# 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. đź™‚

# 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 Î±Â and Î˛

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 bottomÂ halves…

• There exists top halvesÂ Î±+Î˛ and also…Â Î˛+y
• There exists bottom halves x+y but alsoÂ Î±+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 needÂ these equations…

Î±+Î˛ andÂ Î±+x are equivalent (180 degrees), so we can deduce that:

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

*SubtractÂ Î± from both sides of the equation.

Î±+Î˛Â and Î˛+y are equivalent (180 degrees), so we can deduce that:

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

*SubtractÂ Î˛ from both sides of the equation.

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

Â Î˛=x andÂ Î±=y:

# How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

If you look at this diagram carefully, what you will notice is:

• A circle exists and each point on this circle has the coordinate (x, y).
• The centre of the circle can be found at (a, b).
• The circle has a radius ‘r’.
• The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression:

$\left( x-a \right)Â$ or $\left| x-a \right|$

The opposite lengths can be described using the expression:

$\left( y-b \right)Â$ or $\left| y-b \right|Â$

Also, very interestingly:

• Each of the right angled triangles you can think of has a hypotenuse ‘r’.
• ${ \left( x-a \right) Â }^{ 2 }={ \left| x-a \right| Â }^{ 2 }$
• ${ \left( y-b \right) Â }^{ 2 }={ \left| y-b \right| Â }^{ 2 }$

When you combine all the information above, what you get is a neat formula which looks like this:

${ \left( x-a \right) Â }^{ 2 }+{ \left( y-b \right) Â }^{ 2 }={ r }^{ 2 }$

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? đź™‚