# Proving that arg(z_1/z_2)=arg(z_1)-arg(z_2)

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

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Now firstly I will have to say that:

${ z }_{ 1 }={ r }_{ 1 }\left( \cos { { \theta }_{ 1 }+i\sin { { \theta }_{ 1 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }$

And also that:

${ z }_{ 2 }={ r }_{ 2 }\left( \cos { { \theta }_{ 2 }+i\sin { { \theta }_{ 2 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }$

If this is the case, then…

Since this is in the form:

$z=r\left( \cos { \theta +i\sin { \theta } } \right)$

I would have to conclude that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) ={ \theta }_{ 1 }-{ \theta }_{ 2 }=arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Hence I’ve proven that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

# arg(z_1*z_2)=arg(z_1)+arg(z_2) Proof

In this post I’ll be proving why:

$arg\left( { z }_{ 1 }{ z }_{ 2 } \right) =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)$

Let’s say that:

${ z }_{ 1 }={ r }_{ 1 }\left( \cos { \left( { \theta }_{ 1 } \right) +i\sin { \left( { \theta }_{ 1 } \right) } } \right)$

And also that:

${ z }_{ 2 }={ r }_{ 2 }\left( \cos { \left( { \theta }_{ 2 } \right) +i\sin { \left( { \theta }_{ 2 } \right) } } \right)$

This would imply that:

$arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }$

$arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }$

Now if we multiply ${ z }_{ 1 }$ and ${ z }_{ 2 }$ together, we get:

Which is thanks to what we know about trigonometric identities.

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

${ z }_{ 1 }{ z }_{ 2 }={ r }_{ 1 }{ { r }_{ 2 }\left( \cos { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) +i\sin { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) } } \right) }$

And this is in the form of:

$z=r\left( \cos { \left( \theta \right) +i\sin { \left( \theta \right) } } \right)$

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

$arg\left( { z }_{ 1 }{ z }_{ 2 } \right) \\ \\ ={ { \theta } }_{ 1 }+{ { \theta } }_{ 2 }\\ \\ =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)$

Hence, we have our proof.

# Finding the formulas for areas of triangles

In this post I’ll be demonstrating how one can derive the three formulas which can be used to find the areas of triangles.

These formulas are in fact:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

To begin with, let’s start by looking at the diagram below:

Now, if you look at the diagram carefully – you will notice that the area of the triangle is:

$A=\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 }$

This can be simplified into:

$\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { x\cdot CN+\left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { CN\left\{ x+\left( c-x \right) \right\} }{ 2 } \\ \\ =\frac { CN\cdot c }{ 2 }$

Because of SOH CAH TOA, what we can also say is that:

$\sin { \left( A \right) } =\frac { O }{ H } =\frac { CN }{ b } \\ \\ \therefore \quad b\cdot \sin { \left( A \right) } =CN\\ \\ \sin { \left( B \right) =\frac { O }{ H } } =\frac { CN }{ a } \\ \\ \therefore \quad a\cdot \sin { \left( B \right) } =CN$

Now because:

$A=\frac { CN\cdot c }{ 2 }$

This ultimately means that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } \\ \\ A=\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) } \\ \\ \therefore \quad A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) =\frac { 1 }{ 2 } ac } \cdot \sin { \left( B \right) }$

Alright, so far so good… Now we must put the icing on the cake and attach the final piece of the jigsaw puzzle to the formula above. In order to find the three equations which can be used to find the areas of triangles, we must now discover the expression for sin(C). We can discover its expression by first saying that:

$C=\left( 90-A \right) +\left( 90-B \right) \\ \\ =90-A+90-B\\ \\ =180-A-B\\ \\ =180-\left( A+B \right) \\ \\ \therefore \quad \sin { \left( C \right) } =\sin { \left( 180-\left( A+B \right) \right) }$

And if we use the trigonometric identity below:

$\sin { \left( \alpha -\beta \right) } =\sin { \left( \alpha \right) \cdot \cos { { \left( \beta \right) } } -\cos { \left( \alpha \right) \cdot \sin { \left( \beta \right) } } }$

We will reach the conclusion:

$\sin { \left( 180-\left( A+B \right) \right) } =\sin { \left( 180 \right) \cdot \cos { \left( A+B \right) -\cos { \left( 180 \right) \cdot \sin { \left( A+B \right) } } } }$

But because:

$\sin { \left( 180 \right) =0 } ,\quad \cos { \left( 180 \right) =-1 } \\ \\ \sin { \left( 180-\left( A+B \right) \right) =-\left( -1 \right) \cdot \sin { \left( A+B \right) } } \\ \\ \therefore \quad \sin { \left( C \right) =\sin { \left( A+B \right) } }$

