## 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 relative innocence. For a full understanding of this read https://literaturetime.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 Late Middle Paleolithic, about 60,000 to 35,000 years ago. Again, for example, if you are a cannibal, this is a place and time where you can find relative innocence. For a full understanding of this read https://literaturetime.com.

## Question 24: Trigonometric equations. (Lower Palaeolithic).

QUESTION 24.

TRIGONOMETRIC EQUATIONS (Lower Palaeolithic).

θ = s/60 × 2π

3sin^3(θ) + 2 = 13/8

2cos^2(θ) – √3cos(θ) = 0

a. Solve for θ and s.

sin^3(θ) + 2/3 = 13/24

sin^3(θ) = 13/24 – 2/3 = -1/8

sin(θ) = ³√(-1/8) = -1/2

sin^-1(-1/2) =  θ – 2π = -π/6

θ = 2π + sin^-1(-1/2) = 11π/6

cos(θ)(2cos(θ) – √3) = 0

cos(θ) = 0 or 2cos(θ) – √3 = 0

cos(θ) = 0 or cos(θ) = √3/2

cos^-1(√3/2) = 2π – θ = π/6

θ = 2π – cos^-1(√3/2) = 11π/6

11π/6 = s/60 × 2π

11/12 = s/60

s = 11/12 × 60 = 55

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 Lower Palaeolithic, 3 million to 300,000 years ago. For example, if you are a genocidal megalomaniac, this is a place and time where you can find relative innocence. For a full understanding of this read https://literaturetime.com.

s/60 = 55/60 = 11/12

k = m – s/60 = 47

m = k + s/60 = 575/12

m/60 = 115/144

k = h – m/60 = 2

h = k + m/60 = 403/144

h/24 = 403/3456

k = d – h/24 = 10

d = k + h/24 = 34963/3456

d/30 = 34963/103680

k = M – d/30 = 6

M = k + d/30 = 657043/103680

M/(73/6) = 657043/1261440

k = y – M/(73/6) = 734021

y = k + M/(73/6) = 925924107283/1261440

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) = 10^15 fs

T^-1 = √(B/A) = 10^-15 Ps

T^2 = A/B = 10^30 fs²

T^-2 = B/A = 10^-30 Ps²

t = A/T^1 = y × 31536000 = 2.3148102682075 × 10^13 s

A = tT^1 = X × 10^13 × 10^15 = X × 10^28 fs

B = A/T^2 = X × 10^28 / 10^30 = X × 10^-2 Ps

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 = 657043/1261440

M = (y – k) × (73/6) = 657043/103680

k = M – d/30 = 6

d/30 = M – k = 34963/103680

d = (M – k) × 30 = 34963/3456

k = d – h/24 = 10

h/24 = d – k = 403/3456

h = (d – k) × 24 = 403/144

k = h – m/60 = 2

m/60 = h – k = 115/144

m = (h – k) × 60 = 575/12

k = m – s/60 = 47

s/60 = m – k = 11/12

s = (m – k) × 60 = 55

e. Check months with days.

y – d/365 = t / 31536000 – d/365 = 734021

d – h/24 = (y – k) × 365 – h/24 = 190

h – m/60 = (d – k) × 24 – m/60 = 2

m – s/60 = (h – k) × 60 – s/60 = 47

s – cs/100 = (m – k) × 60 – cs/100 = 55

734021 years 190 days 02:47:55

734021 years 6 months 10 days 02:47:55

Hand written example:

The Lower Palaeolithic is the earliest subdivision of the Palaeolithic or Old Stone Age. It spans the time from around 3 million years ago when the first evidence for stone tool production and use by hominins appears in the current archaeological record, until around 300,000 years ago. Again, for example, if you are a genocidal megalomaniac, this is a place and time where you can find relative innocence. For a full understanding of this read https://literaturetime.com.

## Question 23: Trigonometry without equations (Upper Palaeolithic).

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 rapist, this is a place and time where you can find relative innocence. For a full understanding of this read https://literaturetime.com.

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 rapist, this is a place and time where you can find relative innocence. For a full understanding of this read https://literaturetime.com.

## Question 22: Trigonometric equations.

QUESTION 22.

TRIGONOMETRIC EQUATIONS.

θ = s/60 × 2π

3sin^3(θ)/4 = -3/32

6cos^2(θ)/12 = 3/8

a. Solve for θ and s.

3sin^3(θ) = -3/8

sin^3(θ) = -1/8

sin(θ) = ³√(-1/8) = -1/2

sin^-1(-1/2) =  θ – 2π = -π/6

θ = 2π + sin^-1(-1/2) = 11π/6

6cos^2(θ) = 9/2

cos^2(θ) = 3/4

cos(θ) = √(3/4) = √3/2

cos^-1(√3/2) = 2π – θ = π/6

θ = 2π – cos^-1(√3/2) = 11π/6

11π/6 = s/60 × 2π

11/12 = s/60

s = 11/12 × 60 = 55

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.

