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Trigonometry

Angles

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

Arc

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

Radian

  • Radian (θ\theta): central angle θ\theta that subtends (limits by both sides) an arc length ss (curved length) of one radius rr
    • one ‘radius’ worth of angle
    • s=θrs = \theta * r
    • 360°=2π360\degree = 2\pi radrad
      • conversion implies: 1=360°2πrad=180°πrad=πrad180°1=\frac{360\degree}{2\pi rad}=\frac{180\degree}{\pi rad}=\frac{\pi rad}{180\degree}
    • Recall: a circle’s circumference is πd\pi*d or 2π2\pi radrad

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 AA and BB are points on the diameter ofa circle, then point CC on any other point of the circle will form a right triangle between AA,BB, and CC s.t. they form a triangle ABC\triangle ABC.

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

A+B+C=180° \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:

α+β+(α+β)=180° 2α+2β=180° α+β=90° \alpha + \beta + (\alpha + \beta) = 180\degree \ 2\alpha + 2\beta = 180\degree \ \alpha + \beta = 90\degree

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

Q.E.D.\blacksquare Q.E.D.

Pythagorean Theorem

  • For any right triangle with sides aa, bb, and hypotenuse cc : a2+b2=c2a^2 + b^2 = c^2

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

cc=c²c*c = c²

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

4(1/2)ab4*(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:

(ab)2(a - b)^2

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

c2=4(1/2)ab+(ab)2=2ab+a22ab+b2c2=a2+b2+2ab2abc²=a²+b² 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 Q.E.D.\blacksquare Q.E.D.

Application: Distance Between Two Points

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

  • 2D: (x1,x1),(y2,y2)(x_1, x_1), (y_2,y_2) c=(Δx)2+(Δy)2c=\sqrt{(\Delta x)^2 + (\Delta y)^2}
  • 3D: (x1,y1,z1),(x2,y2,z2)(x_1,y_1,z_1), (x_2,y_2,z_2) c=(Δx)2+(Δy)2+(Δz)2c=\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}

Trigonometric Ratios

  • Sine (sinsin), Cosine (coscos), Tangent (tantan)
    • S O/H - C A/H - T O/A
  • Cosecant (csccsc), Secant (secsec), Cotangent (cotcot), are almost an anagram (reciprocals, or 1/f(θ)1/f(\theta)):
    • A/O T (coTcoT) - H/A C (seCseC) - H/O S (cSccSc)

Trigonometric ‘Functions’

  • On a unit circle, the hypotenuse, or radius, is of 1 unit. This means for SOHCAHTOA: sin(θ)=OppositeHypotenuse=Opposite1=Oppositecos(θ)=AdjacentHypotenuse=Opposite1=Adjacent sin(\theta) = \frac{Opposite}{Hypotenuse} = \frac{Opposite}{1} = Opposite \\ cos(\theta) = \frac{Adjacent}{Hypotenuse} = \frac{Opposite}{1} = Adjacent

Identities:

tan1(x)=arctan(x)tan^{-1}(x) = arctan(x)

cos1(x)=arc(x)cos^{-1}(x) = arc(x)

sin1(x)=arcsin(x)sin^{-1}(x) = arcsin(x)

sin2(θ)+cos2(θ)=1sin^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 c2=12c^2 = 1^2 or 11.

Values

  • To obtain values of trig functions along the unit circle (or any circle) along the well known thirds, sixths, halves, and quarters of π\pi, utilize the Large (32\frac{\sqrt{3}}{2}), Medium (22\frac{\sqrt{2}}{2}), and Small (12\frac{\sqrt{1}}{2}) pneumonic.

Period

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

Length

  • Test commit

Identities

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

    Proof:

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