Now, sin(A+B) as a trigonometric identity, is:

$\sin { \left( A+B \right) =\sin { \left( A \right) \cdot \cos { \left( B \right) +\cos { \left( A \right) \cdot \sin { \left( B \right) } } } } }$

And, thanks to SOH CAH TOA…

$\sin { \left( A+B \right) =\sin { \left( C \right) } } \\ \\ \sin { \left( A \right) =\frac { CN }{ b } } \\ \\ \cos { \left( B \right) =\frac { A }{ H } } =\frac { \left( c-x \right) }{ a } \\ \\ \cos { \left( A \right) =\frac { A }{ H } =\frac { x }{ b } } \\ \\ \sin { \left( B \right) =\frac { CN }{ a } }$

Which means that…

$\sin { \left( C \right) =\frac { CN }{ b } \cdot \frac { \left( c-x \right) }{ a } +\frac { x }{ b } \cdot \frac { CN }{ a } } \\ \\ =\frac { CN\left( c-x \right) }{ ab } +\frac { CN\cdot x }{ ab } \\ \\ =\frac { CN\left( c-x \right) +CN\cdot x }{ ab } \\ \\ =\frac { CN\left\{ \left( c-x \right) +x \right\} }{ ab } \\ \\ =\frac { CN\cdot c }{ ab } \\ \\ \therefore \quad ab\cdot \sin { \left( C \right) =CN\cdot c } \\ \\ \therefore \quad \frac { 1 }{ 2 } ab\cdot \sin { \left( C \right) =\frac { CN\cdot c }{ 2 } =A }$

As this is the case, we can conclude that:

$A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }$

# How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } }{ 2 }$

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

$A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

# Proof: Thales’ Theorem

In this post I’ll be demonstrating how you can prove that Thales’ Theorem is true. To follow the steps in this post (11 in total), what you will require is a ruler, pair of compasses and a pencil.

##### Step 11: Prove that the angle at point D is equal to 90 degrees.

Thales’ Theorem is as follows:

Because AC is the diameter of the circle you drew, the angle at the point D (α+β) must be equal to 90 degrees. In more specific and general terms, if you have the points A, C and D lying on a circle – and the line AC is in fact the diameter of this circle – then the angle at point D (α+β) must be a right angle.

Proof (which must be derived using the diagram you’ve created):

All angles within a triangle (in 2 space) must add up to 180 degrees.

Mathematically, this means that:

$\alpha +\alpha +\beta +\beta =180\\ \\ \Rightarrow \quad 2\alpha +2\beta =180\\ \\ \Rightarrow \quad 2\left( \alpha +\beta \right) =180\\ \\ \Rightarrow \quad \frac { 2\left( \alpha +\beta \right) }{ 2 } =\frac { 180 }{ 2 } \\ \\ \therefore \quad \alpha +\beta =90$

And as a result, Thales’ theorem must be true. The angle α+β is the angle at point D.

# How to prove that the two angles below the apex of an isosceles triangle are equivalent

*You will need a pair of compasses, a ruler, pen and pencil to formulate this proof.

How would you go about proving that an isosceles triangle has two angles (below its apex) equal to one another?

Well, first of all – let’s start off by drawing a circle…

Now… We can tell that the circle we’ve just drawn has a centre (point at the centre). Next, what we have to do is add a couple of points to the edge of this circle. Like this…

Let’s name all these points A, B and C…

Now, let’s connect these points together with a few lines – to create an isosceles triangle ABC…

Ok… So far, so good… What you will need to do now is – place the needle of your compass on the point C and your pencil on the point B, like this…

Now spin your compass – and create an arc…

Next, get the needle of your compass and place it on the point B and put your pencil on the point C…

Draw another arc, like this…

Where the two arcs you’ve just drawn intersect, create a point… Call this point D…

Now, draw a line going through the points A and D. Call this line L. Line L will be perpendicular to the line BC…

Where the line L intersects the line BC, create a point E…

Now, it turns out, within the isosceles triangle ABC, we’ve created two right angles… This is because the line L is perpendicular to the line BC. Remember that the line L cuts the isosceles triangle down its centre. Let’s name these right angles big R…

If you look at the diagram above carefully, what you will notice is that the radius of the circle is equal in length to the line AB and also the line AC. Let’s name the lines AB and AC… We’ll call them r.

