Pythagorean Theorem

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February 25th 2010, 04:14 AM
mark090480

Pythagorean Theorem

Cos^2 + Sin^2 = 1

I know that this is just the Pythagorean Theorem, so here we are saying that the radius of the circle is 1? As cos = x ( base) and sin = y ( height ), what happens if the circle's radius is 5. Does this apply only to the unit circle? How can the hypotenuse always be 1?

(x-1)^2 + ( y+2)^2 = 3^2

So again it applies but for a circle at x= 1 and y= -2. So, how does that equation equal 9? Isn't that 1+4=3^2. Please help!
February 25th 2010, 04:37 AM
Archie Meade

Quote:

Originally Posted by mark090480

Cos^2 + Sin^2 = 1

I know that this is just the Pythagorean Theorem, so here we are saying that the radius of the circle is 1? As cos = x ( base) and sin = y ( height ), what happens if the circle's radius is 5. Does this apply only to the unit circle? How can the hypotenuse always be 1?

(x-1)^2 + ( y+2)^2 = 3^2

So again it applies but for a circle at x= 1 and y= -2. So, how does that equation equal 9? Isn't that 1+4=3^2. Please help!

hi mark090480,

The circle centre is (1,-2).

But that locates only the point at which you can place the tip of your compass.
You can now draw a circle any size you want.
The one in question has a radius of 9.

Remember, the distance formula is also Pythagoras' Theorem.
A circle is the set of all points the same distance from a fixed one.

$distance=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

$(distance)^2=(x-x_1)^2+(y-y_1)^2$

If you take any right angled triangle

$SinA=\frac{opp}{hyp},\ CosA=\frac{adj}{hyp}$

$Sin^2A+Cos^2A=\frac{opp^2+adj^2}{hyp^2}=\frac{hyp^ 2}{hyp^2}=1$
February 25th 2010, 04:50 AM
mark090480

"The one in question has a radius of 9."

Actually that equation says 0^2 + 0^2 = 9. I can draw any size but where does that 9 come from. In this equation? So, this equation should not be taken as an equation?
February 25th 2010, 05:13 AM
Archie Meade

No, the centre is the fixed point (1,-2).

The circle equation is, from the distance formula, or Pythagoras' Theorem..

$(x-x_c)^2+(y-y_c)^2=R^2=3^2$

The (x,y) values are the co-ordinates of every point on the circle circumference.

$(x-1)^2+(y-[-2])^2=9,\ for\ (x_1,y_1)=(x_c,y_c)=(1,-2)=centre$
February 25th 2010, 09:47 AM
mark090480

OK thanks a lot; it is clear now. So, you cannot work out the radius of that circle unless you know at least one coordinate. That is what confused me.
February 25th 2010, 10:23 AM
Archie Meade

1 Attachment(s)

Quote:

Originally Posted by mark090480

OK thanks a lot; it is clear now. So, you cannot work out the radius of that circle unless you know at least on coordinate. That is what confused me.

Hi mark,

well, it really depends on the information given.

You are in fact given both the centre and radius, in this case...

$(x-1)^2+(y+2)^2=3^2$

When dealing with co-ordinates, we rewrite Pythagoras' Theorem
to reflect the fact that the right-angled triangle side lengths are obtained
from the difference between x on the circle and the x co-ordinate of the centre, also the difference between y on the circle circumference and y at the centre.

To give you a picture of this, check the attachment.

The circle centre co-ordinates are the "negatives" of the values in brackets.
If we are given the circle equation as

$(x+g)^2+(y+h)^2=k^2$

then the centre is (-g,-h) and the radius is k.

In the attachment

$a^2+b^2=r^2$

$a=x-x_c,\ b=y-y_c$

$(x-x_c)^2+(y-y_c)^2=r^2$

which is the same as the distance formula $r=\sqrt{(x-x_c)^2+(y-y_c)^2}$
February 25th 2010, 09:34 PM
mark090480

Hi and thanks for the help,

Just one more thing ;) Still not sure why cos^2 + sin^2 must equal one. I understand that sin = opp/hyp etc.. But why can't it equal, say 1.3? What is it about those ratios that mean it must equal 1?
February 26th 2010, 02:32 AM
Archie Meade

Quote:

Originally Posted by mark090480

Hi and thanks for the help,

Just one more thing ;) Still not sure why cos^2 + sin^2 must equal one. I understand that sin = opp/hyp etc.. But why can't it equal, say 1.3? What is it about those ratios that mean it must equal 1?

Hi mark,

this is because, although the side lengths of a right-angled triangle
can be much greater then 1, the hypotenuse is always the longest side,
so the ratios Sin(angle) and Cos(angle) are somewhere between 0 and 1.

$SinA=\frac{y}{z},\ y \le z$

$CosA=\frac{x}{z},\ x \le z$

Pythagoras' Theorem then shows that the square of the two perpendicular sides equals the square of the hypotenuse.

If you write the opposite and adjacent using Sin(angle) and Cos(angle)
we get

$x^2+y^2=z^2$

where z is the hypotenuse, x is the horizontal side, the "adjacent" to angle A,
y is the vertical side, the "opposite" to angle A.

Therefore $SinA=\frac{y}{z}\ \Rightarrow\ y=zSinA$

$CosA=\frac{x}{z}\ \Rightarrow\ x=zCosA$

Then $x^2+y^2=z^2$ may be written $(zCosA)^2+(zSinA)^2=z^2$

$z^2Cos^2A+z^2Sin^2A=z^2$

$z^2\left(Cos^2A+Sin^2A\right)=z^2$

This means that $Cos^2A+Sin^2A=1$

The same can be shown for the circle as it's only a slight variation.
February 26th 2010, 09:45 AM
mark090480

Hi, thanks for all this help, I follow it all.

Are there any explanations as to why the Pythagorean Theorem works? I know why it works and it is easy to prove by drawing squares on the outside of the triangle. When using graph paper just count the squares. Are there math reasons as to why a right angle triangle would do that?
February 26th 2010, 10:16 AM
Archie Meade

1 Attachment(s)

Hi Mark,

Drawing squares on the 3 sides needs further geometry to be classed
as a proof.
That is more of an "illustration".

There are 2 ways that I use to prove Pythagoras' theorem.

One involves using semicircles inside a square whose side length
is the hypotenuse.

A simple way is to split the right angled triangle into 2 right-angled triangles.

In the attachment, if we draw the blue line perpendicular to the hypotenuse,
then, since A+B=90 degrees, E=B, F=A

Carefully stand all three triangles so that angle A is at the lower left and the right-angle is at the lower right.

Then we have 3 similar triangles (magnifications and reductions of the same basic triangle).

We want to examine what $x^2+y^2$ is, so we use ratios

with x and y in the numerator and denominator positions.

Hence $\frac{z}{x}=\frac{x}{z_1}$

and $\frac{z}{y}=\frac{y}{z_2}$

Therefore $x^2=(z)z_1,\ y^2=(z)z_2$

$x^2+y^2=z(z_1+z_2)$

but $z=z_1+z_2$

hence $x^2+y^2=z^2$
February 26th 2010, 10:32 AM
mark090480

Hi,

I will work through that and see if I get it. My main question was more to do with that 90 degrees. So it works for 90 degrees but not with 89.9 or 90.1, why is that? Maybe you just answered it, if so ignore this. So there seems to be some magic with 90 degrees that lets Base^2 + Height^2 = Hyp^2

Thanks again, so the internet can be useful ;)
February 26th 2010, 10:44 AM
Archie Meade

Quote:

Originally Posted by mark090480

Hi,

I will work through that and see if I get it. My main question was more to do with that 90 degrees. So it works for 90 degrees but not with 89.9 or 90.1, why is that? Maybe you just answered it, if so ignore this. So there seems to be some magic with 90 degrees that lets Base^2 + Height^2 = Hyp^2

Thanks again, so the internet can be useful ;)

Ah yes Mark!!

absolutely!!

The triangle must be right-angled.
One of the 3 angles must be 90 degrees.

Once this is clear,
you will notice that "workarounds" are used to solve non-right angled triangles.

These are typically the "triangle area" calculation,
the Sine Rule (Law of Sines) and the Cosine Rule (Law of Cosines).

You need to understand how Pythagoras' Theorem works (sides only)
along with Sine, Cosine and Tangent (Sin, Cos, Tan) for sides and angles
for the right-angled triangle.

Then you move on from there to discover how the Sine and Cosine Rules
solve other triangles and then move onto circles and more complex geometry and trigonometry.
February 26th 2010, 10:57 AM
mark090480

"The triangle must be right-angled."

Why does 90 degrees hold such magic? Glad it does, this seems to pop up everywhere.

But why? I have looked for the magic trick to be explained but don't know what it is. Not even my math books say. They just say how things work not why. I can use the equation without problems, even with circles but I don't know why all this stuff works.
February 26th 2010, 11:10 AM
Archie Meade

The 90 degree angle is quite convenient as a starting point,
as we can resolve a geometric situation into
up-down and left-right measurements.

We can write the ratio of the sides of similar triangles.
Corresponding ratios will always be the same
if the triangles have the same 3 angles.

Unfortunately, angle measure is continuous,
so there would be too many ratios to deal with with.

The side ratios of a right-angled triangle have been named Sine, Cosine and Tangent.

We then simplify other triangles using those as a basis.

Otherwise calculations would be "primitive".

Sine, Cosine and Tangent are ratios defined in the right-angled triangle.
Those are the names given to the 3 ratios of the sides of a right-angled triangle.

You will get used to it all as you study.
Your questions are teasing out why all this works, for your own understanding.

Pythagoras' theorem would work even if Pythagoras himself hadn't discovered it.
To understand my proof of Pythagoras theorem,
you must first understand that if something stays the same shape
and it has angles, then those angles stay the same
and the side ratios stay the same if you magnify or reduce the shape.

that is why Sin, Cos, Tan have been defined as such
and also why Pythagoras theorem is a mathematical law.

It's all based on....

Same shape, same angles, same ratios irrespective of size.