Question 25: Trigonometric equations. (Late Middle Palaeolithic).

QUESTION 25.

TRIGONOMETRIC EQUATIONS (Late Middle Palaeolithic).

θ = s/60 × 2π

8sin^2(θ) + 2sin(θ) – 1 = 0

2cos^2(θ) + √3cos(θ) = 0

a. Solve for θ and s. (Note: there are several possible θ values, find the specific angle that corresponds to both sine and cosine).

8sin^2(θ) + 4sin(θ) – 2sin(θ) – 1 = 0

4sin(θ)(2sin(θ) + 1) – 1(2sin(θ) + 1) = 0

(2sin(θ) + 1) (4sin(θ) – 1) = 0

2sin(θ) + 1 = 0

2sin(θ) = -1

sin(θ) = -1/2

4sin(θ) – 1 = 0

4sin(θ) = 1

sin(θ) = 1/4

cos(θ)(2cos(θ) + √3) = 0

cos(θ) = 0 or 2cos(θ) + √3 = 0

cos(θ) = 0 or cos(θ) = -√3/2

sin^-1(-1/2) = π – θ = π/6

θ = π – sin^-1(-1/2) = 7π/6

cos^-1(-√3/2) = 2π – θ = 5π/6

θ = 2π – cos^-1(-√3/2) = 7π/6

7π/6 = s/60 × 2π

7/12 = s/60

s = 7/12 × 60 = 35

b. Re-divide s by 60 to get s/60. Then choose k for minutes (m) and add k + s/60 to get m. Then divide m by 60 and choose k for hours (h) then add k + m/60 to get h. Then divide h by 24 and choose k for days (d) then add k + h/24 to get d. Then divide d by 30 and choose k for months (M) then add k + d/30 to get M. Then divide M by (73/6) and choose k for years (y) then add k + M/(73/6) to get y.

Tip: choose a meaningful k for years (y), especially if you have sinned, such as the Late Middle Palaeolithic, 60,000 to 35,000 years ago. For example, if you are a cannibal, this is a place and ‘time’ where you can find acceptance and forgiveness. For a full understanding of this read https://crimerelativity.com.

s/60 = 35/60 = 7/12

k = m – s/60 = 39

m = k + s/60 = 475/12

m/60 = 95/144

k = h – m/60 = 19

h = k + m/60 = 2831/144

h/24 = 2831/3456

k = d – h/24 = 23

d = k + h/24 = 82319/3456

d/30 = 82319/103680

k = M – d/30 = 7

M = k + d/30 = 808079/103680

M/(73/6) = 808079/1261440

k = y – M/(73/6) = 50350

y = k + M/(73/6) = 63514312079/1261440

c. Choose and workout the magnitudes of T^1, T^-1, T^2 and T^-2, also workout t, A and B.

T^1 = √(A/B) = 10^18 as

T^-1 = √(B/A) = 10^-18 Es

T^2 = A/B = 10^36 as²

T^-2 = B/A = 10^-36 Es²

t = A/T^1 = y × 31536000 = 1.587857801975 × 10^12 s

A = tT^1 = X × 10^12 × 10^18 = X × 10^30 as

B = A/T^2 = X × 10^30 / 10^36 = X × 10^-6 Es

d. Although you have done it forward or ascending, undo or reverse-workout the integers and decimals of the different time units backward or descending.

M/(73/6) = y – k = 808079/1261440

M = (y – k) × (73/6) = 808079/103680

k = M – d/30 = 7

d/30 = M – k = 82319/103680

d = (M – k) × 30 = 82319/3456

k = d – h/24 = 23

h/24 = d – k = 2831/3456

h = (d – k) × 24 = 2831/144

k = h – m/60 = 19

m/60 = h – k = 95/144

m = (h – k) × 60 = 475/12

k = m – s/60 = 39

s/60 = m – k = 7/12

s = (m – k) × 60 = 35

e. Check months with days.

y – d/365 = t / 31536000 – d/365 = 50350

d – h/24 = (y – k) × 365 – h/24 = 233

h – m/60 = (d – k) × 24 – m/60 = 19

m – s/60 = (h – k) × 60 – s/60 = 39

s – cs/100 = (m – k) × 60 – cs/100 = 35

50350 years 233 days 19:39:35

50350 years 7 months 23 days 19:39:35

Hand written example:

The Middle Paleolithic (or Middle Palaeolithic) is the second subdivision of the Paleolithic or Old Stone Age as it is understood in Europe, Africa and Asia. The term Middle Stone Age is used as an equivalent or a synonym for the Middle Paleolithic in African archeology. The Middle Paleolithic broadly spanned from 300,000 to 30,000 years ago. The Late Middle Paleolithic as about 60,000 to 35,000 years ago.

https://en.wikipedia.org/wiki/Middle_Paleolithic

78FA82CD-C27A-4F7D-8A29-9F1A4B265AF7
Neanderthals in summer, artwork. Like modern humans, Neanderthals (Homo neanderthalensis) are members of the Homo genus. They inhabited Europe and western Asia between 230,000 and 29,000 years ago. https://www.thoughtco.com/beginners-guide-to-the-middle-paleolithic-171839

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