Solving Simple Arcsine and Arccosine Equations

 Solving Simple Arcsine and Arccosine Equations

Angle Measure

This document will focus on angle measurement in degrees. For radians and grads, please use the appropriate measurement.

90° = π/2 rad = 100 grad

180° = π rad = 200 grad

Simple Arcsine Equations

The calculator arcsine function gives: Domain: -1 ≤ x ≤ 1, Range: -90° ≤ Θ ≤ 90°

Note that for any angle x: sin(180° - x) = sin(x), sin(x) = sin(x ± 360°*z) (z is an integer)

Given n, solve for Θ:

n = sin(Θ) = sin(180° - Θ)

Base Solution 1:

n = sin(Θ)

⇒ Θ = arcsin(n)

Base Solution 2:

n = sin(180° - Θ)

arcsin(n) = 180° - Θ

⇒ Θ = 180° - arcsin(n)

Example:

0.67 = sin(Θ)

Base Solution 1: Θ = arcsin(0.67) ≈ 42.0670648025°

Base Solution 2: Θ = 180° - arcsin(0.67) ≈ 137.932935198°

Given n and α, solve for Θ:

n = sin(α + Θ)

Base Solution 1:

n = sin(α + Θ)

arcsin(n) = α + Θ

⇒ Θ = arcsin(n) – α

Base Solution 2:

n = sin(180° - (α + Θ))

n = sin(180° - α – Θ)

arcsin(n) = 180° - α – Θ

⇒ Θ = 180° - α – arcsin(n)

Example:

0.7757 = sin(Θ + 76°)

Base Solution 1: Θ = arcsin(0.7757) – 76° ≈ -25.1314549842°

Base Solution 2: Θ = 180° - 76° - arcsin(0.7757) = 104° - arcsin(0.7757) ≈ 53.131459842°

To get all the possible angles, add and subtract multiples of 360°.

Solving Simple Arccosine Equations

The calculator arccosine function gives: Domain: -1 ≤ x ≤ 1, Range: -90° ≤ Θ ≤ 90°

Note that for any angle x: cos(180° - x) = -cos(x), cos(x) = cos(x ± 360°*z) (z is an integer)

Given n, solve for Θ:

n = cos(Θ), n = cos(-Θ)

Base Solution 1:

n = cos(Θ)

⇒ Θ = arccos(n)

Base Solution 2:

n = cos(-Θ)

⇒ Θ = -arccos(n)

Example:

0.58 = cos(Θ)

Base Solution 1: Θ = arccos(0.58) ≈ 54.54945736°

Base Solution 2: Θ = -arccos(0.58) ≈ -54.54945736°

Given n and α, solve for Θ:

n = cos(α + Θ)

Base Solution 1:

n = cos(α + Θ)

arccos(n) = α + Θ

⇒ Θ = arccos(n) – α

Base Solution 2:

n = cos(-(α + Θ))

n = cos(-α – Θ)

arccos(n) = -α – Θ

-arccos(n) = α + Θ

⇒ Θ = -arccos(n) – α

Example:

0.6 = cos(35° + Θ)

Base Solution 1: Θ = arccos(0.6) – 35° ≈ 18.13012035°

Base Solution 2: Θ = -arccos(0.6) – 35° ≈ -88.13010235°

To get all the possible angles, add and subtract multiples of 360°.

I hope you find this useful and helpful,

Eddie

All original content copyright, © 2011-2026. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Arcsine and Arccosine in terms of Arctangent

Arcsine and Arccosine in terms of Arctangent

Some Motivation

I recall reading about the Sinclair Scientific Programmable, a vintage scientific calculator that was introduced in 1975.  This calculator just had three trigonometric functions:  sine, cosine, and arctangent.  It is the aim of this blog to fill in the blanks.  

You can find more information here:  

http://www.vintagecalculators.com/html/scientific_prog_.html

In this blog entry, angles will have radian measure.  

For tangent, it's pretty easy use of the trig identity:

tan x = sin x ÷ cos x

Determining Arcsine 

Imagine the right triangle shown below:

Then:

sin Θ = x

Θ = arcsin x

By the Pythagorean Theorem:

t^2 + x^2 = 1

t^2 = 1 - x^2

t = √(1 - x^2)

tan Θ = x ÷ t

tan Θ = x ÷ √(1 - x^2)

Θ = arctan(x ÷ √(1 - x^2))

Then:

arcsin x = arctan(x ÷ √(1 - x^2))

Determining Arccosine

Most calculators for the arccos function have the range [0, π].  To accomplish this, we are going to use the identity

cos Θ = sin(π÷2 - Θ).

Let w = cos Θ

Then:  

w = cos Θ

w = sin(π÷2 - Θ)

arcsin w = π÷2 - Θ

Θ = π÷2 - arcsin w

Θ = π÷2 - arctan(w ÷ √(1 - w^2))

For any angle w:

arccos w = π÷2 - arctan(w ÷ √(1 - w^2))

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

An Alternative Way of Finding the Angle in a Rectangular to Polar Conversion

An Alternative Way of Finding the Angle in a Rectangular to Polar Conversion

Welcome to a special Monday edition of Eddie’s Math and Calculator Blog. 

The Traditional Method

Often we are required to find polar coordinates of a given point (x,y).   Finding the radius, r, is fairly simple:

r = √(x^2 + y^2) 

When talking about complex numbers, r represents the absolute value of x + yi where i = √-1.

Finding the angle, θ, often uses the formula:

θ = atan(y/x)

In complex numbers, θ represents the argument (arg) function.

On a scientific calculator the range of the arctangent function is ( -90°, 90° ).  (open interval).   In finding the true angle, adjustments will be required:

Let a = atan(y/x). Then: 

Quadrant I (x and y are both positive):  θ = a

Quadrant II (x is negative, y is positive): θ = a + 180°

Quadrant III (x and y are both negative):  θ = a - 180°

Quadrant IV (x is positive, y is negative):  θ = a 

If you are working with radian angle measures, know that 90° = π/2, and 180° = π.

This does not take into consideration situations where either x or y is 0:

If x > 0 and y = 0:  θ = 0°

If x = 0 and y > 0:  θ = 90°

If x < 0 and y = 0:  θ = 180°

If x = 0 and y < 0:  θ = -90°

Is there a shorter way to calculate θ?  

The Vector Method

Consider the point (x, y) as a vector [x, y].   Now draw another vector [x, 0].  In a regular Cartesian coordinate system, angles are measured from the x-axis counter clockwise.  

Let a and b represent two vectors.  Then the angle between two vectors are:

cos θ = (a ● b) / ( ||a|| ||b|| ) = dot(a,b) / ( norm(a) * norm(b) )

with:

dot(a,b) = a1 * b1 + a2 * b2

norm(a) = √(a1^2 + a2^2)

norm(b) = √(b1^2 + b2^2)

Let a = [x, y] and b = [x, 0].  Then:

cos θ = (x^2) / (√(x^2 + y^2) * √(x^2))

cos θ = (x^2) / (√(x^2 + y^2) * x)

cos θ = x / √(x^2 + y^2)

θ = acos( x / √(x^2 + y^2) )

The range of the arccosine function of a calculator is [ 0°, 180° ].  

If y < 0, the angle would be measured clockwise, and therefore I would make the adjustment:

θ = -acos( x / √(x^2 + y^2) )

In summary:

If y ≥ 0, θ = acos( x / √(x^2 + y^2) )

If y < 0, then θ = -acos( x / √(x^2 + y^2) )

Examples:

Find the angle, in degrees, in a rectangular to polar conversions:

Quadrant I  (2, 4):  y ≥ 0:   θ = acos( 2 / √(2^2 + 4^2) ) ≈ 63.43494882°

Quadrant II (-2, 4):  y ≥ 0:    θ = acos( -2 / √((-2)^2 + 4^2) ) ≈ 116.5650512°

Quadrant II (-2, -4):  y < 0:  θ = -acos( (-2) / √((-2)^2 + (-4)^2) ) ≈ -116.5650512°

Quadrant IV (2, -4):  y < 0:  θ = -acos( 2) / √(2^2 + (-4)^2) ) ≈ -63.43494882°

Vector Method for Navigation

In navigation, angles start from true North (up) and rotate clockwise towards East (right).  Angles are measured from 0° to 360°.

Use the vectors [0, N] and [E, N], then the angle between these vectors are:

If E ≥ 0,  θ = acos( N / √(E^2 + N^2))

If E < 0, θ = 360° - acos( N / √(E^2 + N^2))

Note:  This is the first blog entry that I have typed on Google Docs.  I have been using Windows app Wordpad for the last three years.   I am testing Google apps as I am considering buying a Chromebook.

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.