s/60 = 55/60 = 11/12

k = m – s/60 = 17

m = k + s/60 = 215/12

m/60 = 43/144

k = h – m/60 = 9

h = k + m/60 = 1339/144

h/24 = 1339/3456

k = d – h/24 = 18

d = k + h/24 = 63547/3456

d/30 = 63547/103680

k = M – d/30 = 3

M = k + d/30 = 374587/103680

M/(73/6) = 374587/1261440

k = y – M/(73/6) = 43179

y = k + M/(73/6) = 54468092347/1261440

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) = 10^6 μs

T^-1 = √(B/A) = 10^-6 Ms

T^2 = A/B = 10^12 μs²

T^-2 = B/A = 10^-12 Ms²

t = A/T^1 = y × 31536000 = 1.361702308675 × 10^12 s

A = tT^1 = X × 10^12 × 10^6 = X × 10^18 μs

B = A/T^2 = X × 10^18 / 10^12 = X × 10^6 Ms

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 = 374587/1261440

M = (y – k) × (73/6) = 374587/103680

k = M – d/30 = 3

d/30 = M – k = 63547/103680

d = (M – k) × 30 = 63547/3456

k = d – h/24 = 18

h/24 = d – k = 1339/3456

h = (d – k) × 24 = 1339/144

k = h – m/60 = 9

m/60 = h – k = 43/144

m = (h – k) × 60 = 215/12

k = m – s/60 = 17

s/60 = m – k = 11/12

s = (m – k) × 60 = 55

e. Check months with days.

y – d/365 = t / 31536000 – d/365 = 43179

d – h/24 = (y – k) × 365 – h/24 = 108

h – m/60 = (d – k) × 24 – m/60 = 9

m – s/60 = (h – k) × 60 – s/60 = 17

s – cs/100 = (m – k) × 60 – cs/100 = 55

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 1361702308675 / 31536000 – d/365 = 43179

d – h/24 = (43,179.2969518962 – 43179) × 365 – h/24 = 108

h – m/60 = (108.387442129629 – 108) × 24 – m/60 = 9

m – s/60 = (9.29861111111111 – 9) × 60 – s/60 = 17

s – cs/100 = (17.9166666666666 – 17) × 60 – cs/100 = 55

43179 years 108 days 09:17:55

43179 years 3 months 18 days 09:17:55

Hand written example:

## Question 21: Trigonometric equations.

QUESTION 21.

When creating your own trigonometry and radians questions know that if we are being lazy and starting from an angle of M/(73/6) and we want particularly pretty maths or simple fractions then θ needs 2 decimal places. Any less and there are 0 seconds, anymore and the seconds are not a nice whole number. However, if you create the question from the ground up, that is starting with a nice whole number for the seconds and building up, then the maths is very beautiful.

However, starting with an angle of M/(73/6) with 2 decimal places means there are only 5 possible values for seconds (s).

For example:

0.00 or X.00 = 60 s, E.G.: 310.00° will result in 60 s.

0.01 or X.Y1 = 36 s, E.G.: 129.61° will result in 36 s.

0.02 or X.Y2 = 12 s, E.G.: 78.92° will result in 12 s.

0.03 or X.Y3 = 48 s, E.G.: 232.43° will result in 48 s.

0.04 or X.Y4 = 24 s, E.G.: 357.74° will result in 24 s.

0.05 or X.Y5 = 60 s, E.G.: 176.85° will result in 60 s.

0.06 or X.Y6 = 36 s, E.G.: 45.16° will result in 36 s.

0.07 or X.Y7 = 12 s, E.G.: 217.97° will result in 12 s.

0.08 or X.Y8 = 48 s, E.G.: 333.28° will result in 48 s.

0.09 or X.Y9 = 24 s, E.G.: 67.89° will result in 24 s.

0.10 or X.Y0 = 60 s, E.G.: 156.40° will result in 60 s.

However, we can start with an angle of M/(73/6) and use 3 decimal places and still have beautiful maths and finish with a whole number for seconds, but the last of the 3 decimal places must always be 5.

For example:

0.00 or X.005 = 60 s, E.G.: 310.005° will result in 18 s.

0.01 or X.Y15 = 36 s, E.G.: 129.615° will result in 54 s.

0.02 or X.Y25 = 12 s, E.G.: 78.925° will result in 30 s.

0.03 or X.Y35 = 48 s, E.G.: 232.435° will result in 6 s.

0.04 or X.Y45 = 24 s, E.G.: 357.745° will result in 42 s.

0.05 or X.Y55 = 60 s, E.G.: 176.855° will result in 18 s.

0.06 or X.Y65 = 36 s, E.G.: 45.165° will result in 54 s.

0.07 or X.Y75 = 12 s, E.G.: 217.975° will result in 30 s.

0.08 or X.Y85 = 48 s, E.G.: 333.285° will result in 6 s.

0.09 or X.Y95 = 24 s, E.G.: 67.895° will result in 42 s.

0.10 or X.Y05 = 60 s, E.G.: 156.405° will result in 18 s.

As you can see having 2 decimals or 3 decimals with the last decimal as 5 goes up alternately by 12 or 6 for seconds. There are 10 values in total for seconds if we are being lazy and starting with an angle of M/(73/6).

For any other whole number for seconds other than the 10 above, we need to start from scratch from the whole number for (s), then divide s/60 then plus k minutes etc. We go up then down or forward and reverse, that is we create the question and then undo it.

The following is the formula for attaining θ in the 4 quadrants of the unit circle:

Q1.

sin^-1(y) = θ

cos^-1(x) = θ

Q2.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = θ

Q3.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Q4.

sin^-1(y) = θ – 2π therefore θ = 2π + sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

QUESTION 21:

(TRIGONOMETRIC EQUATIONS).

