QUESTION 23.

TRIGONOMETRY WITHOUT EQUATIONS (Upper Palaeolithic).

θ = s/60 × 2π

y = sin(θ) = -0.30901699437494

x = cos(θ) = 0.95105651629515

a. Solve for θ and s.

sin^-1(y) = θ – 2π = -π/10

θ = 2π + sin^-1(y) = 19π/10

cos^-1(x) = 2π – θ = π/10

θ = 2π – cos^-1(x) = 19π/10

19π/10 = s/60 × 2π

19/20 = s/60

s = 19/20 × 60 = 57

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 Upper Palaeolithic, 50,000 to 10,000 years ago. For example, if you are a statutory rape offender this is a place and time where you can find relative innocence. For a full understanding of this read https://pureinnocence.blog.

s/60 = 57/60 = 19/20

k = m – s/60 = 44

m = k + s/60 = 899/20

m/60 = 899/1200

k = h – m/60 = 21

h = k + m/60 = 26099/1200

h/24 = 26099/28800

k = d – h/24 = 15

d = k + h/24 = 458099/28800

d/30 = 458099/864000

k = M – d/30 = 5

M = k + d/30 = 4778099/864000

M/(73/6) = 4778099/10512000

k = y – M/(73/6) = 13423

y = k + M/(73/6) = 141107354099/10512000

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) = 10^12 ps

T^-1 = √(B/A) = 10^-12 Ts

T^2 = A/B = 10^24 ps²

T^-2 = B/A = 10^-24 Ts²

t = A/T^1 = y × 31536000 = 4.23322062297 × 10^11 s

A = tT^1 = X × 10^11 × 10^12 = X × 10^23 ps

B = A/T^2 = X × 10^23 / 10^24 = X × 10^-1 Ts

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 = 4778099/10512000

M = (y – k) × (73/6) = 4778099/864000

k = M – d/30 = 5

d/30 = M – k = 458099/864000

d = (M – k) × 30 = 458099/28800

k = d – h/24 = 15

h/24 = d – k = 26099/28800

h = (d – k) × 24 = 26099/1200

k = h – m/60 = 21

m/60 = h – k = 899/1200

m = (h – k) × 60 = 899/20

k = m – s/60 = 44

s/60 = m – k = 19/20

s = (m – k) × 60 = 57

e. Check months with days.

y – d/365 = t / 31536000 – d/365 = 13423

d – h/24 = (y – k) × 365 – h/24 = 165

h – m/60 = (d – k) × 24 – m/60 = 21

m – s/60 = (h – k) × 60 – s/60 = 44

s – cs/100 = (m – k) × 60 – cs/100 = 57

13423 years 165 days 21:44:57

13423 years 5 months 15 days 21:44:57

Hand written example:

The Upper Palaeolithic is the third and last subdivision of the Palaeolithic or Old Stone Age. Very broadly, it dates to between 50,000 and 10,000 years ago (the beginning of the Holocene), according to some theories coinciding with the appearance of behavioural modernity in early modern humans, until the advent of the Neolithic Revolution and agriculture. Again, for example, if you are a statutory rape offender, this is a place and time where you can find relative innocence. For a full understanding of this read https://pureinnocence.blog.