Trigonometrical Addition Theorems and Repercussions

Perl Regular Expression Tester

45/52-e/pi = 0.00012863595235031A long-time obsession of mine has been transcendental algebraic equations, which despite sounding a long way away from the real world, threw up a real gem of a function the other night. I have a function plotter that will map out phenomenally complex equations of the single variant variety. It looks like this (

Trying out a few randomisations, I began to recognise patterns and groups of curves and asymptotes. Then a level 9 function produced this :-

The amazing thing about this curve is that it seems to go along smoothly, then flips out completely, but finitely, and then continues along smoothly in the same direction and from the same area as where the flip-out started. I call this the first portrayal of a mathematically rigorous representation of a quantum jump. For example, consider an electron orbiting a nucleus. Prior to the point of flip-out (or supposed asymptote), the electron is in a steadily-changing energy field which is drawing the electron out of its current orbit, then a sudden plummeting followed by an even more sudden rise, as the electron joins its new orbit, which continues on smoothly. Many other representations may be inferred from this graph, but it's just one of many that can be generated. This is my "pet" application and it was ported from a programmable Casio calculator. Here is another one portraying Electron Shell Mappings :-

It can be seen from the diagram that the areas of triangles 1 and 2 add up to the area of triangle 3. It can be also seen that the angles in all 3 triangles are equal to each other (look at the angles θ and σ in the diagram) and therefore, the triangles are mathematically similar to each other. This means that each triangle can be scaled to exactly fit either of the other 2 triangles by multiplying each of the sides by a fixed factor. This also applies to the triangles' heights. Let's call the height of triangle 1's right-angle above its base, h. Then, using scaling, triangle 2's height would be :-

hb/a

and triangle 3's height would be :-

hc/a

In terms of areas of triangles (half base times height), we would have :-

1) Triangle 1 Area = ½ah

2) Triangle 2 Area = ½bhb/a

3) Triangle 3 Area = ½chc/a

But we already know that (1)+(2)=(3) (triangle areas 1 and 2 add up to 3), so adding equations and doubling, we get :-

ah + b²h/a = c²h/a

Dividing by h throughout and multiplying both sides by a, we get :-

Once you have accepted that a² + b² = c² in the case of a right-angled triangle where c is the hypotenuse, it becomes quite interesting how things turn out for the relationships between the angles and sides of a triangle. The following diagram is representative of a couple of simple, but sweet theorems, which lead nicely on to the next section on sine and cosine laws.

Using Pythagoras we see that

c² + d² = 1

Using definitions of sine and cosine we get

1) c = cos B

2) d = sin B

Hence

Continuing, we get

3) a + b = c cos A = cos B cos A from (1) and

4) b = d sin A = sin B sin A from (2).

We also know

cos (A + B) = a = (a + b) - b

Hence from (3) and (4), we get

Similarly,

1) A + B = 180° - C

2) cos (180° - C) = - cos C

3) f = b sin A

4) f = a sin B

5) d = b cos A

6) e = a cos B

(3) and (4) yield the sine law in

Again, using Pythagoras, we can see from the diagram that

8) d² + f² = b²

9) e² + f² = a²

Adding (8) and (9), we get

a² + b² = d² + e² + 2 f²

a² + b² = (d + e)² - 2 d e + 2 f²

Using (4), (5), and (6), we get

a² + b² = c² - 2 a b cos A cos B + 2 a² sin² B

Using a rearranged (7), so that a sin B = b sin A, we get

a² + b² = c² - 2 a b (cos A cos B - sin A sin B)

Now, we use the Addition Theorem above for cosine to simply the bracket contents and rearrange to get

c² = a² + b² + 2 a b cos (A + B)

Putting (1) and (2) together, we get cos (A + B) = - cos C, giving the final equation :-

This looks like a crooked variety of Pythagoras, bent by the fact that there may not be a right angle. When C is 90°, cos 90° is zero, so it is Pythagoras.

The angle A in the above and below diagrams, is exactly twice the size of angle B, wherever you draw the chord or the point on the circumference (as long as it's above the chord). B is the angle subtended by the chord at the centre of the circle, and A is the angle subtended at the circumference. Lines with a fleck through them are equal in length and are radii of the circle. To prove this, we use a property of isosceles triangles : the angles opposite the equal sides are also equal. And that the angles of a triangle add up to 180°.

1) A = x + y

x + x + y + y + z + z = 180°

Therefore, x + y + z = 90°

Rearranging, 2) z = 90° - x - y

Now, B = 180° - z - z

So from (2), B = 180° - 90° + x + y - 90° + x + y

Simplifying, B = 2(x + y)

Using (1), we get

In this case, angle A has been subtended at a point swung round the circumference to the left. Despite this, we can still prove the same, no matter how small angle w becomes.

A + x = y

Hence, 3) y - x = A

Also, B + 2z = 180°

4) 2z = 180° - B

Now, 5) A + y + z + w = 180°

But, z = x + w

6) w = z - x

So, using (6) in (5), we get A + y + z + z - x = 180°

7) A + y - x + 2z = 180°

Using (4) in (7), we get A + y - x + 180° - B = 180°

A + y - x = B

Using (3), we obtain

This is true even as the angle w tends to zero, which is surprising because one would expect angle A to tend to 0 rather than remain fixed. The reason is because the point on the circumference gets ever closer to the end of the subtending chord, and so the line rendering the angle gets ever shorter, hence w can become really small, but angle A remains the same.