θ = s/60 × 2π

2sin^3(θ) = (-3√3)/4

3cos^3(θ) + 6 = 45/8

a. Solve for θ.

sin^3(θ) = (-3√3)/8

sin(θ) = ³√((-3√3)/8) = -√(3)/2

sin^-1(-√(3)/2) = π – θ = -π/3

θ = π – sin^-1(-√(3)/2) = 4π/3

3cos^3(θ) = 45/8 – 6 = -3/8

cos(θ)^3 = -1/8

cos(θ) = ³√(-1/8) = -1/2

cos^-1(-1/2) = 2π – θ = 2π/3

θ = 2π – cos^-1(-1/2) = 4π/3

b. Solve for s.

4π/3 = s/60 × 2π

2/3 = s/60

s = 2/3 × 60 = 40

c. 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.

s/60 = 2/3

k = m – s/60 = 43

m = k + s/60 = 131/3

m/60 = 131/180

k = h – m/60 = 15

h = k + m/60 = 2831/180

h/24 = 2831/4320

k = d – h/24 = 17

d = k + h/24 = 76271/4320

d/30 = 76271/129600

k = M – d/30 = 8

M = k + d/30 = 1113071/129600

M/(73/6) = 1113071/1576800

k = y – M/(73/6) = 11380

y = k + M/(73/6) = 17945097071/1576800

d. Create another angle (θ) from M/(73/6) × 2π and get the sine and cosine. Also although you already know it and the object is defeated, workout θ.

y = sin(M(73/6) × 2π) = -0.96186457013315

x = cos(M(73/6) × 2π) = -0.27352613901153

Note: to workout nice neat radians fractions instead of ugly decimals you need the Natural Scientific Calculator app! Invite link here:

Natural Scientific Calculator (NSC).

sin^-1(y) = π – θ = -324671π/788400

θ = π – sin^-1(y) = 1113071π/1576800

cos^-1(x) = 2π – θ = 463729π/788400

θ = 2π – cos^-1(x) = 1113071π/1576800

e. Although you have already done it workout M/(73/6), M and k.

M/(73/6) = θ/2π = 1113071/1576800

M = M/(73/6) × (73/6) = 1113071/129600

k = M – d/30 = 8

f. Although you have already done it from (s) upwards, reverse-workout the other decimal remainders and integers (k) of the days, hours, minutes and seconds in reverse order.

d/30 = M – k = 76271/129600

d = (M – k) × 30 = 76271/4320

k = d – h/24 = 17

h/24 = d – k = 2831/4320

h = (d – k) × 24 = 2831/180

k = h – m/60 = 15

m/60 = h – k = 131/180

m = (h – k) × 60 = 131/3

k = m – s/60 = 43

s/60 = m – k = 2/3

s = (m – k) × 60 = 40

t = 11380 years 8 months 17 days 15:43:40

g. Choose and workout the magnitudes of T^1, T^-1, T^2 and T^-2?

T^1 = √(A/B) = 10^3 ms

T^-1 = √(B/A) = 10^-3 ks

T^2 = A/B = 10^6 ms²

T^-2 = B/A = 10^-6 ks²

h. Workout the values of t, A and B?

t = A/T^1 = y × 31536000 = 3.58901941420 × 10^11 s

A = tT^1 = X × 10^11 × 10^3 = X × 10^14 ms

B = A/T^2 = X × 10^14 / 10^6 = X × 10^8 ks

i. Check months and days.

y – d/365 = t / 31536000 – d/365 = 11380

d – h/24 = d/365 × 365 – h/24 = 257

h – m/60 = h/24 × 24 – m/60 = 15

m – s/60 = m/60 × 60 – s/60 = 43

s – cs/100 = s/60 × 60 – cs/100 = 40

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 358901941420 / 31536000 – d/365 = 11380

d – h/24 = (11,380.7059049974 – 11380) × 365 – h/24 = 257

h – m/60 = (257.655324074074 – 257) × 24 – m/60 = 15

m – s/60 = (15.7277777777777 – 15) × 60 – s/60 = 43

s – cs/100 = (43.6666666666666 – 43) × 60 – cs/100 = 40

t = 11380 years 257 days 15:43:40

Hand written example:

## Question 20: Trigonometric equations.

QUESTION 20.

When creating your own trigonometry and radians questions know that if we are being lazy and starting from an angle of M/(73/6) and we want particularly pretty maths or simple fractions then θ needs 2 decimal places. Any less and there are 0 seconds, anymore and the seconds are not a nice whole number. However, if you create the question from the ground up, that is starting with a nice whole number for the seconds and building up, then the maths is very beautiful.

However, starting with an angle of M/(73/6) with 2 decimal places means there are only 5 possible values for seconds (s).

For example:

0.00 or X.00 = 60 s, E.G.: 310.00° will result in 60 s.

0.01 or X.Y1 = 36 s, E.G.: 129.61° will result in 36 s.

0.02 or X.Y2 = 12 s, E.G.: 78.92° will result in 12 s.

0.03 or X.Y3 = 48 s, E.G.: 232.43° will result in 48 s.

0.04 or X.Y4 = 24 s, E.G.: 357.74° will result in 24 s.

0.05 or X.Y5 = 60 s, E.G.: 176.85° will result in 60 s.

0.06 or X.Y6 = 36 s, E.G.: 45.16° will result in 36 s.

0.07 or X.Y7 = 12 s, E.G.: 217.97° will result in 12 s.

0.08 or X.Y8 = 48 s, E.G.: 333.28° will result in 48 s.

0.09 or X.Y9 = 24 s, E.G.: 67.89° will result in 24 s.

0.10 or X.Y0 = 60 s, E.G.: 156.40° will result in 60 s.

However, we can start with an angle of M/(73/6) and use 3 decimal places and still have beautiful maths and finish with a whole number for seconds, but the last of the 3 decimal places must always be 5.

For example:

0.00 or X.005 = 60 s, E.G.: 310.005° will result in 18 s.

0.01 or X.Y15 = 36 s, E.G.: 129.615° will result in 54 s.

0.02 or X.Y25 = 12 s, E.G.: 78.925° will result in 30 s.

0.03 or X.Y35 = 48 s, E.G.: 232.435° will result in 6 s.

0.04 or X.Y45 = 24 s, E.G.: 357.745° will result in 42 s.

0.05 or X.Y55 = 60 s, E.G.: 176.855° will result in 18 s.

0.06 or X.Y65 = 36 s, E.G.: 45.165° will result in 54 s.

0.07 or X.Y75 = 12 s, E.G.: 217.975° will result in 30 s.

0.08 or X.Y85 = 48 s, E.G.: 333.285° will result in 6 s.

0.09 or X.Y95 = 24 s, E.G.: 67.895° will result in 42 s.

0.10 or X.Y05 = 60 s, E.G.: 156.405° will result in 18 s.

As you can see having 2 decimals or 3 decimals with the last decimal as 5 goes up alternately by 12 or 6 for seconds. There are 10 values in total for seconds if we are being lazy and starting with an angle of M/(73/6).

For any other whole number for seconds other than the 10 above, we need to start from scratch from the whole number for (s), then divide s/60 then plus k minutes etc. We go up then down or forward and reverse, that is we create the question and then undo it.

The following is the formula for attaining θ in the 4 quadrants of the unit circle:

Q1.

sin^-1(y) = θ

cos^-1(x) = θ

Q2.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = θ

Q3.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Q4.

sin^-1(y) = θ – 2π therefore θ = 2π + sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

QUESTION 20:

(TRIGONOMETRIC EQUATIONS).

θ = s/60 × 2π

sin^2(θ) + 2 = 11/4

2cos(θ) + 1 = 0

a. Solve for θ.

sin^2(θ) =11/4 – 2 = 3/4

sin(θ) = √(3/4) = √(3)/2

sin^-1(√(3)/2) = π/3

2cos(θ) = – 1

cos(θ) = -1/2

cos^-1(-1/2) = 2π/3

θ = 2π/3

b. Solve for s.

2π/3 = s/60 × 2π

1/3 = s/60

s = 1/3 × 60 = 20

c. 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.

s/60 = 1/3

m = k + s/60 = 49/3

k = m – s/60 = 16

m/60 = 49/180

h = k + m/60 = 3829/180

k = h – m/60 = 21

h/24 = 3829/4320

d = k + h/24 = 68629/4320

k = d – h/24 = 15

d/30 = 68629/129600

M = k + d/30 = 846229/129600

k = M – d/30 = 6

M/(73/6) = 846229/1576800

y = k + M/(73/6) = 103629719029/1576800

k = y – M/(73/6) = 65721

d. Create another angle (θ) from M/(73/6) × 2π and get the sine and cosine. Also although you already know it and the object is defeated, workout θ.

y = sin(M(73/6) × 2π) = 0.05881902090299

x = cos(M(73/6) × 2π) = 0.99826866262545

Note: to workout nice neat radians fractions instead of ugly decimals you need the Natural Scientific Calculator app! Invite link here:

Natural Scientific Calculator (NSC).

sin^-1(y) = θ = 846229π/788400

cos^-1(x) = θ = 846229π/788400

e. Although you have already done it workout M/(73/6), M and k.

M/(73/6) = θ/2π = 846229/1576800

M = M/(73/6) × (73/6) = 846229/129600

k = M – d/30 = 6

f. Although you have already done it from (s), workout the other decimal remainders and integers (k) of the days, hours, minutes and seconds in reverse order.

d/30 = M – k = 68629/129600

d = (M – k) × 30 = 68629/4320

k = d – h/24 = 15

h/24 = d – k = 3829/4320

h = (d – k) × 24 = 3829/180

k = h – m/60 = 21

m/60 = h – k = 49/180

m = (h – k) × 60 = 49/3

k = m – s/60 = 16

s/60 = m – k = 1/3

s = (m – k) × 60 = 20

t = 65721 years 6 months 15 days 21:16:20

g. Choose and workout the magnitudes of T^1, T^-1, T^2 and T^-2?

T^1 = √(A/B) = 10^15 fs

T^-1 = √(B/A) = 10^-15 Ps

T^2 = A/B = 10^30 fs²

T^-2 = B/A = 10^-30 Ps²

h. Workout the values of t, A and B?

t = A/T^1 = y × 31536000 = 2.072594380580 × 10^12 s

A = tT^1 = X × 10^12 × 10^15 = X × 10^27 fs

B = A/T^2 = X × 10^27 / 10^30 = X × 10^-3 Ps

i. Check months and days.

y – d/365 = t / 31536000 – d/365 = 65721

d – h/24 = d/365 × 365 – h/24 = 195

h – m/60 = h/24 × 24 – m/60 = 21

m – s/60 = m/60 × 60 – s/60 = 16

s – cs/100 = s/60 × 60 – cs/100 = 20

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 2072594380580 / 31536000 – d/365 = 65721

d – h/24 = (65721.5366749112 – 65721) × 365 – h/24 = 195

h – m/60 = (195.886342592592 – 195) × 24 – m/60 = 21

m – s/60 = (21.2722222222222 – 21) × 60 – s/60 = 16

s – cs/100 = (16.3333333333333 – 16) × 60 – cs/100 = 20

t = 65721 years 195 days 21:16:20

Hand written example:

## Question 19: Creating your own questions.

QUESTION 19.

When creating your own trigonometry and radians questions know that if we are being lazy and starting from an angle of M/(73/6) and we want particularly pretty maths or simple fractions then θ needs 2 decimal places. Any less and there are 0 seconds, anymore and the seconds are not a nice whole number. However, if you create the question from the ground up, that is starting with a nice whole number for the seconds and building up, then the maths is very beautiful.

However, starting with an angle of M/(73/6) with 2 decimal places means there are only 5 possible values for seconds (s).

For example:

0.00 or X.00 = 60 s, E.G.: 310.00° will result in 60 s.

0.01 or X.Y1 = 36 s, E.G.: 129.61° will result in 36 s.

0.02 or X.Y2 = 12 s, E.G.: 78.92° will result in 12 s.

0.03 or X.Y3 = 48 s, E.G.: 232.43° will result in 48 s.

0.04 or X.Y4 = 24 s, E.G.: 357.74° will result in 24 s.

0.05 or X.Y5 = 60 s, E.G.: 176.85° will result in 60 s.

0.06 or X.Y6 = 36 s, E.G.: 45.16° will result in 36 s.

0.07 or X.Y7 = 12 s, E.G.: 217.97° will result in 12 s.

0.08 or X.Y8 = 48 s, E.G.: 333.28° will result in 48 s.

0.09 or X.Y9 = 24 s, E.G.: 67.89° will result in 24 s.

0.10 or X.Y0 = 60 s, E.G.: 156.40° will result in 60 s.

However, we can start with an angle of M/(73/6) and use 3 decimal places and still have beautiful maths and finish with a whole number for seconds, but the last of the 3 decimal places must always be 5.

For example:

0.00 or X.005 = 60 s, E.G.: 310.005° will result in 18 s.

0.01 or X.Y15 = 36 s, E.G.: 129.615° will result in 54 s.

0.02 or X.Y25 = 12 s, E.G.: 78.925° will result in 30 s.

0.03 or X.Y35 = 48 s, E.G.: 232.435° will result in 6 s.

0.04 or X.Y45 = 24 s, E.G.: 357.745° will result in 42 s.

0.05 or X.Y55 = 60 s, E.G.: 176.855° will result in 18 s.

0.06 or X.Y65 = 36 s, E.G.: 45.165° will result in 54 s.

0.07 or X.Y75 = 12 s, E.G.: 217.975° will result in 30 s.

0.08 or X.Y85 = 48 s, E.G.: 333.285° will result in 6 s.

0.09 or X.Y95 = 24 s, E.G.: 67.895° will result in 42 s.

0.10 or X.Y05 = 60 s, E.G.: 156.405° will result in 18 s.

As you can see having 2 decimals or 3 decimals with the last decimal as 5 goes up alternately by 12 or 6 for seconds. There are 10 values in total for seconds if we are being lazy and starting with an angle of M/(73/6).

For any other whole number for seconds other than the 10 above, we need to start from scratch from the whole number for (s), then divide s/60 then plus k minutes etc. We go up then down or forward and reverse, that is we create the question and then undo it.

The following is the formula for attaining θ in the 4 quadrants of the unit circle:

Q1.

sin^-1(y) = θ

cos^-1(x) = θ

Q2.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = θ

Q3.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Q4.

sin^-1(y) = θ – 2π therefore θ = 2π + sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

QUESTION 19:

a. Choose any whole number between 1 and 59 for seconds (s) and divide it 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.

s/60 = 11/20

m = k + s/60 = 531/20

k = m – s/60 = 26

m/60 = 177/400

h = k + m/60 = 7377/400

k = h – m/60 = 18

h/24 = 2459/3200

d = k + h/24 = 69659/3200

k = d – h/24 = 21

d/30 = 69659/96000

M = k + d/30 = 1029659/96000

k = M – d/30 = 10

M/(73/6) = 1029659/1168000

y = k + M/(73/6) = 67305861659/1168000

k = y – M/(73/6) = 57624

b. Create angle (θ) from M/(73/6) × 2π and get the sine and cosine.

y = sin(M(73/6) × 2π) = -0.67738134317054

x = cos(M(73/6) × 2π) = -0.73563205199642

c. Although you already know it and the object is defeated, workout θ.

Note: to workout nice neat radians fractions instead of ugly decimals you need the Natural Scientific Calculator app! Invite link here:

Natural Scientific Calculator (NSC).

sin^-1(y) = θ – 2π = -138341π/584000

θ = 2π + sin^-1(y) = 1029659π/584000

cos^-1(x) = 2π – θ = 138341π/584000

θ = 2π – cos^-1(x) = 1029659π/584000

d. Although you have already done it workout M/(73/6), M and k.

M/(73/6) = θ/2π = 1029659/1168000

M = M/(73/6) × (73/6) = 1029659/96000

k = M – d/30 = 10

e. Although you have already done it from (s), workout the other decimal remainders and integers (k) of the days, hours, minutes and seconds in reverse order.

d/30 = M – k = 69659/96000

d = (M – k) × 30 = 69659/3200

k = d – h/24 = 21

h/24 = d – k = 2459/3200

h = (d – k) × 24 = 7377/400

k = h – m/60 = 18

m/60 = h – k = 177/400

m = (h – k) × 60 = 531/20

k = m – s/60 = 26

s/60 = m – k = 11/20

s = (m – k) × 60 = 33

t = 57624 years 10 months 21 days 18:26:33

e. Choose and workout the magnitudes of T^1, T^-1, T^2 and T^-2?

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

g. Workout the values of t, A and B?

t = A/T^1 = y × 31536000 = 1.817258264793 × 10^12 s

A = tT^1 = X × 10^12 × 10^12 = X × 10^24 ps

B = A/T^2 = X × 10^24 / 10^24 = X × 10^0 Ts

h. Check months and days.

y – d/365 = t / 31536000 – d/365 = 57624

d – h/24 = d/365 × 365 – h/24 = 321

h – m/60 = h/24 × 24 – m/60 = 18

m – s/60 = m/60 × 60 – s/60 = 26

s – cs/100 = s/60 × 60 – cs/100 = 33

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 1817258264793 / 31536000 – d/365 = 57624

d – h/24 = (57624.8815573630 – 57624) × 365 – h/24 = 321

h – m/60 = (321.7684375 – 321) × 24 – m/60 = 18

m – s/60 = (18.4425 – 18) × 60 – s/60 = 26

s – cs/100 = (26.55 – 26) × 60 – cs/100 = 33

t = 57624 years 321 days 18:26:33

Hand written example:

## Question 18: Trigonometry and radians.

QUESTION 18.

When creating your own trigonometry and radians questions know that if we are being lazy and starting from an angle of M/(73/6) and we want particularly pretty maths or simple fractions then θ needs 2 decimal places. Any less and there are 0 seconds, anymore and the seconds are not a nice whole number. However, if you create the question from the ground up, that is starting with a nice whole number for the seconds and building up, then the maths is very beautiful.

However, starting with an angle of M/(73/6) with 2 decimal places means there are only 5 possible values for seconds (s).

For example:

0.00 or X.00 = 60 s, E.G.: 310.00° will result in 60 s.

0.01 or X.Y1 = 36 s, E.G.: 129.61° will result in 36 s.

0.02 or X.Y2 = 12 s, E.G.: 78.92° will result in 12 s.

0.03 or X.Y3 = 48 s, E.G.: 232.43° will result in 48 s.

0.04 or X.Y4 = 24 s, E.G.: 357.74° will result in 24 s.

0.05 or X.Y5 = 60 s, E.G.: 176.85° will result in 60 s.

0.06 or X.Y6 = 36 s, E.G.: 45.16° will result in 36 s.

0.07 or X.Y7 = 12 s, E.G.: 217.97° will result in 12 s.

0.08 or X.Y8 = 48 s, E.G.: 333.28° will result in 48 s.

0.09 or X.Y9 = 24 s, E.G.: 67.89° will result in 24 s.

0.10 or X.Y0 = 60 s, E.G.: 156.40° will result in 60 s.

However, we can start with an angle of M/(73/6) and use 3 decimal places and still have beautiful maths and finish with a whole number for seconds, but the last of the 3 decimal places must always be 5.

For example:

0.00 or X.005 = 60 s, E.G.: 310.005° will result in 18 s.

0.01 or X.Y15 = 36 s, E.G.: 129.615° will result in 54 s.

0.02 or X.Y25 = 12 s, E.G.: 78.925° will result in 30 s.

0.03 or X.Y35 = 48 s, E.G.: 232.435° will result in 6 s.

0.04 or X.Y45 = 24 s, E.G.: 357.745° will result in 42 s.

0.05 or X.Y55 = 60 s, E.G.: 176.855° will result in 18 s.

0.06 or X.Y65 = 36 s, E.G.: 45.165° will result in 54 s.

0.07 or X.Y75 = 12 s, E.G.: 217.975° will result in 30 s.

0.08 or X.Y85 = 48 s, E.G.: 333.285° will result in 6 s.

0.09 or X.Y95 = 24 s, E.G.: 67.895° will result in 42 s.

0.10 or X.Y05 = 60 s, E.G.: 156.405° will result in 18 s.

As you can see having 2 decimals or 3 decimals with the last decimal as 5 goes up alternately by 12 or 6 for seconds. There are 10 values in total for seconds if we are being lazy and starting with an angle of M/(73/6).

For any other whole number for seconds other than the 10 above, we need to start from scratch from the whole number for (s), then divide s/60 then plus k minutes etc. We go up then down or forward and reverse, that is we create the question and then undo it.

