Trigonometry

Opposite = height Hypotenuse = diagonal Adjacent = flat line $(\theta)$ = angle degrees

Sine & Arcsin

$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

• What it finds: The Height (Vertical).
• Use this when you know the Diagonal length and the angle.
• Real World Example: A kid is flying a kite. He let out 100 feet of string (Hypotenuse). The string is at a $30^\circ$ angle. How high is the kite?
• to find the Height, you move the Diagonal to the other side: $\text{Height} = 100 \times \sin(30^\circ)$ $100 \times 0.5 = \mathbf{50 \text{ feet high}}$

If you already have the hypotenuse and the opposite (vertical) side, you are usually trying to find one of two things: the Angle or the Adjacent (horizontal) side.

arcsin

$\theta = \arcsin\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right)$

Finding Adjacent

$a^2 + b^2 = c^2$

$\text{Adjacent} = \sqrt{\text{Hyp}^2 - \text{Opp}^2}$

Cosine & Arccos

$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

• What it finds: The Base (Horizontal/Flat line).
• Use this when you know the Diagonal length and want to know how far it travels along the ground.
• Real World Example: A 12-foot ladder is leaning against a wall at a $60^\circ$ angle from the ground. How far is the base of the ladder from the wall? $\text{Distance} = 12 \times \cos(60^\circ)$ $12 \times 0.5 = \mathbf{6 \text{ feet from the wall}}$

Arccos

$\theta = \arccos\left(\frac{\text{Adjacent}}{\text{Hypotenuse}}\right)$

$\text{If } \cos(y) = x, \text{ then } y = \arccos(x)$

Tangent & Arctan

$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

• What it finds: The Height (if you don’t have the diagonal).
• Use this when you are standing on the ground and looking up at something tall.
• Real World Example: You are standing 50 feet away from a tree. You look at the top at a $45^\circ$ angle. How tall is the tree? $\text{Height} = 50 \times \tan(45^\circ)$ $50 \times 1 = \mathbf{50 \text{ feet tall}}$

Arctan or $\tan^{-1}$

$\theta = \arctan\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)$

• It finds the Angle of elevation or depression.
• Use this when you have the width and height, but no diagonal.
• Example: You are building a wheelchair ramp. The ramp must rise 2 feet (Opposite) over a horizontal distance of 24 feet (Adjacent). What is the angle of the ramp? \tan(\theta) = \frac{2}{24} = 0.0833$$$$\theta = \arctan(0.0833)$$$$\theta \approx \mathbf{4.76^\circ}

Unit circle

Because it is a circle with radius $r = 1$, its equation is derived from the Pythagorean theorem:

$x^2 + y^2 = 1$

Because the Hypotenuse ($H$) is 1, the SOH CAH TOA formulas turn into direct values:

$\text{Opposite} = H \times \sin(\theta) = 1 \times \sin(\theta) = \mathbf{\sin(\theta)}$

$\text{Adjacent} = H \times \cos(\theta) = 1 \times \cos(\theta) = \mathbf{\cos(\theta)}$

The radius ending at (0, 1) on the positive vertical axis has a 90 degrees angle with the positive horizontal axis.

Radian

radian is a measurement of the angle (the “opening” between two lines

Arc length

Arc length is the distance along the curved edge (the “crust” of the pizza slice).

Coterminal Angle

Keep subtracting full circles ($360^\circ$) until we get a number we recognize between $0^\circ$ and $360^\circ$.

• First Lap: $900^\circ - 360^\circ = 540^\circ$ (Still bigger than a circle).
• Second Lap: $540^\circ - 360^\circ = \mathbf{180^\circ}$.

Reference Angle

How far is the angle from the x-axis?

The “Height”: $\sin \theta = 0.6$. This tells us that the point on the unit circle is exactly $0.6$ units above the center line.

The “Location”: $\frac{\pi}{2} < \theta < \pi$. This tells us the angle is between $90^\circ$ and $180^\circ$. Looking at the circle, this means the point is in the top-left section (Quadrant II).

Since it’s a unit circle, we know the Hypotenuse is 1. As we discussed earlier, every point $(x, y)$ must follow the rule $x^2 + y^2 = 1^2$.

$\text{(Horizontal Side)}^2 + \text{(Vertical Side)}^2 = \text{Total}^2$

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

Because $x$ is $\cos \theta$ and $y$ is $\sin \theta$, the problem uses this formula:

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

1. $(\cos \theta)^2 + (0.6)^2 = 1$.
2. $(0.6) \times (0.6) = 0.36$.
3. $(\cos \theta)^2 = 1 - 0.36$, which equals $0.64$.
4. To find $\cos \theta$, you take the square root of $0.64$. This gives you two possible answers: $0.8$ or $-0.8$.