Let’s also name the line BC… We’ll call it x. This means that the line BE is equal to half of x, and because of this, the line CE must also be equal to half of x…

Finally (I know you must be tired of drawing), let’s call the angle ABC alpha and the angle ACB beta…

With our diagram complete, we can now prove that alpha and beta are equivalent to each other.

*You will need to know a bit of trigonometry to pass this point. SOH CAH TOA rules to be precise.

It turns out that:

$\cos { \left( \alpha \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

And also:

$\cos { \left( \beta \right) } =\frac { A }{ H } =\frac { \frac { x }{ 2 } }{ r } =\frac { x }{ 2r }$

This means that:

$\cos { \left( \alpha \right) } =\cos { \left( \beta \right) } \\ \\ \therefore \quad \alpha =\beta$

Hence, we’ve proven that an isosceles triangle has two angles (below its apex) equal to one another.

# Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle…

We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)?

Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because:

$\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn…

As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β

Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that:

$\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y

It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that:

$x+y=180$

So, it turns out we now have two useful equations:

$\alpha +\beta =180$

$x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves…

• There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial:

$\beta +y=180$

$\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof.

$\alpha +\beta =180$

$x+y=180$

$\beta +y=180$

$\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y:

# How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically

In this post I’ll be demonstrating how one can prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically…

First of all, let me show you this diagram…

sin(A-B)=sin(A)cos(B)-cos(A)sin(B) proof

*If you click on the diagram, you will be able to see its full size version.

Now, to begin with, I will have to write about some of the properties related to the diagram…

Property 1:

Angle B + (A – B) = B + A – B = A

Therefore, angle POR = A.

Property 2:

Angle OPS = 90 degrees

Property 3:

Length OS = 1

Also note:

All angles within a triangle on a flat plane should add up to 180 degrees. If you understand this rule, you will be able to discover why the angles shown on the diagram are correct. Angles which are 90 degrees are shown on the diagram too.

PROVING THAT SIN(A-B)=SIN(A)COS(B)-COS(A)SIN(B)

Since I’ve noted down some of the important properties related to the diagram, I can now focus on demonstrating why the formula above is true. I will demonstrate why the formula above is true using mathematics and the SOH CAH TOA rule…

$\sin { \left( A-B \right) } =\frac { O }{ H } =\frac { ST }{ 1 } =ST$

But it turns out that…

$ST=PR-PQ$

Because:

$QR=ST$

Now, what is PR and what is PQ?

$\sin { \left( B \right) } =\frac { O }{ H } =\frac { PS }{ 1 } =PS\\ \\ \cos { \left( B \right) } =\frac { A }{ H } =\frac { OP }{ 1 } =OP\\ \\ \sin { \left( A \right) } =\frac { O }{ H } =\frac { PR }{ \cos { \left( B \right) } } \quad \\ \\ \therefore \quad \sin { \left( A \right) } \cos { \left( B \right) } =PR\\ \\ \cos { \left( A \right) } =\frac { A }{ H } =\frac { PQ }{ \sin { \left( B \right) } } \\ \\ \therefore \quad \cos { \left( A \right) } \sin { \left( B \right) } =PQ$

And finally, to sum it all up:

$ST=PR-PQ\\ \\ \therefore \quad \sin { \left( A-B \right) =\sin { \left( A \right) } \cos { \left( B \right) } -\cos { \left( A \right) } \sin { \left( B \right) } }$

Need a better explanation? Watch this video…

Related Videos:

Related posts:

Simple But Elegant Way To Prove That sin(A+B)=sinAcosB+cosAsinB (Edexcel Proof Simplified)

# Vector Proof – Angle Between Two Vectors

Prove that:

$cos\theta =\frac { \underline { a } \cdot \underline { b } }{ \left| \underline { a } \right| \left| \underline { b } \right| }$

Firstly, look at the image below:

Also know that:

$cosC=\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab }$

$\left| \underline { a } \right| \left| \underline { a } \right| ={ \underline { a } }^{ 2 }\\ \\ \left| \underline { b } \right| \left| \underline { b } \right| ={ \underline { b } }^{ 2 }\\ \\ \left| \underline { b } -\underline { a } \right| \left| \underline { b } -\underline { a } \right| ={ \left( \underline { b } -\underline { a } \right) }^{ 2 }$

From the image, you’ll be able to see that:

$a=\left| \underline { b } \right| \\ \\ b=\left| \underline { a } \right| \\ \\ c=\left| \underline { b } -\underline { a } \right| \\ \\ cosC=cos\theta ,\quad \therefore \quad C=\theta$

Now: