In this post I’ll be proving to you that:

Firstly, I’ll say that:

And also that:

If this is the case, then:

And as this is in the form:

I would have to conclude that:

Hence, I have my proof.

In this post I’ll be proving to you that:

Firstly, I’ll say that:

And also that:

If this is the case, then:

And as this is in the form:

I would have to conclude that:

Hence, I have my proof.

In this post I’ll be proving why:

Let’s say that:

And also that:

This would imply that:

Now if we multiply and together, we get:

Which is thanks to what we know about **trigonometric identities**.

As we can see above, we’ve formed another complex number:

And this is in the form of:

And because of the rules of complex numbers, we can say that:

Hence, we have our proof.

In this post I’ll be showing you how to prove that:

Firstly, let’s say that:

If this is the case, then according to the rules of complex numbers:

Secondly, let’s determine what is…

As you can see, we get the result above – which is another complex number.

This means that:

Therefore we’ve proven that:

You can watch a video related to this proof below…

Hello. In this post I’ll be showing you how to derive sin(0°), sin(15°), sin(30°), sin(45°), sin(60°), sin(75°), sin(90°), cos(0°), cos(15°), cos(30°), cos(45°), cos(60°), cos(75°), cos(90°), tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°) from absolute scratch.

Now, I’ll first start off by showing you how to derive sin(30°), sin(60°), cos(30°) and cos(60°) with the use of an **equilateral triangle** (image above). This equilateral triangle has lengths equal to 2. If you look at the diagram above and its properties carefully, you should conclude that:

Alright, so far so good. Next, have a look at this **isosceles triangle** (image above). If you take its properties into consideration – you’ll discover that:

Ok, so I’ve already shown you how to derive sin(30°), sin(45°), sin(60°), cos(30°), cos(45°) and cos(60°) using simple diagrams. It turns out that **with the information above** and also some **trigonometric identities** – we can derive sin(15°), sin(75°), cos(15°) and cos(75°). Let me show you what I mean…

sin(0°), sin(90°), cos(0°) and cos(90°) are values you should already know, so I won’t be demonstrating how to derive them. If you have studied the **unit circle** – you’ll know that:

These values are fairly easy to find.

So, this is the moment you’ve been waiting for… The complete set of derivations I said I’d give you. Although it may seem hard to derive tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°) from absolute scratch, or like a tedious task – we have already done most of the hard work. All these tangent values can be derived using the information we’ve already accumulated, because:

Therefore:

And now, the set of derivations is complete. 😀

On this website I previously showed you **why the formulas used to complete the square work** – and how they can be used to **derive formulas such as the quadratic equation.** Now, I’ll be doing something different, but related… On this post, I’ll be showing you **how to come up with the formulas (2 in total) used to complete the square – geometrically**.

In the diagram above, what we can see is that:

This means that:

In the diagram above, what we can see is that:

This means that:

I hope these **geometrical proofs** have helped you better understand why the formulas we use to complete the square are in existence. Thanks for reading! 😀

In this blog post I’ll be revealing more ways (4 in fact) in which to express or come up with the value of the **golden ratio**…

**Number One:**

**Number Two:**

**Number Three:**

**Number Four:**

And check out this calculator trick…

If you’re not satisfied with what I’ve already produced, then you can have a go at proving that…

Without using the phi (φ) symbol.

Enjoy!!! 😀

In this post I’m going to be proving that…

So, here I go…

Wait for it…

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

So, you have the **equation of the ellipse** but **you need to completely isolate y**. How would you go about doing this? Well, here is a fantastic example…

This will come in handy if you’re trying to **derive the area of an ellipse from absolute scratch**.

Thanks for reading! 🙂

If you’re trying to find the **area of a circle** using **integration methods**, then these **trigonometric formulas** are going to be very useful:

**First formulas:**

**Second formulas:**

These formulas are to be used when you have to transform the expression:

You can either make:

Or…

The choice is yours. 🙂