# 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. 🙂

# 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 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.

# 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)

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation

# How to differentiate y=arccosx

Below I’ll be demonstrating how to differentiate y=arccosx using implicit differentiation…

$y=\arccos { x } \\ \\ \cos { y=x } \\ \\ -\sin { y\cdot \frac { dy }{ dx } } =1\\ \\ \frac { dy }{ dx } =-\frac { 1 }{ \sin { y } }$

But…

$\sin ^{ 2 }{ y+\cos ^{ 2 }{ y=1 } } \\ \\ \sin ^{ 2 }{ y } =1-\cos ^{ 2 }{ y } \\ \\ \sin { y=\sqrt { 1-\cos ^{ 2 }{ y } } }$

Therefore…

$\frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-\cos ^{ 2 }{ y } } } \\ \\ \therefore \quad \frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

# How to differentiate y=arcsinx

Below I’ll be demonstrating how to differentiate y=arcsinx using implicit differentiation…

$y=\arcsin { x } \\ \\ \sin { y } =x\\ \\ \cos { y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \cos { y } }$

But…

$\sin ^{ 2 }{ y+\cos ^{ 2 }{ y } } =1\\ \\ \cos ^{ 2 }{ y=1-\sin ^{ 2 }{ y } } \\ \\ \cos { y=\sqrt { 1-\sin ^{ 2 }{ y } } }$

Therefore:

$\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ y } } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

# How to differentiate y=arctanx

Below I’m going to demonstrate how to integrate y=arctanx…

Firstly, we need to know that:

$\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1\\ \\ \frac { \sin ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } +\frac { \cos ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } =\frac { 1 }{ \cos ^{ 2 }{ y } } \\ \\ \tan ^{ 2 }{ y } +1=\sec ^{ 2 }{ y }$

We also need to know that:

$x=\tan { y } \\ \\ x=\frac { \sin { y } }{ \cos { y } } \\ \\ x\cdot \cos { y } =\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } +x\cdot \left( -\sin { y } \right) =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } -x\sin { y } =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +x\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ \frac { 1 }{ \cos { y } } \cdot \frac { dx }{ dy } \cdot \cos { y } =\frac { 1 }{ \cos { y } } \left( \cos { y+\frac { \sin ^{ 2 }{ y } }{ \cos { y } } } \right) \\ \\ \frac { dx }{ dy } =1+\tan ^{ 2 }{ y } \\ \\ \therefore \quad \frac { dx }{ dy } =\sec ^{ 2 }{ y }$

And finally:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Now, using implicit differentiation:

$y=\arctan { x } \\ \\ \tan { y } =x\\ \\ \sec ^{ 2 }{ y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \sec ^{ 2 }{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \tan ^{ 2 }{ y+1 } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ { x }^{ 2 }+1 }$