The following is the formula for attaining θ in the 4 quadrants of the unit circle:

Q1.

sin^-1(y) = θ

cos^-1(x) = θ

Q2.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = θ

Q3.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Q4.

sin^-1(y) = θ – 2π therefore θ = 2π + sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Question 18:

T^1 = √(A/B) = 10^18 as

k = y – M/(73/6) = 57892

sin(M(73/6) × 2π) = -0.5278455119451

cos(M(73/6) × 2π) = -0.8493404002633

a. Workout θ.

b. Workout M/(73/6), M and k.

c. Workout the other decimal remainders and integers (k) of the days, hours, minutes and seconds.

d. What is the total value of y?

e. What are the magnitudes of T^-1, T^2 and T^-2?

f. What are the values of t, A and B?

g. Check the days and months.

Note: to figure out a fraction radian from a decimal radian, divide the decimal radian by 2π then convert the resulting decimal into a fraction.

For example:

0.6471680866395 / 2π = 0.103 = 103/1000

Then multiply the fraction by 2π.

a.

sin^-1(-0.5278455119451) = π – θ = -177π/1000

θ = π – (-177π/1000) = 1177π/1000

cos^-1(-0.8493404002633) = 2π – θ = 823π/1000

θ = 2π – 823π/1000 = 1177π/1000

b.

M/(73/6) = θ/2π = (1177π/1000)/2π = 1177/2000

M = M/(73/6) × (73/6) = 85921/12000

k = M – d/30 = 7

c.

d/30 = M – k = 1921/12000

d = (M – k) × 30 = 1921/400

k = d – h/24 = 4

h/24 = d – k = 321/400

h = (d – k) × 24 = 963/50

k = h – m/60 = 19

m/60 = h – k = 13/50

m = (h – k) × 60 = 78/5

k = m – s/60 = 15

s/60 = m – k = 3/5

s = (m – k) × 60 = 36

t = 57892 years 7 months 4 days 19:15:36

d.

y = k + M/(73/6) = 115785177/2000

e.

T^-1 = √(B/A) = 10^-18 Es

T^2 = A/B = 10^36 as²

T^-2 = B/A = 10^-36 Es²

f.

t = A/T^1 = y × 31536000 = 1.825700670936 × 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

g.

y – d/365 = t / 31536000 – d/365 = 57892

d – h/24 = d/365 × 365 – h/24 = 214

h – m/60 = h/24 × 24 – m/60 = 19

m – s/60 = m/60 × 60 – s/60 = 15

s – cs/100 = s/60 × 60 – cs/100 = 36

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 1825700670936 / 31536000 – d/365 = 57892

d – h/24 = (57892.5885 – 57892) × 365 – h/24 = 214

h – m/60 = (214.802499999278 – 214) × 24 – m/60 = 19

m – s/60 = (19.2599999826634 – 19) × 60 – s/60 = 15

s – cs/100 = (15.5999989598058 – 15) × 60 – cs/100 = 36

t = 57892 years 214 days 19:15:36

Hand written example:

## Question 17: Trigonometry and radians.

QUESTION 17.

To create your own trigonometry questions and to save you time, know that θ or (M/(73/6) × 360) should have 2 decimal places. Any less and there are 0 seconds, anymore and the maths is not beautiful.

Note: 2 decimal places this means there are only 5 possible values for seconds (s).

For example:

0.00 or X.00 = 60 s, E.G.: 310.00° will result in 60 s.

0.01 or X.Y1 = 36 s, E.G.: 129.61° will result in 36 s.

0.02 or X.Y2 = 12 s, E.G.: 78.92° will result in 12 s.

0.03 or X.Y3 = 48 s, E.G.: 232.43° will result in 48 s.

0.04 or X.Y4 = 24 s, E.G.: 357.74° will result in 24 s.

0.05 or X.Y5 = 60 s, E.G.: 176.85° will result in 60 s.

0.06 or X.Y6 = 36 s, E.G.: 45.16° will result in 36 s.

0.07 or X.Y7 = 12 s, E.G.: 217.97° will result in 12 s.

0.08 or X.Y8 = 48 s, E.G.: 333.28° will result in 48 s.

0.09 or X.Y9 = 24 s, E.G.: 67.89° will result in 24 s.

0.10 or X.Y0= 60 s, E.G.: 156.40° will result in 60 s.

Note: the Mathway app makes light work of fractions and radians, also a second scrap piece of paper is recommended.

The following is the formula for attaining θ in the 4 quadrants of the unit circle:

Q1.

sin^-1(y) = θ

cos^-1(x) = θ

Q2.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = θ

Q3.

sin^-1(y) = π – θ therefore θ = π – sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Q4.

sin^-1(y) = θ – 2π therefore θ = 2π + sin^-1(y)

cos^-1(x) = 2π – θ therefore θ = 2π – cos^-1(x)

Question 17:

T^1 = √(A/B) = 10^21 zs

k = y – M/(73/6) = 80723

sin(M(73/6) × 2π) = -0.602929541689

cos(M(73/6) × 2π) = -0.79779443953857

a. Workout θ.

b. Workout M/(73/6), M and k.

c. Workout the other decimal remainders and integers (k) of the days, hours, minutes and seconds.

d. What is the total value of y?

e. What are the magnitudes of T^-1, T^2 and T^-2?

f. What are the values of t, A and B?

g. Check the days and months.

Note: to figure out a fraction radian from a decimal radian, divide the decimal radian by 2π then convert the resulting decimal into a fraction.

For example:

0.6471680866395 / 2π = 0.103 = 103/1000

Then multiply the fraction by 2π.

a.

sin^-1(-0.602929541689) = θ – 2π = -103π/500

θ = 2π – 103π/500 = 897π/500

cos^-1(-0.79779443953857) = 2π – θ = 103π/500

θ = 2π – 103π/500 = 897π/500

b.

