Trigonometry

Trigonometric Functions

Right Triangle Trigonometry for any angle, say $\theta$
Pneumonic for it is: SOHCAHTOA and cosecHOsecHAcotanAO

$(1.)\;\; \sin \theta = \dfrac{opp}{hyp} \\[7ex] (2.)\;\; \cos \theta = \dfrac{adj}{hyp} \\[7ex] (3.)\;\; \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[5ex] = \sin \theta \div \cos \theta \\[3ex] = \dfrac{opp}{hyp} \div \dfrac{adj}{hyp} \\[5ex] = \dfrac{opp}{hyp} * \dfrac{hyp}{adj} \\[5ex] \therefore \tan \theta = \dfrac{opp}{adj} \\[7ex] (4.)\;\; \csc \theta = \dfrac{1}{\sin \theta} \\[5ex] = 1 \div \sin \theta \\[3ex] = 1 \div \dfrac{opp}{hyp} \\[5ex] \therefore \csc \theta = \dfrac{hyp}{opp} \\[7ex] (5.)\;\; \sec \theta = \dfrac{1}{\cos \theta} \\[5ex] = 1 \div \cos \theta \\[3ex] = 1 \div \dfrac{adj}{hyp} \\[5ex] \therefore \sec \theta = \dfrac{hyp}{adj} \\[7ex] (6.)\;\; \cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta} \\[5ex] = \cos \theta \div \sin \theta \\[3ex] = \dfrac{adj}{hyp} \div \dfrac{opp}{hyp} \\[5ex] = \dfrac{adj}{hyp} * \dfrac{hyp}{opp} \\[5ex] \therefore \cot \theta = \dfrac{adj}{opp}$

Summary: The Unit Circle

Trigonometric Identities

$$\begin{array}{c | c} II & I \\ \hline III & IV \end{array}$$ = $$\begin{array}{c | c} S & A \\ \hline T & C \end{array}$$ = $$\begin{array}{c | c} Sine\:\: is\:\: positive & All\:\: are\:\: positive \\ \hline Tangent\:\: is\:\: positive & Cosine\:\: is\:\: positive \end{array}$$

The pneumonic to remember it:
First Quadrant to Fourth Quadrant (clockwise direction): A-C-T-S (ACTS of the Apostles)

First Quadrant to Fourth Quadrant (counter-clockwise direction): A-S-T-C: ADD - SUGAR - TO - COFFEE

Fourth Quadrant to Third Quadrant (counter-clockwise direction): C-A-S-T (Simon Peter, CAST your net into the sea.)

Second Quadrant to Third Quadrant (clockwise direction): S-A-C-T

Cofunction Identities (Identities of Complements)
First Quadrant Identities
First Quadrant: All (sine, cosine, tangent) are positive
This implies that cosecant, secant, and cotangent are also positive.

Given: two angles say: $\alpha$ and $\beta$;
If $\alpha$ and $\beta$ are complementary, then the:
Sine function and the Cosine functions are cofunctions

$\sin \alpha = \cos \beta \\[3ex] \cos \alpha = \sin \beta \\[3ex]$ Tangent function and the Cotangent functions are cofunctions

$\tan \alpha = \cot \beta \\[3ex] \cot \alpha = \tan \beta \\[3ex]$ Secant function and the Cosecant functions are cofunctions

$\sec \alpha = \csc \beta \\[3ex] \csc \alpha = \sec \beta \\[3ex]$ Given: one angle say: $\theta$;
First Quadrant Identities or Cofunction Identities or Identities of Complements

$0 \lt \theta \lt 90 ...Angle\:\: in\:\: Degrees \\[3ex]$ Reference Angle = $\theta$ ... Angle in Degrees

$0 \lt \theta \lt \dfrac{\pi}{2} ...Angle\:\: in\:\: Radians \\[5ex]$ Reference Angle = $\theta$ ... Angle in Radians

First Quadrant: sine, cosine, tangent are positive
This implies that cosecant, secant, and cotangent are also positive

Complement of $\theta$ = $90 - \theta$ where $\theta$ is in degrees:

Complement of $\theta$ = $\dfrac{\pi}{2} - \theta$ where $\theta$ is in radians:

$(1.)\:\: \sin \theta = \cos (90 - \theta) \\[3ex] (2.)\:\: \sin \theta = \cos \left(\dfrac{\pi}{2} - \theta \right) \\[5ex] (3.)\:\: \cos \theta = \sin (90 - \theta) \\[3ex] (4.)\:\: \cos \theta = \sin \left(\dfrac{\pi}{2} - \theta \right) \\[5ex] (5.)\:\: \tan \theta = \cot (90 - \theta) \\[3ex] (6.)\:\: \tan \theta = \cot \left(\dfrac{\pi}{2} - \theta \right) \\[5ex] (7.)\:\: \cot \theta = \tan (90 - \theta) \\[3ex] (8.)\:\: \cot \theta = \tan \left(\dfrac{\pi}{2} - \theta \right) \\[5ex] (9.)\:\: \sec \theta = \csc (90 - \theta) \\[3ex] (10.)\:\: \sec \theta = \csc \left(\dfrac{\pi}{2} - \theta \right) \\[5ex] (11.)\:\: \csc \theta = \sec (90 - \theta) \\[3ex] (12.)\:\: \csc \theta = \sec \left(\dfrac{\pi}{2} - \theta \right) \\[7ex]$ Second Quadrant Identities or Identities of Supplements

$90 \lt \theta \lt 180$ ... Angle in Degrees
Reference Angle = $180 - \theta$ ... Angle in Degrees
$\dfrac{\pi}{2} \lt \theta \lt \pi$ ... Angle in Radians
Reference Angle = $\pi - \theta$ ... Angle in Radians
Second Quadrant: sine is positive
This implies that cosecant is also positive
Symmetric across the y-axis

Given: one angle say: $\theta$;
Supplement of $\theta$ = $180 - \theta$ where $\theta$ is in degrees:
Supplement of $\theta$ = $\pi - \theta$ where $\theta$ is in radians:

$(1.)\;\; \sin \theta = \sin (180 - \theta) \\[3ex] (2.)\;\; \sin (180 - \theta) = \sin \theta \\[3ex] (3.)\;\; \sin \theta = \sin (\pi - \theta) \\[3ex] (4.)\;\; \sin (\pi - \theta) = \sin \theta \\[3ex] (5.)\;\; \cos \theta = -\cos (180 - \theta) \\[3ex] (6.)\;\; \cos (180 - \theta) = -\cos \theta \\[3ex] (7.)\;\; \cos \theta = -\cos (\pi - \theta) \\[3ex] (8.)\;\; \cos (\pi - \theta) = -\cos \theta \\[3ex] (9.)\;\; \tan \theta = -\tan (180 - \theta) \\[3ex] (10.)\;\; \tan (180 - \theta) = -\tan \theta \\[3ex] (11.)\;\; \tan \theta = -\tan (\pi - \theta) \\[3ex] (12.)\;\; \tan (\pi - \theta) = -\tan \theta \\[5ex]$ Third Quadrant Identities
$180 \lt \theta \lt 270$ ... Angle in Degrees
Reference Angle = $\theta - 180$ ... Angle in Degrees
$\pi \lt \theta \lt \dfrac{3\pi}{2}$ ... Angle in Radians
Reference Angle = $\theta - \pi$ ... Angle in Radians
Third Quadrant: tangent is positive
This implies that cotangent is also positive
Symmetric across the origin

Given: one angle say: $\theta$;