Trigonometric Functions of Real Numbers

On the unit circle, the primary functions are defined as:

• Sine: $\sin(\theta) = y$
• Cosine: $\cos(\theta) = x$
• Tangent: $\tan(\theta) = \frac{y}{x}$ (where $x \neq 0$)

Reciprocal Trigonometric Formulas

These functions are the “flipped” versions of the primary ones:

• Cosecant: $\csc(\theta) = \frac{1}{y}$ (where $y \neq 0$)
• Secant: $\sec(\theta) = \frac{1}{x}$ (where $x \neq 0$)
• Cotangent: $\cot(\theta) = \frac{x}{y}$ (where $y \neq 0$)

Domain of trigonometric function

imagine a “laser pointer” rotating from the center of the circle, there are two specific moments where the Tangent or other value simply doesn’t exist.

Remember that Tangent is the ratio of Vertical ($y$) to Horizontal ($x$):

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

As you rotate the radius, you are building that rectangle we talked about earlier.

At $90^\circ$ (The top edge): Your point is at $(0, 1)$. The horizontal width ($x$) is zero.

The Math: $\tan(90^\circ) = \frac{1}{0}$.

you cannot divide by zero. At this exact angle, the Tangent “breaks” and becomes undefined.

https://www.desmos.com/calculator/ajd5qthpcw

Cosecant

$\csc(\theta) = \frac{\text{Hypotenuse}}{\text{Opposite}}$ (The flip of Sine)

• What it finds: The Diagonal length.
• Use this when you know how high you need to go and want to find the diagonal path.
• Real World Example: You want to build a zip-line. The platform is 20 feet high. You want the line at a $30^\circ$ angle. How much cable do you need to buy?$\text{Cable} = 20 \times \csc(30^\circ)$ (Note: $\csc$ is $1/0.5 = 2$) $20 \times 2 = \mathbf{40 \text{ feet of cable}}$

Secant

$\sec(\theta) = \frac{\text{Hypotenuse}}{\text{Adjacent}}$ (The flip of Cosine)

• What it finds: The Diagonal length.
• Use this when you know the flat distance and need to find the diagonal.
• Real World Example: An architect is designing a bridge. The gap to cross is 100 feet wide. The support beam must be at a $30^\circ$ angle. How long is the beam? \text{Beam} = 100 \times \sec(30^\circ)$$$$100 \times 1.15 = \mathbf{115 \text{ feet long}}

Cotangent

$\cot(\theta) = \frac{\text{Adjacent}}{\text{Opposite}}$ (The flip of Tangent)

• What it finds: The Flat distance.
• Use this when you know the Height and want to find the ground distance.
• Real World Example: A lighthouse is 100 feet tall. It spots a boat at a $10^\circ$ angle from the top. How far away is the boat on the water? $\text{Distance} = 100 \times \cot(10^\circ)$ The Answer: $100 \times 5.67 = \mathbf{567 \text{ feet away}}$

Transformation of Trigonometric Functions

The standard sine wave $y = \sin(x)$ can be stretched, squished, and moved. The general formula is:

$y = A \sin(B(x - C)) + D$

Amplitude ($A$)

• What it does: Vertical Stretch.
• Visual: Controls the height of the wave peaks.
• Physics: In sound, this is Volume (Loudness).
• Formula: $|A|$ is the distance from the center line to the peak.

Period ($B$)

• What it does: Horizontal Squish/Stretch.
• Visual: How fast the wave repeats.
• Physics: In sound, this is Frequency (Pitch). High $B$ = High Pitch.
• Formula: The new Period is $\frac{2\pi}{B}$.

Phase Shift ($C$)

• What it does: Horizontal Shift.
• Visual: Moving the start of the wave Left or Right.
• Physics: This is Timing. (e.g., Does the sound start now, or 1 second later?)

Vertical Shift ($D$)

• What it does: Vertical Shift.
• Visual: Moving the entire wave Up or Down.
• Math: It changes the “center line” (midline) of the oscillation.

Example:

$y = 3 \sin(2(x - \pi)) + 5$

• Amplitude: 3 (The wave goes 3 units up and down from the center).
• Period: $\frac{2\pi}{2} = \pi$ (It oscillates twice as fast as normal).
• Phase Shift: Right by $\pi$.
• Vertical Shift: Up 5 (The center line is now at $y=5$).

The Law of Sines and the Law of Cosines

These two laws allow you to solve triangles that are not right-angled triangles (Oblique Triangles).

The Law of Sines

Use this when you know a “pair” (an angle and the side opposite to it) and one other piece of information.

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

Two fire towers are 10 miles apart. Tower A spots a fire at a $40^\circ$ angle. Tower B spots the fire at a $60^\circ$ angle. How far is the fire from Tower A?

1. Find the third angle: $180^\circ - 40^\circ - 60^\circ = 80^\circ$ (Angle C).
2. Set up the ratio: $\frac{x}{\sin(60^\circ)} = \frac{10}{\sin(80^\circ)}$
3. Solve: $x = \frac{10 \times 0.866}{0.985} \approx \mathbf{8.79 \text{ miles}}$.

The Law of Cosines

Use this when you don’t have a matching pair. Usually when you have SSS (Side-Side-Side) or SAS (Side-Angle-Side). This is basically the Pythagorean Theorem adjusted for non-90-degree angles.

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

Real World Example: You walk 3 miles North, turn $120^\circ$, and walk 4 miles. How far are you from your starting point?

$c^2 = 3^2 + 4^2 - 2(3)(4)\cos(120^\circ)$ $c^2 = 9 + 16 - 24(-0.5)$ $c^2 = 25 + 12 = 37$ $c = \sqrt{37} \approx \mathbf{6.08 \text{ miles}}$

Identities

Use identities to rewrite complex expressions in a simpler form

Reciprocal and Quotient Identities

These define how the six functions relate to one another:

Quotient: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ and $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ Reciprocal: \csc(\theta) = \frac{1}{\sin(\theta)}$$$$\sec(\theta) = \frac{1}{\cos(\theta)}and $\cot(\theta) = \frac{1}{\tan(\theta)}$

Pythagorean Identities

Based on the Pythagorean theorem ($a^2 + b^2 = c^2$) applied to the unit circle:

$\sin^2(\theta) + \cos^2(\theta) = 1$ $1 + \tan^2(\theta) = \sec^2(\theta)$ $1 + \cot^2(\theta) = \csc^2(\theta)$

example

$f(x) = (\sin x + \cos x)^2 - 2\sin x \cos x$

First, we use basic algebra to expand the squared part $(a + b)^2 = a^2 + 2ab + b^2$:

$(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x$

Now, put that back into the full equation:

$f(x) = (\sin^2 x + 2\sin x \cos x + \cos^2 x) - 2\sin x \cos x$

1. Notice $+2\sin x \cos x$ and a $-2\sin x \cos x$. They cancel each other out completely.
2. Pythagorean Identity: You are left with $\sin^2 x + \cos^2 x$. As we know from the fundamental identity, this always equals 1.

Result: $f(x) = 1$

Beyond the basics, there are several dozen “advanced” identities used in calculus, physics, and complex engineering:

• Even/Odd Identities (6): Based on the symmetry of the circle (e.g., $\sin(-x) = -\sin x$).
• Cofunction Identities (6): Relating sines to cosines of complementary angles (e.g., $\sin(\frac{\pi}{2} - x) = \cos x$).
• Sum and Difference (6): Formulas for $\sin(A \pm B)$, etc.
• Double-Angle (5+): Formulas for $\sin(2x)$, $\cos(2x)$, and $\tan(2x)$.
• Half-Angle (3): Formulas for $\sin(\frac{x}{2})$, etc.
• Product-to-Sum & Sum-to-Product (8): Used heavily in audio and signal processing.

Inverse Identities

Composition Identities (Undoing the function)

If you take the Sine of an Arcsine, they cancel out, leaving just the number.

$\sin(\arcsin(x)) = x$ $\cos(\arccos(x)) = x$ $\tan(\arctan(x)) = x$

Complementary Identities

Since the two non-right angles in a right triangle always add up to $90^\circ (\frac{\pi}{2})$ radians, their inverse functions are linked.

$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$ $\arctan(x) + arccot(x) = \frac{\pi}{2}$ $\text{sec}^{-1}(x) + \csc^{-1}(x) = \frac{\pi}{2}$

Negative Argument Identities

What happens if you plug a negative number into the inverse?

• $\arcsin(-x) = -\arcsin(x)$ (The negative moves outside)
• $\arctan(-x) = -\arctan(x)$ (The negative moves outside)
• $\arccos(-x) = \pi - \arccos(x)$ (Special Case: You must subtract from 180 degrees)

Addition and Subtraction Formulas

These allow you to calculate the sine or cosine of a sum of two angles. This is useful for finding the exact value of angles like $75^\circ$ (which is just $45^\circ + 30^\circ$).

The Formulas

• Sine: $\sin(A + B) = \sin A \cos B + \cos A \sin B$ $\sin(A - B) = \sin A \cos B - \cos A \sin B$
• Cosine: (Remember: signs flip!) $\cos(A + B) = \cos A \cos B - \sin A \sin B$ $\cos(A - B) = \cos A \cos B + \sin A \sin B$
• Tangent: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Double-Angle and Half-Angle Formulas

These are used to simplify calculus problems or calculate exact values for angles without a calculator.

Double-Angle Formulas

Used to find the value of $2\theta$ when you know $\theta$.

Sine: $\sin(2\theta) = 2\sin\theta\cos\theta$ Cosine: $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ Alt form: $2\cos^2\theta - 1$ Alt form: $1 - 2\sin^2\theta$ Tangent: $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$

Half-Angle Formulas

Used to find the value of $\frac{\theta}{2}$ when you know $\theta$. The $\pm$ depends on which quadrant the new half-angle lands in.

• Sine: $\sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{2}}$
• Cosine: $\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$
• Tangent: $\tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta}$