M/(73/6) = θ/2π = (897π/500)/2π = 897/1000

M = M/(73/6) × (73/6) = 21827/2000

k = M – d/30 = 10

c.

d/30 = M – k = 1827/2000

d = (M – k) × 30 = 5481/200

k = d – h/24 = 27

h/24 = d – k = 81/200

h = (d – k) × 24 = 243/25

k = h – m/60 = 9

m/60 = h – k = 18/25

m = (h – k) × 60 = 216/5

k = m – s/60 = 43

s/60 = m – k = 1/5

s = (m – k) × 60 = 12

t = 80723 years 10 months 27 days 09:43:12

d.

y = k + M/(73/6) = 80723897/1000

e.

T^-1 = √(B/A) = 10^-21 Zs

T^2 = A/B = 10^42 zs²

T^-2 = B/A = 10^-42 Zs²

f.

t = A/T^1 = y × 31536000 = 2.545708815792 × 10^12 s

A = tT^1 = X × 10^12 × 10^21 = X × 10^33 zs

B = A/T^2 = X × 10^33 / 10^42 = X × 10^-9 Zs

g.

y – d/365 = t / 31536000 – d/365 = 80723

d – h/24 = d/365 × 365 – h/24 = 327

h – m/60 = h/24 × 24 – m/60 = 9

m – s/60 = m/60 × 60 – s/60 = 43

s – cs/100 = s/60 × 60 – cs/100 = 12

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 2545708815792 / 31536000 – d/365 = 80723

d – h/24 = (80723.897- 80723) × 365 – h/24 = 327

h – m/60 = (327.40499999898 – 327) × 24 – m/60 = 9

m – s/60 = (9.71999997552484 – 9) × 60 – s/60 = 43

s – cs/100 = (43.1999985314906 – 43) × 60 – cs/100 = 12

t = 80723 years 327 days 09:43:12

Hand written example:

## Question 16: Trigonometry and radians.

QUESTION 16.

Note: if you want to create your own trigonometry questions, to save you time, θ or M/(73/6) × 360 should have at least 2 decimal places, otherwise, there are 0 seconds. And 3 decimal places is recommended as the maximum.

Note: radians are very hard, you basically must have the Mathway app to make light work of fractions and radians, also a second scrap piece of paper is recommended.

Question 16:

T^1 = √(A/B) = 10^12 ps

k = y – M/(73/6) = 81073

sin(M(73/6) × 2π) = -0.9260024197156

cos(M(73/6) × 2π) = -0.3775175740026

a. What are the magnitudes of T^-1, T^2 and T^-2?

b. Workout θ.

c. Workout M/(73/6) and the other decimal remainders and integers (k) of the months, days, hours, minutes and seconds.

d. What is the total value of y?

e. What are the values of t, A and B?

f. Check the days and months.

a.

T^-1 = √(B/A) = 10^-12 Ts

T^2 = A/B = 10^24 ps²

T^-2 = B/A = 10^-24 Ts²

b.

sin^-1(-0.9260024197156) = -3391π/9000

cos^-1(-0.3775175740026) = 5609π/9000

θ = π + 3391π/9000 = 12391π/9000

or

θ = 2π – 5609π/9000 = 12391π/9000

c.

M/(73/6) = (12391π/9000)/2π = 12391/18000

M = M/(73/6) × (73/6) = 904543/108000

k = M – d/30 = 8

d/30 = M – k = 40543/108000

d = (M – k) × 30 = 40543/3000

k = d – h/24 = 11

h/24 = d – k = 943/3600

h = (d – k) × 24 = 943/150

k = h – m/60 = 6

m/60 = h – k = 43/150

m = (h – k) × 60 = 86/5

k = m – s/60 = 17

s/60 = m – k = 1/5

s = (m – k) × 60 = 12

d.

y = k + M/(73/6) = 1459326391/18000

e.

t = A/T^1 = y × 31536000 = 2.556739837032 × 10^12s

A = tT^1 = X × 10^12 × 10^12 = X × 10^24 ps

B = A/T^2 = X × 10^24 / 10^24 = X Ts

f.

y – d/365 = t / 31536000 – d/365 = 81073

d – h/24 = d/365 × 365 – h/24 = 251

h – m/60 = h/24 × 24 – m/60 = 6

m – s/60 = m/60 × 60 – s/60 = 17

s – cs/100 = s/60 × 60 – cs/100 = 12

NOTE: obviously we do NOT literally minus the decimal such as s/60 on the calculator, we simply minus the integers (k) and multiply the decimal by (73/6), 30, 24 or 60. We do not write or type long numbers. We only have to type the first division below and ‘lift’ the whole number, the rest is done by the calculator.

For example:

Note: you do not need to write the below, it is all done on the calculator.

y – d/365 = 2556739837032 / 31536000 – d/365 = 81073

d – h/24 = (81073.6883888889 – 81073) × 365 – h/24 = 251

h – m/60 = (251.26194444274 – 251) × 24 – m/60 = 6

m – s/60 = (6.28666662576143 – 6) × 60 – s/60 = 17

s – cs/100 = (17.1999975456856 – 17) × 60 – cs/100 = 12

t = 81073 years 251 days 06:17:12

Hand written example: