Trigonometry

Angles

• Right angle is $90\degree$
• Complementary angles are two positive angles that add up to $90\degree$
• Supplementary angles are two positive angles that add up to $180\degree$

Arc

• 1 degree = 60 minutes of arc : 1$\degree$ = 60'
• 1 minute = 60 seconds of arc : 1’ = 60''

Radian

• Radian ($\theta$): central angle $\theta$ that subtends (limits by both sides) an arc length $s$ (curved length) of one radius $r$
  • one ‘radius’ worth of angle
  • $s = \theta * r$
  • $360\degree = 2\pi$ $rad$
    • conversion implies: $1=\frac{360\degree}{2\pi rad}=\frac{180\degree}{\pi rad}=\frac{\pi rad}{180\degree}$
  • Recall: a circle’s circumference is $\pi*d$ or $2\pi$ $rad$

Triangles

• Obtuse: Largest angle is >90$\degree$
• Acute: Largest angles are acute
• Congruent: Corresponding parts are congruent (CPCTC)
  • implies 3 angles and 3 sides are congruent
• Similar: Corresponding angles are congruent, and sides are proportional
  • implies 3 angles are congruent, but sides are not

Thales’ Theorem

• If points $A$ and $B$ are points on the diameter ofa circle, then point $C$ on any other point of the circle will form a right triangle between $A$,$B$, and $C$ s.t. they form a triangle $\triangle ABC$.

$Proof$: Construct an arbitrary circle such that there are two points $A$ and $B$ on the diameter with radius $r$. Add an arbitrary point $C$ to any other point of the diameter on the circle. Construct a triangle with angles at each point with $\angle$A, $\angle$B, and $\angle$C. Draw a line within this arbitrary triangle such that the line connects the center, to $C$, and subdivides the arbitrary triangle. The two resulting triangles will be one equilateral triangle with length of radius $r$, and an obtuse triangle also with radial sides $r$ and a length from $\overline{AC}$. label the angles in the equilateral triangle $\measuredangle\alpha$. Label the obtuse triangle such that the smaller angles are $\measuredangle\beta$. We can infer that the remaining angle is $\measuredangle\alpha$. Since the sum of the internal angles of every triangle is 180$\degree$, recall our arbitrary triangle will have internal measurements:

$$ \angle A + \angle B + \angle C = 180\degree $$

By equivalence, we can substitute the angles with the measurements of the smaller triangles within, such that:

$$ \alpha + \beta + (\alpha + \beta) = 180\degree \ 2\alpha + 2\beta = 180\degree \ \alpha + \beta = 90\degree $$

Recall that angle $\alpha + \beta$ is the same as $\angle C$, making the arbitrary triangle a right triangle.

$\blacksquare Q.E.D.$

Pythagorean Theorem

• For any right triangle with sides $a$, $b$, and hypotenuse $c$ : $$a^2 + b^2 = c^2$$

$Proof$: c, the hypotenuse, is given if we know the area of the bigger square, or:

$$c*c = c²$$

The area of the triangles are formed by finding the area of 4 of our original triangle, or:

$$4*(1/2)ab$$

What about that inner square? Well let’s consider what relations we can deduce, then we would be able to equate the area of a c² square with the parts that make it up. Well we know it’s just a square of (a - b) edge lengths, by looking at where the triangles touch each other, so it would have an area of:

$$(a - b)^2$$

The area of the entire square of c² is given by the area of the triangles and inner square, or:

$$ c^2 = 4*(1/2)ab + (a - b)^2 \\ = 2ab + a^2 - 2ab + b^2 \\ \rightarrow c^2 = a^2 + b^2 + 2ab - 2ab \\ \rightarrow c² = a² + b² $$ $\blacksquare Q.E.D.$

Application: Distance Between Two Points

For the distance between two points $c$ in 2 and 3-dimensional space:

• 2D: $(x_1, x_1), (y_2,y_2)$ $$c=\sqrt{(\Delta x)^2 + (\Delta y)^2}$$
• 3D: $(x_1,y_1,z_1), (x_2,y_2,z_2)$ $$c=\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$$

Trigonometric Ratios

• Sine ($sin$), Cosine ($cos$), Tangent ($tan$)
  • S O/H - C A/H - T O/A
• Cosecant ($csc$), Secant ($sec$), Cotangent ($cot$), are almost an anagram (reciprocals, or $1/f(\theta)$):
  • A/O T ($coT$) - H/A C ($seC$) - H/O S ($cSc$)

Trigonometric ‘Functions’

• On a unit circle, the hypotenuse, or radius, is of 1 unit. This means for SOHCAHTOA: $$ sin(\theta) = \frac{Opposite}{Hypotenuse} = \frac{Opposite}{1} = Opposite \\ cos(\theta) = \frac{Adjacent}{Hypotenuse} = \frac{Opposite}{1} = Adjacent $$

Identities:

$tan^{-1}(x) = arctan(x)$

$cos^{-1}(x) = arc(x)$

$sin^{-1}(x) = arcsin(x)$

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

• Apply the pythagorean theorem to a triangle in the unit circle. Notice that when we square non-hypotenuse sides, it equals to $c^2 = 1^2$ or $1$.

Values

• To obtain values of trig functions along the unit circle (or any circle) along the well known thirds, sixths, halves, and quarters of $\pi$, utilize the Large ($\frac{\sqrt{3}}{2}$), Medium ($\frac{\sqrt{2}}{2}$), and Small ($\frac{\sqrt{1}}{2}$) pneumonic.

Period

• Length of a functions cycle, typically peak to peak.

Length

• Test commit

Identities

• $cos(A-B) = cos(A)cos(B) + sin(A)sin(B)$

  Proof: