# 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.

# sin(0°) to sin(90°), cos(0°) to cos(90°) and tan(0°) to tan(90°) derivations. 15° steps.

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.

#### sin(30°), sin(60°), cos(30°) and cos(60°): 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: $\sin { \left( { 30 } \right) } =\frac { O }{ H } =\frac { 1 }{ 2 } \\ \\ \sin { \left( { 60 } \right) } =\frac { O }{ H } =\frac { \sqrt { 3 } }{ 2 } \\ \\ \cos { \left( { 30 } \right) } =\frac { A }{ H } =\frac { \sqrt { 3 } }{ 2 } \\ \\ \cos { \left( { 60 } \right) } =\frac { A }{ H } =\frac { 1 }{ 2 }$

#### sin(45°) and cos(45°): 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: $\sin { \left( 45 \right) =\frac { O }{ H } } =\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \cdot \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 2 } }{ 2 } \\ \\ \cos { \left( 45 \right) =\frac { A }{ H } } =\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \cdot \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 2 } }{ 2 }$

#### sin(15°), sin(75°), cos(15°) and cos(75°):

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… $\sin { \left( 15 \right) } \\ \\ =\sin { \left( 45-30 \right) } \\ \\ =\sin { \left( 45 \right) \cos { \left( 30 \right) -\cos { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } -\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 }$ $\sin { \left( 75 \right) } \\ \\ =\sin { \left( 45+30 \right) } \\ \\ =\sin { \left( 45 \right) \cos { \left( 30 \right) +\cos { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } +\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 }$ $\cos { \left( 15 \right) } \\ \\ =\cos { \left( 45-30 \right) } \\ \\ =\cos { \left( 45 \right) \cos { \left( 30 \right) +\sin { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } +\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 }$ $\cos { \left( 75 \right) } \\ \\ =\cos { \left( 45+30 \right) } \\ \\ =\cos { \left( 45 \right) \cos { \left( 30 \right) -\sin { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } -\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 }$

#### sin(0°), sin(90°), cos(0°) and cos(90°):

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: $\sin { \left( 0 \right) } =0\\ \\ \sin { \left( 90 \right) } =1\\ \\ \cos { \left( 0 \right) =1 } \\ \\ \cos { \left( 90 \right) } =0$

These values are fairly easy to find.

#### tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°):

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: $\tan { \left( \theta \right) } =\frac { \sin { \left( \theta \right) } }{ \cos { \left( \theta \right) } }$

Therefore: $\tan { \left( 0 \right) } =\frac { \sin { \left( 0 \right) } }{ \cos { \left( 0 \right) } } =\frac { 0 }{ 1 } =0$ $\tan { \left( 15 \right) } \\ \\ =\frac { \sin { \left( 15 \right) } }{ \cos { \left( 15 \right) } } \\ \\ =\frac { \frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 } }{ \frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } } \\ \\ =\frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 4 } \cdot \frac { 4 }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \\ \\ =\frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \cdot \frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ \left( \sqrt { 6 } -\sqrt { 2 } \right) } \\ \\ =\frac { 6-\sqrt { 12 } -\sqrt { 12 } +2 }{ 6-\sqrt { 12 } +\sqrt { 12 } -2 } \\ \\ =\frac { 8-2\sqrt { 12 } }{ 4 } \\ \\ =\frac { 8-2\sqrt { 4 } \sqrt { 3 } }{ 4 } \\ \\ =\frac { 8-4\sqrt { 3 } }{ 4 } \\ \\ =\frac { 4\left( 2-\sqrt { 3 } \right) }{ 4 } \\ \\ =2-\sqrt { 3 }$ $\tan { \left( 30 \right) } \\ \\ =\frac { \sin { \left( 30 \right) } }{ \cos { \left( 30 \right) } } \\ \\ =\frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \\ \\ =\frac { 1 }{ 2 } \cdot \frac { 2 }{ \sqrt { 3 } } \\ \\ =\frac { 1 }{ \sqrt { 3 } } \\ \\ =\frac { 1 }{ \sqrt { 3 } } \cdot \frac { \sqrt { 3 } }{ \sqrt { 3 } } \\ \\ =\frac { \sqrt { 3 } }{ 3 }$ $\tan { \left( 45 \right) } \\ \\ =\frac { \sin { \left( 45 \right) } }{ \cos { \left( 45 \right) } } \\ \\ =\frac { \frac { \sqrt { 2 } }{ 2 } }{ \frac { \sqrt { 2 } }{ 2 } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 2 }{ \sqrt { 2 } } \\ \\ =1$ $\tan { \left( 60 \right) } \\ \\ =\frac { \sin { \left( 60 \right) } }{ \cos { \left( 60 \right) } } \\ \\ =\frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 1 }{ 2 } } \\ \\ =\frac { \sqrt { 3 } }{ 2 } \cdot \frac { 2 }{ 1 } \\ \\ =\sqrt { 3 }$ $\tan { \left( 75 \right) } \\ \\ =\frac { \sin { \left( 75 \right) } }{ \cos { \left( 75 \right) } } \\ \\ =\frac { \frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } }{ \frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 } } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } \cdot \frac { 4 }{ \sqrt { 6 } -\sqrt { 2 } } \\ \\ =\frac { \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ \left( \sqrt { 6 } -\sqrt { 2 } \right) } \cdot \frac { \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \\ \\ =\frac { 6+\sqrt { 12 } +\sqrt { 12 } +2 }{ 6+\sqrt { 12 } -\sqrt { 12 } -2 } \\ \\ =\frac { 8+2\sqrt { 12 } }{ 4 } \\ \\ =\frac { 8+2\sqrt { 4 } \sqrt { 3 } }{ 4 } \\ \\ =\frac { 8+4\sqrt { 3 } }{ 4 } \\ \\ =\frac { 4\left( 2+\sqrt { 3 } \right) }{ 4 } \\ \\ =2+\sqrt { 3 }$ $\tan { \left( 90 \right) } \\ \\ =\frac { \sin { \left( 90 \right) } }{ \cos { \left( 90 \right) } } \\ \\ =\frac { 1 }{ 0 } \\ \\ =undefined$

