negative below)

5.3 The Other Trigonometric Functions

Tangent and cotangent

\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}

Secant and cosecant

\sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}

Reciprocal pairs (product form)

\tan\theta \cdot \cot\theta = 1, \quad \sec\theta \cdot \cos\theta = 1, \quad \csc\theta \cdot \sin\theta = 1

Each pair multiplies to 1 — useful for simplifying expressions

Pythagorean identity (tangent form)

1 + \tan^2\theta = \sec^2\theta

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

Pythagorean identity (cotangent form)

1 + \cot^2\theta = \csc^2\theta

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

Key Points

All six trig functions come from sin and cos: $\tan = \frac{\sin}{\cos}$ , $\sec = \frac{1}{\cos}$ , $\csc = \frac{1}{\sin}$ , $\cot = \frac{\cos}{\sin}$
Reciprocal pairs: sec goes with cos, csc goes with sin — the names are counterintuitive
Three Pythagorean identities: $\sin^2 + \cos^2 = 1$ , $1 + \tan^2 = \sec^2$ , $1 + \cot^2 = \csc^2$
When stuck simplifying, rewrite everything in terms of sine and cosine first

Watch Out For

Mixing up reciprocal pairs: $\sec$ goes with $\cos$ (not $\sin$ ) and $\csc$ goes with $\sin$ (not $\cos$ ) — the names are counterintuitive
Forgetting to rationalize denominators — e.g., writing $\frac{1}{\sqrt{3}}$ instead of $\frac{\sqrt{3}}{3}$

5.4 Right Triangle Trigonometry

SOH-CAH-TOA

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

Cofunction identity (sine–cosine)

\sin\theta = \cos\!\left(\frac{\pi}{2} - \theta\right), \quad \cos\theta = \sin\!\left(\frac{\pi}{2} - \theta\right)

Cofunction identity (tangent–cotangent)

\tan\theta = \cot\!\left(\frac{\pi}{2} - \theta\right), \quad \cot\theta = \tan\!\left(\frac{\pi}{2} - \theta\right)

Cofunction identity (secant–cosecant)

\sec\theta = \csc\!\left(\frac{\pi}{2} - \theta\right), \quad \csc\theta = \sec\!\left(\frac{\pi}{2} - \theta\right)

Pythagorean theorem

a^2 + b^2 = c^2

where $c$ is the hypotenuse

Key Points

SOH-CAH-TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent
"Opposite" and "adjacent" are relative to the angle you're working with — always label the triangle first
Cofunction identities: $\sin\theta = \cos(90° - \theta)$ — the 'co' in cosine literally means 'complement'
Angle of elevation (looking up) equals angle of depression (looking down) by alternate interior angles

Watch Out For

Labeling 'opposite' and 'adjacent' from the wrong angle — these labels change depending on which angle you're working with; always ask 'opposite to *which* angle?'
Using the wrong trig ratio — double-check that the two sides you're connecting match SOH, CAH, or TOA before solving

6.1 Graphs of Sine & Cosine

General Sinusoidal Form

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

A, B, C, D are real constants with B > 0

Amplitude

\text{Amplitude} = |A|

The height from the midline to a peak. If A is negative, the graph is reflected vertically.

Period

\text{Period} = \frac{2\pi}{|B|}

The horizontal length of one full cycle

Phase Shift

\text{Phase Shift} = \frac{C}{B}

Positive means shift right, negative means shift left

Vertical Shift

\text{Vertical Shift} = D

The midline of the graph is y = D

Key Points

General form: $y = A\sin(Bx - C) + D$ — amplitude $|A|$ , period $\frac{2\pi}{|B|}$ , phase shift $\frac{C}{B}$ , vertical shift $D$
Phase shift is $\frac{C}{B}$ , not just $C$ — this is the most common mistake
Negative $A$ flips the graph vertically, but amplitude is always $|A|$ (positive)
Range of a sinusoidal function: $[D - |A|,\, D + |A|]$

Watch Out For

Confusing phase shift with $C$ — the phase shift is $\frac{C}{B}$ , not just $C$ (e.g., in $y = \sin(2x - \pi)$ , the shift is $\frac{\pi}{2}$ , not $\pi$ )
Saying amplitude is negative — amplitude is always $|A|$ , a positive number; a negative $A$ flips the graph but doesn't make amplitude negative

6.2 Graphs of Other Trig Functions

Period of Tangent / Cotangent

\text{Period} = \frac{\pi}{|B|}

Half the period of sine and cosine for the same B value

Period of Secant / Cosecant

\text{Period} = \frac{2\pi}{|B|}

Same period formula as sine and cosine

Tangent Asymptotes (standard)

x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}

Where cos(x) = 0

Cotangent Asymptotes (standard)

x = n\pi, \quad n \in \mathbb{Z}

Where sin(x) = 0

General Tangent Form

y = A\tan(Bx - C) + D

Phase shift = C/B, vertical stretch = |A|, vertical shift = D

Key Points

Tangent/cotangent period: $\frac{\pi}{|B|}$ ; secant/cosecant period: $\frac{2\pi}{|B|}$ — don't mix them up
Asymptotes come from zeros in the denominator: tan/sec blow up where $\cos = 0$ ; cot/csc blow up where $\sin = 0$
Tangent increases through each branch; cotangent decreases — they go in opposite directions
Phase shifts move the asymptotes too — set the transformed argument equal to the base asymptote values

Watch Out For

Using the sine/cosine period formula $\frac{2\pi}{|B|}$ for tangent or cotangent — tan and cot use $\frac{\pi}{|B|}$ instead
Forgetting that asymptotes shift when there's a phase shift — set the full transformed argument equal to the base asymptote values

6.3 Inverse Trigonometric Functions

Arcsine

y = \arcsin(x): \quad \text{Domain } [-1, 1], \quad \text{Range } \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Arccosine

y = \arccos(x): \quad \text{Domain } [-1, 1], \quad \text{Range } [0, \pi]

Arctangent

y = \arctan(x): \quad \text{Domain } (-\infty, \infty), \quad \text{Range } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Cancellation (inner inverse)

\sin(\arcsin(x)) = x \;\text{ for } x \in [-1,1]; \quad \arcsin(\sin(x)) = x \;\text{ for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Same pattern holds for cos/arccos and tan/arctan with their respective domains

Composition via Triangle

\sin(\arccos(x)) = \sqrt{1 - x^2}, \quad \cos(\arcsin(x)) = \sqrt{1 - x^2}

Derived from a right triangle with hypotenuse 1

Key Points

Restricted ranges: $\arcsin$ returns $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\arccos$ returns $[0, \pi]$ , $\arctan$ returns $(-\frac{\pi}{2}, \frac{\pi}{2})$
$\arcsin(\sin(x)) = x$ only when $x$ is already in the restricted range — otherwise find the equivalent angle
For compositions like $\sin(\arccos(x))$ : draw a right triangle, label sides, use Pythagorean theorem
Input to inverse trig is a ratio (number); output is an angle in the restricted range

Watch Out For

Assuming $\arcsin(\sin(x)) = x$ always — it only equals $x$ when $x$ is already in the restricted range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
Returning an angle outside the restricted range — e.g., giving $\frac{5\pi}{6}$ for $\arcsin(\frac{1}{2})$ instead of $\frac{\pi}{6}$

7.1 Trig Identities

Pythagorean Identity

\sin^2(\theta) + \cos^2(\theta) = 1

The most important identity. Rearranges to give you sin² or cos² alone.

Pythagorean Identity (tan/sec)

1 + \tan^2(\theta) = \sec^2(\theta)

Divide the main Pythagorean identity by cos²(θ) to get this one.

Pythagorean Identity (cot/csc)

1 + \cot^2(\theta) = \csc^2(\theta)

Divide the main Pythagorean identity by sin²(θ) to get this one.

Quotient Identities

\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}

Reciprocal Identities

\csc(\theta) = \frac{1}{\sin(\theta)}, \quad \sec(\theta) = \frac{1}{\cos(\theta)}, \quad \cot(\theta) = \frac{1}{\tan(\theta)}

Key Points

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ is the most-used identity — know its rearranged forms ( $\sin^2 = 1 - \cos^2$ and $\cos^2 = 1 - \sin^2$ )
To simplify: rewrite everything in terms of $\sin$ and $\cos$ , then look for cancellations or Pythagorean patterns
To verify an identity: work on one side only — never move terms across the equals sign
Common strategies: factor, multiply by a conjugate ( $1 + \sin\theta$ ), or convert to sin/cos

Watch Out For

Working on both sides of an identity at once — when verifying, only transform one side; treat the equals sign as a wall you can't cross
Moving terms across the equals sign — this assumes the identity is true, which is exactly what you're trying to prove

7.2 Sum & Difference Identities

Sine of a Sum

\sin(A + B) = \sin A \cos B + \cos A \sin B

Sine of a Difference

\sin(A - B) = \sin A \cos B - \cos A \sin B

Cosine of a Sum

\cos(A + B) = \cos A \cos B - \sin A \sin B

Watch the sign — it's minus for the sum, which is the opposite of what you might guess.

Cosine of a Difference

\cos(A - B) = \cos A \cos B + \sin A \sin B

Tangent of a Sum/Difference

\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

The sign in the denominator is opposite to the sign in the numerator.

Key Points

$\sin(A + B) \neq \sin A + \sin B$ — you must use the formula: $\sin A \cos B + \cos A \sin B$
Cosine formulas have the opposite sign: $\cos(A - B)$ has a plus, $\cos(A + B)$ has a minus
Key decompositions to memorize: $75° = 45° + 30°$ , $15° = 45° - 30°$ , $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$
If given $\sin A$ and $\cos B$ with quadrant info, find the missing values via $\sin^2 + \cos^2 = 1$ first

Watch Out For

Thinking $\sin(A + B) = \sin A + \sin B$ — this is wrong; you must use the full formula $\sin A\cos B + \cos A\sin B$
Getting the sign wrong in the cosine formula — $\cos(A + B)$ uses minus and $\cos(A - B)$ uses plus, which is the opposite of what you'd guess

7.3 Double-Angle & Half-Angle Formulas

Sine Double-Angle

\sin(2\theta) = 2\sin\theta\cos\theta

Cosine Double-Angle (3 forms)

\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta

Choose the form that best fits what you already know (sin only, cos only, or both).

Tangent Double-Angle

\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

Half-Angle: Sine

\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Half-Angle: Cosine

\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Power-Reducing: Sine

\sin^2\theta = \frac{1 - \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Power-Reducing: Cosine

\cos^2\theta = \frac{1 + \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Half-Angle: Tangent

\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 + \cos\alpha}

Two equivalent forms — pick whichever avoids a zero denominator.

Key Points

$\sin(2\theta) = 2\sin\theta\cos\theta$ — you need both sin and cos, so find the missing one first
$\cos(2\theta)$ has three forms — pick the one matching what you know: sin only → $1 - 2\sin^2\theta$ , cos only → $2\cos^2\theta - 1$
Half-angle formulas have a $\pm$ that depends on the quadrant of the half-angle, not the original angle
Power-reducing formulas turn squares into first-power expressions: $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$

Watch Out For

Writing $\sin(2\theta) = 2\sin\theta$ instead of $2\sin\theta\cos\theta$ — you need both sine AND cosine
Picking the wrong form of $\cos(2\theta)$ — use $1 - 2\sin^2\theta$ when you only know sine, $2\cos^2\theta - 1$ when you only know cosine

7.5 Solving Trigonometric Equations

General Solution (Sine/Cosine)

x = x_0 + 2n\pi, \quad n \in \mathbb{Z}

Sine and cosine repeat every 2π.

General Solution (Tangent)

x = x_0 + n\pi, \quad n \in \mathbb{Z}

Tangent repeats every π.

Zero Product Property

AB = 0 \implies A = 0 \text{ or } B = 0

Factor and set each factor to zero — never divide by a trig expression.

Quadratic Substitution

a\sin^2(x) + b\sin(x) + c = 0 \implies \text{let } u = \sin(x)

Treat it like au² + bu + c = 0, solve for u, then find x.

Key Points

Strategy: get everything down to one trig function, then solve — use identities, factoring, or substitution
Never divide by a trig expression — factor instead, or you'll lose solutions where that expression equals zero
Don't forget $\pm$ when taking a square root — $\sin^2 x = \frac{1}{4}$ gives $\sin x = \pm\frac{1}{2}$
General solutions: add $+ 2n\pi$ for sin/cos equations, $+ n\pi$ for tangent equations ( $n \in \mathbb{Z}$ )

Watch Out For

Dividing both sides by $\sin(x)$ or $\cos(x)$ instead of factoring — this loses solutions where that function equals zero
Forgetting the $\pm$ when taking a square root: $\sin^2(x) = \frac{1}{4}$ means $\sin(x) = \frac{1}{2}$ OR $\sin(x) = -\frac{1}{2}$ (four solutions, not two)

⚡ Quick Review / Cram Sheet

5.1 Angles & Radian Measure

5.2 Unit Circle — Sine & Cosine

5.3 The Other Trigonometric Functions

5.4 Right Triangle Trigonometry

6.1 Graphs of Sine & Cosine

6.2 Graphs of Other Trig Functions

6.3 Inverse Trigonometric Functions

7.1 Trig Identities

7.2 Sum & Difference Identities

7.3 Double-Angle & Half-Angle Formulas

7.5 Solving Trigonometric Equations