$(1.)\;\; \sin \theta = -\sin (\theta - 180) \\[3ex] (2.)\;\; \sin (\theta - 180) = -\sin \theta \\[3ex] (3.)\;\; \sin \theta = -\sin (\theta - \pi) \\[3ex] (4.)\;\; \sin (\theta - \pi) = -\sin \theta \\[3ex] (5.)\;\; \cos \theta = -\cos (\theta - 180) \\[3ex] (6.)\;\; \cos (\theta - 180) = -\cos \theta \\[3ex] (7.)\;\; \cos \theta = -\cos (\theta - \pi) \\[3ex] (8.)\;\; \cos (\theta - \pi) = -\cos \theta \\[3ex] (9.)\;\; \tan \theta = \tan (\theta - 180) \\[3ex] (10.)\;\; \tan (\theta - 180) = \tan \theta \\[3ex] (11.)\;\; \tan \theta = \tan (\theta - \pi) \\[3ex] (12.)\;\; \tan (\theta - \pi) = \tan \theta \\[5ex]$ Fourth Quadrant Identities
$270 \lt \theta \lt 360$ ... Angle in Degrees
Reference Angle = $360 - \theta$ ... Angle in Degrees
$\dfrac{3\pi}{2} \lt \theta \lt 2\pi$ ... Angle in Radians
Reference Angle = $2\pi - \theta$ ... Angle in Radians
Fourth Quadrant: cosine is positive
This implies that secant is also positive
Symmetric across the x-axis

Given: one angle say: $\theta$;

$(1.)\;\; \sin \theta = -\sin (360 - \theta) \\[3ex] (2.)\;\; \sin (360 - \theta) = -\sin \theta \\[3ex] (3.)\;\; \sin \theta = -\sin (2\pi - \theta) \\[3ex] (4.)\;\; \sin (2\pi - \theta) = -\sin \theta \\[3ex] (5.)\;\; \cos \theta = \cos (360 - \theta) \\[3ex] (6.)\;\; \cos (360 - \theta) = \cos \theta \\[3ex] (7.)\;\; \cos \theta = \cos (2\pi - \theta) \\[3ex] (8.)\;\; \cos (2\pi - \theta) = \cos \theta \\[3ex] (9.)\;\; \tan \theta = -\tan (360 - \theta) \\[3ex] (10.)\;\; \tan (360 - \theta) = -\tan \theta \\[3ex] (11.)\;\; \tan \theta = -\tan (2\pi - \theta) \\[3ex] (12.)\;\; \tan (2\pi - \theta) = -\tan \theta \\[3ex]$ Reciprocal Identities

$(1.)\;\; \csc \theta = \dfrac{1}{\sin \theta} \\[3ex] (2.)\;\; \sec \theta = \dfrac{1}{\cos \theta} \\[3ex] (3.)\;\; \cot \theta = \dfrac{1}{\tan \theta} \\[3ex]$ From Reciprocal Identities

$(1.)\;\; \sin \theta = \dfrac{1}{\csc \theta} \\[3ex] (2.)\;\; \cos \theta = \dfrac{1}{\sec \theta} \\[3ex] (3.)\;\; \tan \theta = \dfrac{1}{\cot \theta} \\[5ex]$ Quotient Identities
$(1.)\;\; \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[3ex] (2.)\;\; \cot \theta = \dfrac{\cos \theta}{\sin \theta} \\[3ex]$ As you can see, $\cot \theta$ has two formulas

$\cot \theta = \dfrac{1}{\tan \theta} \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} \\[5ex]$ Sometimes, we shall use the first one. Sometimes, we shall use the second one.
We have to use the formula that will not give us an undefined value (where the denominator is zero)

Even Identities
Recall: a function, say $f(x)$ is even if $f(-x) = f(x)$
In Trigonometry, the cosine function and the secant function are even functions.

$(1.)\;\; \cos (-\theta) = \cos \theta \\[3ex] (2.)\;\; \sec (-\theta) = \sec \theta \\[5ex]$ Odd Identities
Recall: a function, say $f(x)$ is odd if $f(-x) = -f(x)$
In Trigonometry, the sine function, the cosecant function, the tangent function, and the cotangent function are odd functions.