And now, the set of derivations is complete. 😀

# Useful trigonometric formulas for finding areas of circles

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: $\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\cos ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }$

Second formulas: $\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) }$

These formulas are to be used when you have to transform the expression: ${ y }=\sqrt { { r }^{ 2 }-{ x }^{ 2 } }$

You can either make: $x=r\sin { \theta }$

Or… $x=r\cos { \theta }$

The choice is yours. 🙂

# tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Prove that: $tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

Firstly: $tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta }$

Also, remember that: $tan\left( -\theta \right) =-tan\theta$

So: $\beta =-\phi \\ \\ tan\left( \alpha +\left( -\phi \right) \right) =\frac { tan\alpha +tan\left( -\phi \right) }{ 1-tan\alpha tan\left( -\phi \right) } \\ \\ =\frac { tan\alpha -tan\phi }{ 1+tan\alpha tan\phi } =tan\left( \alpha -\phi \right) \\ \\ \therefore \quad tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

# sin(A-B)=sinAcosB-cosAsinB

Prove that: $sin(A-B)=sinAcosB-cosAsinB$ $sin\left( \alpha +\beta \right) =sin\alpha cos\beta +cos\alpha sin\beta \\ \\ But:\quad \\ \\ sin(-\theta )=-sin\theta \\ cos(-\theta )=cos\theta \\ tan(-\theta )=-tan\theta \\ \\ Say:\quad \beta =-\phi \\ \\ \therefore \quad sin\left( \alpha +\left( -\phi \right) \right) =sin\alpha cos\left( -\phi \right) +cos\alpha sin\left( -\phi \right) \\ \\ =sin\alpha cos\phi -cos\alpha sin\phi =sin\left( \alpha -\phi \right) \\ \\ If\quad \alpha =A\quad and\quad \phi =B\\ \\ sin\left( A-B \right) =sinAcosB-cosAsinB.$

# tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Prove that: $tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB }$ $LHS=tan(A+B)=\frac { sin(A+B) }{ cos(A+B) } \\ \\ =\frac { sinAcosB+cosAsinB }{ cosAcosB-sinAsinB } =\frac { \frac { sinA }{ 1 } \cdot \frac { sinB }{ tanB } +\frac { sinA }{ tanA } \cdot \frac { sinB }{ 1 } }{ \frac { sinA }{ tanA } \cdot \frac { sinB }{ tanB } -\frac { sinAsinB }{ 1 } } \\ \\ =\frac { \frac { sinAsinB }{ tanB } +\frac { sinAsinB }{ tanA } }{ \frac { sinAsinB }{ tanAtanB } -\frac { sinAsinB }{ 1 } } =\frac { sinAsinB\left( \frac { 1 }{ tanB } +\frac { 1 }{ tanA } \right) }{ sinAsinB\left( \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } \right) } \\ \\ =\frac { \frac { 1 }{ tanB } +\frac { 1 }{ tanA } }{ \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } } =\frac { \frac { tanA+tanB }{ tanAtanB } }{ \frac { 1-tanAtanB }{ tanAtanB } } =\frac { tanA+tanB }{ tanAtanB } \cdot \frac { tanAtanB }{ 1-tanAtanB } \\ \\ =\frac { tanA+tanB }{ 1-tanAtanB } =RHS$