Sine and Cosine Rule

In this post, I will discuss the derivations for the sine and cosine rule. Unlike right-angled trigonometry, these two rules are applicable to all triangles, regardless of lengths or angles.

From the Pythagorean Theorem:

And we know that x and y are:

And y:

remember that c=y+adj; adj is short for the unlabelled side that is adjacent to angle 

We can now substitute these formulae into the formula for a:

Expanding the brackets:

Factorising the b squared terms:

From the Pythagorean Identity:

Which is the cosine rule. It doesn't matter which side is a,b or c providing the angle A is opposite side a, the angle B opposite side b etcetera. The formula states that the side opposite to an angle can be calculated from the angle itself and the two adjacent sides.

For the sine rule, we'll use the same triangle changed a bit:

From the triangle we can see (from line y):

Which implies:

And from line x:

So:

Combining the two equations, we have:

Taking the reciprocal of each term also gives:

I hope you have understood these derivations of the sine and cosine rules. Don't be afraid to use the links if you didn't, and keep referring back to the triangle if need be. With thanks to desmos.-naBla5040

Proof for the Area of a Circle

The area of a circle can easily be calculated, it is a formula we all know of, but very few think of why it is true. We can derive the formula for the area of a circle using trigonometry and Archimedes' Method of Exhaustion. First we will start with regular polygons:

As you can see the number of triangles a polygon can be divided up into is equal to its number of sides (for example, a square can be split into 4 triangles; and a pentagon can be split into 5).

If I asked you to find the area of a regular hexagon you will most probably find it quite difficult unless you did the trick of splitting it into 6 different triangles. To find the area of any n sided polygon we just need to find the area of the consecutive triangles that make it up and multiply it by the number of triangles. So:

remember n is the number of triangles, which is the number of sides; and the A subscript t is just the area of each triangle

To find the area of each triangle, we can use:

From the image above:

The triangles above are isosceles so a and b are equal, above I've shortened this to r squared (a and b are both r). Also, the angle theta is the same for all the triangles (as the triangles are all the same). So, as there are 360 degrees in a full circle, we just divide this by the number of triangles or the number of sides to find theta. So we have:

Here we have found a formula to find the area of any regular polygon as a function of its number of sides and the distance to the vertex (point). However, a circle isn't a polygon, but we can pretend it is... 

Archimedes' Method of Exhaustion:

As you can see, as we add more and more sides to a regular polygon, it becomes closer and closer to a circle, we can say: as the number of sides approaches infinity the area of the polygon approaches the area of the circle. So:

A subscript c means the area of the circle, the limit just means what the equation on the right approaches as the number of sides gets bigger and bigger

We will isolate this part, as this is what we'll be focusing on for now:

Now we have to find what the function above approaches as n gets closer and closer to infinity.

n is on the x-axis (made on desmos)

Hopefully, you can see the graph levelling out. This shows us the graph approaches a certain value. We can roughly see it is 6, if you were to go out even further you would see it approaches 6.283. Which interestingly is approximately 2 pi. If you would like to stop here, you can take my word it approaches 2 pi; and therefore:

Let's assume the function converges to a limit:

I have switched to radians here

The limit of the function in this case is the minimum value L takes where there is no longer any solution to the equation in terms of n. For there to be a solution for n to the equation below L has to be less than 2 pi.

As if L is 2 pi than the only solution to the equation would be n is infinity, or no solution (as 2 pi divided by n would have to be 0). Anything less than 2 pi can have a solution, but anything greater cannot. Since L is 2 pi:

We can see the formula for the area of a circle.

I'm not sure if my reasoning is sound for the last part, so please keep that in mind. Thanks for reading-naBla5040

Circle Theorems

I have had a few requests for circle theorems from people I know. While I will be happy to write about this topic, I feel especially at a GCSE level, the circle theorem questions can often be more difficult compared to the rest of the paper. Spotting what theorem to use can be extremely difficult to do, even if you know the theorems inside out. So before I get started, I would like you to know that knowing the theorems is not enough. You need to practice spotting the theorems and getting used to the questions. No teacher on the planet can show you how to do this: they can show you the theorems, but you need to be able to apply them through practice.

There are 9 theorems to know about; I will group them into three categories: the obvious ones, the ones that need learning and the more obscure ones. However, what goes into what category is entirely subjective, so please keep that in mind.

The Obvious Ones:

The radius and the tangent are at 90 degrees to each other, this should seem quite obvious as they are clearly perpendicular to one another.

the green is the radius, and the blue the tangent

Two radii form an isosceles triangle; this should also be obvious as an isosceles triangle has two sides of the same length and the radius is the same all around the circle.

Tangents drawn from the same point meet at the circumference with the same length, this can be hard to understand in words, but I'm basically saying the two tangents below are equal in length from the point they meet to where they each meet the circumference.

the green and black lines are tangents

The Ones That Need Learning:

The angle in a semicircle is 90 degrees. People often forget this one, remember it! However, this is providing the line goes through the centre of the circle (the blue line).

The angle at the centre is twice the size of the angle at the circumference.

Also, angles opposite the same segment are always equal.

The Obscure Ones:

The alternate segment theorem states that the angle between the tangent and the chord is equal to the angle in the opposite segment. It's difficult to explain, but hopefully the diagram helps!

the chord is the black line, the tangent is purple

Cyclic quadrilaterals: both pairs of opposite angles add to 180 degrees.

So:

Now there is another theorem, stating that the bisector of a chord passes through the centre of the circle: this is simple, however, providing you understand that the bisector is the perpendicular line through the mid-point of the chord.

the green line is the chord, and the blue line the bisector-which passes through the centre

I have briefly gone through all 9 theorems, however I haven't discussed any example problems, although I would be happy to in the future. To get the images I used the graphing software desmos. If you would like me to briefly go through anything else I would be happy to, please leave a comment. Thanks for reading-naBla5040

Shortest Distance Between Two Points on a Sphere

Imagine a sphere in three-dimensional space (also known as Euclidean Space), and imagine adding two points on the surface of the sphere (with cartesian coordinates; aka (x,y,z) coordinates for the two points). How would we find the shortest distance along the surface of the sphere between the two points? Like the distance between A and B along the surface of the sphere.

This involves nothing more than trigonometry, Pythagoras' Theorem and the equation for the length of the arc.

So we know from Pythagoras:

And for three dimensions:

To put this in terms of the coordinates of A and B:

remember x is the difference between the x coordinates of the 2 points and the same with y and z

d is the straight-line distance between the 2 points, not the distance along the surface

Now if you were to place a single point on a sphere and another point on the sphere, you could rotate the sphere to align the two points along the same axis, and if you were to follow the line along the sphere to the other point, well that should give you the shortest distance.

For example:

The shortest distance between A and B is the line highlighted in black. This shows us that the arc for the shortest distance is the arc of the circle with the same radius as the sphere; this is due to the radius of the circle shown (A and B) being equal to the radius of the sphere.

And the arc length is given by:

where theta is the angle between the two points from the centre, in the example above it is 90 degrees

Since we know the direct distance between A and B (d), we can form an isosceles triangle with the radii connecting the points to the centre, and the direct distance between the 2 points. Like:

We can find the angle theta from the cosine rule now:

We can now substitute this into the equation for arc length, and we can substitute the earlier formula for the direct distance into the one above, and the resulting formula looks like this:

Now, this is our final formula, for the shortest distance between two points along the surface of the sphere as a function of the (x,y,z) coordinates for both points and the radius of the sphere.

I hope you've enjoyed the post, I would like to credit GeoGebra and the website desmos, as they have proved very helpful over the last posts. Be sure to check them out!-naBla5040 

Differentiation from First Principles

     For those unfamiliar with calculus, I want to shed a bit of light onto the subject. Starting with basic differentiation and how it comes about. I hope you enjoy reading this as a potential first introduction. 

    I am presuming you're familiar with linear (straight line) graphs and how they have a constant gradient, and how they can be referred to as functions, like below:

    However, you are probably familiar with graphs like these:

     And with that, I ask you what is the gradient of these graphs?

If you rightly concluded, that well there is no single answer then you are right, as the gradient is constantly changing. This then leads us to the question of how do we figure out the gradient at a specific point.

Now, we're not apes, so we're not going to draw a tangent and find the gradient of that: instead we're going to use our brains!

Let's look closely at the following function (this will be the function we will be focusing on):

We can find the gradient of the line connecting the two points on the graph above:

Now imagine moving that line so that it passes through the point (1.5,2.25):

The gradient becomes:

Now, imagine the point getting even closer to (1,1), you can clearly see how as the point gets closer and closer to (1,1) the gradient between the 2 points gets closer and closer to the tangent gradient at (1,1). Remember the gradient of the tangent at the point (1,1) is equal to the gradient at the point (1,1).

To find the gradient at the point exactly we need to bring the point infinitely close to the point (1,1), ie right next to it. The gradient at the point is approximately given by:

where the change in x (difference) is close to 0

if you're unsure about the function notation: functions introduction

This can be represented also with:

where the triangle symbol means 'change in'

You can find the actual gradient at a point by finding the limit as delta x approaches 0 (as the 2 points get closer together):

remember: delta x is the difference between the 2 points

the use of different notation here indicates that the equation represents an infinitely small change in y divided by an infinitely small change in x

Usually, in calculus the gradient function (formula to find the gradient) is represented like this:

Ok, so we now have our formula, let's try an example function:

We now have a formula for the gradient of the function x squared, so we can find the gradient at certain points. When x is 0, for example, the gradient is also 0 (if you look on the graph this makes sense); and when x is 1 (the point 1,1) the gradient is 2.

I hope you have enjoyed reading this post, stay tuned and follow.-naBla5040