$(1.)\;\; \sin (-\theta) = -\sin \theta \\[3ex] (2.)\;\; \csc (-\theta) = -\csc \theta \\[3ex] (3.)\;\; \tan (-\theta) = -\tan \theta \\[3ex] (4.)\;\; \cot (-\theta) = -\cot \theta \\[5ex]$ Pythagorean Identities

$(1.)\;\; \sin^2 \theta + \cos^2 \theta = 1 \\[3ex] (2.)\;\; \tan^2 \theta + 1 = \sec^2 \theta \\[3ex] (3.)\;\; \cot^2 \theta + 1 = \csc^2 \theta \\[3ex]$ From Pythagorean Identities

$(1.)\;\; \sin \theta = \pm \sqrt{1 - \cos^2 \theta} \\[3ex] (2.)\;\; \cos \theta = \pm \sqrt{1 - \sin^2 \theta} \\[3ex] (3.)\;\; \tan \theta = \pm \sqrt{\sec^2 \theta - 1} \\[3ex] (4.)\;\; \sec \theta = \pm \sqrt{\tan^2 \theta + 1} \\[3ex] (5.)\;\; \cot \theta = \pm \sqrt{\csc^2 \theta - 1} \\[3ex] (6.)\;\; \csc \theta = \pm \sqrt{\cot^2 \theta + 1}$

Trigonometric Formulas

Sum and Difference Formulas

$(1.)\;\; \sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\[3ex] (2.)\;\; \cos (\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\[3ex] (3.)\;\; \tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} \\[5ex]$ Half-Angle Formulas

$(1.)\;\; \sin \dfrac{\theta}{2} = \pm \sqrt{\dfrac{1 - \cos \theta}{2}} \\[5ex] (2.)\;\; \cos {\theta \over 2} = \pm \sqrt{\dfrac{1 + \cos \theta}{2}} \\[5ex] (3.)\;\; \tan {\theta \over 2} = \pm \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}} \\[5ex] (4.)\;\; \tan {\theta \over 2} = \dfrac{\sin \theta}{1 + \cos \theta} \\[5ex] (5.)\;\; \tan {\theta \over 2} = \dfrac{1 - \cos \theta}{\sin \theta} \\[5ex]$ Formulas from Half-Angle Formulas

$(1.)\;\; \sin^2 \dfrac{\theta}{2} = \dfrac{1 - \cos \theta}{2} \\[5ex] (2.)\;\; \cos^2 \dfrac{\theta}{2} = \dfrac{1 + \cos \theta}{2} \\[5ex] (3.)\;\; \tan^2 \dfrac{\theta}{2} = \dfrac{1 - \cos \theta}{1 + \cos \theta} \\[7ex]$ Double-Angle Formulas

$(1.)\;\; \sin (2\theta) = 2 \sin \theta \cos \theta \\[3ex] (2.)\;\; \cos (2\theta) = \cos^2 \theta - \sin^2 \theta \\[3ex] (3.)\;\; \cos (2\theta) = 1 - 2\sin^2 \theta \\[3ex] (4.)\;\; \cos (2\theta) = 2\cos^2 \theta - 1 \\[3ex] (5.)\;\; \tan (2\theta) = \dfrac{2\tan \theta}{1 - \tan^2 \theta} \\[5ex]$ Formulas from Double-Angle Formulas

$(1.)\;\; \sin^2 \theta = \dfrac{1 - \cos(2\theta)}{2} \\[5ex] (2.)\;\; \cos^2 \theta = \dfrac{1 + \cos(2\theta)}{2} \\[5ex] (3.)\;\; \tan^2 \theta = \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \\[7ex]$ Triple-Angle Formulas

$(1.)\;\; \sin (3\theta) = 3 \sin \theta - 4\sin^3 \theta \\[3ex] (2.)\;\; \cos (3\theta) = 4 \cos^3 \theta - 3 \cos \theta \\[3ex] (3.)\;\; \tan (3\theta) = \dfrac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \\[7ex]$ Sum-to-Product Formulas

$(1.)\;\; \sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right) \\[5ex] (2.)\;\; \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right) \\[5ex] (3.)\;\; \cos \alpha + \cos \beta = 2 \cos \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right) \\[5ex] (4.)\;\; \cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right) \\[5ex]$ Ask students to write the compact form / shortened form of the first two Sum-to-Product Formulas.

Sum-to-Product Formulas (Compact Form of the First Two Formulas)

$(1.)\;\; \sin \alpha \pm \sin \beta = 2 \sin \dfrac{\alpha \pm \beta}{2} \cos \dfrac{\alpha \mp \beta}{2} \\[7ex]$ Product-to-Sum Formulas

$(1.) \sin \alpha * \sin \beta = \dfrac{1}{2} [\cos(\alpha - \beta) - \cos(\alpha + \beta)] \\[5ex] (2.) \cos \alpha * \cos \beta = \dfrac{1}{2} [\cos(\alpha - \beta) + \cos(\alpha + \beta)] \\[5ex] (3.) \sin \alpha * \cos \beta = \dfrac{1}{2} [\sin(\alpha + \beta) + \sin(\alpha + \beta)] \\[5ex]$

Factoring Formulas

x is any trigonometric ratio
y is any trigonometric ratio

$\underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[5ex]$

Triangle Laws

Pythagorean Theorem:
Applies only to right triangles.
In a right triangle, the square of the length of the hypotenuse is the sum of the squares of the other two sides.

$hyp^2 = leg^2 + leg^2$

x is any trigonometric ratio
y is any trigonometric ratio

Sine Law:
Applies to all triangles.
It states that the ratio of the side length of any triangle to the sine of the angle measure opposite that side is the same for the three sides of the triangle.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\[5ex]$ OR

The ratio of the angle measure of any triangle to the side length opposite that angle is the same for the three angles of the triangle.

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} \\[5ex]$ Cosine Law:
Applies to all triangles.
It states that the square of a side of a triangle is the difference between the sum of the squares of the other two sides and twice the product of the two sides and the included angle.

$a^2 = b^2 + c^2 - 2bc \cos A \\[3ex] \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \\[5ex] A = \cos^{-1} \left(\dfrac{b^2 + c^2 - a^2}{2bc}\right) \\[5ex] b^2 = a^2 + c^2 - 2ac \cos B \\[3ex] \cos B = \dfrac{a^2 + c^2 - b^2}{2ac} \\[5ex] B = \cos^{-1} \left(\dfrac{a^2 + c^2 - b^2}{2ac}\right) \\[5ex] c^2 = a^2 + b^2 - 2ab \cos C \\[3ex] \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} \\[5ex] C = \cos^{-1} \left(\dfrac{a^2 + b^2 - c^2}{2ab}\right) \\[5ex]$

Circle Formulas

Except stated otherwise, use:

$d = diameter \\[3ex] r = radius \\[3ex] L = arc\:\:length \\[3ex] A = area\;\;of\;\;sector \\[3ex] P = perimeter\;\;of\;\;sector \\[3ex] \theta = central\:\:angle \\[3ex] \pi = \dfrac{22}{7} \\[5ex] RAD = radians \\[3ex] ^\circ = DEG = degrees \\[7ex] \underline{\theta\;\;in\;\;DEG} \\[3ex] L = \dfrac{2\pi r\theta}{360} \\[5ex] \theta = \dfrac{180L}{\pi r} \\[5ex] r = \dfrac{180L}{\pi \theta} \\[5ex] A = \dfrac{\pi r^2\theta}{360} \\[5ex] \theta = \dfrac{360A}{\pi r^2} \\[5ex] r = \dfrac{360A}{\pi\theta} \\[5ex] P = \dfrac{r(\pi\theta + 360)}{180} \\[7ex] \underline{\theta\;\;in\;\;RAD} \\[3ex] L = r\theta \\[5ex] \theta = \dfrac{L}{r} \\[5ex] r = \dfrac{L}{\theta} \\[5ex] A = \dfrac{r^2\theta}{2} \\[5ex] \theta = \dfrac{2A}{r^2} \\[5ex] r = \sqrt{\dfrac{2A}{\theta}} \\[5ex] P = r(\theta + 2) \\[7ex] Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex] L = \dfrac{2A}{r} \\[5ex] r = \dfrac{2A}{L} \\[5ex] A = \dfrac{Lr}{2}$