Please see below.
Let us have the two circles centered at
Let the two tangents be
The line joining
Subtracting (B) from (A), we get
Applying componendo and dividendo
Comparing with the available data of radius of the earth I assume that earth's radius
So we can write
Transforming the angle in radian
So we get
( It is comparable to available figure as shown below)
Here is a plot of
Using the corresponding inverse on each equation:
Because the inverses have two possible values, each of the two equations separates into two more equations and each repeats at integer multiples of
We have 4 combinations ,
Add equation [1.2a] to equation [2.2a]:
Subtract equation [1.2a] from equation [2.2a]:
The editor is complaining about how long my answer is getting so I shall decrease the amount of detail.
The most practical way to solve this equation is by using a graphing utility, visually determining the approximate roots, and then employing an iterative process to get as close as possible to the exact roots. Let's take a look at the graph of one period of the function:
As you can see, we have three roots that are close to:
By trying values to the left and right of each, we can get to the actual values with great accuracy.
But if we wanted to actually solve for
Using the following identity:
Using the identities:
Multiplying both sides by
Rearranging terms, we get:
In order to solve for
Using the identity:
We can solve for
Multiplying both sides by
Let's figure out what
After foiling and combining like terms, we end up with:
Let's plug this in:
After foiling and combining like terms, we get:
For simplicity, let's let
This is a polynomial of degree
The other method is the Newton-Raphson method which says:
Start out with
Continuing this process, you can find the roots. It is a lengthy process and would exceed what Socratic would allow me to input. But you can look it up in calculus books and follow it.
Back to our equation, it turns out it has
The real roots are where it crosses the
Within one period of the function, we get three real values for
which are very close to what we observed at the beginning from the graph. You can try each of these values in the original problem equation and see that they work.
The minimum value of numerator is =0,
Denominator is always
The value of denominator
Graph shows that
This is an idea i have heard about a while ago...
Lets say we take a piece of slack string, and hold it, so it is still slack, and so it hangs vertically downward, or even a peice of chain, tethered to two posts, and hangs downward, both of these can be directly modelled by the hyperbolic cosine,
Or in general the function is:
So in this image, it models a peice of string bieng held at
This is the curve:
Where this curve can be parametrically defined as:
Where also arcs and archways can also be modelled by the caternary curve, where this has the properties of having a very strong foundation and infrastucture...
I hope this is a small insight to how hyperbolic functions can be used in the real world!
Proof by conventional trigonometric identities...
We can prove the follwing statment by trig identities...
We can use this identity...
Again we can use the fact that
Inserting these values in the given relation.
This means the triangle is right angled.
1st part of LHS
2nd part of LHS
Adding we get
We can find the angle between vectors using the Dot Product
The dot product states that for vectors a and b:
The dot product is sometimes called the inner product, because of the way the vectors a and b are multiplied and summed.
In algebra we are used to multiplying brackets in the following way.
In the case of the dot product we multiply the vectors in the following way.
So we are just multiplying corresponding components and then adding them together.
Now to our example:
First find the product of:
Now find the magnitudes of A and C:
So we have for:
( 2 .d.p.)
So the angle between vectors A and C is
Notice from the diagram that the angle found by the dot product, is the angle between the vectors where they are pointing in the same relative direction.