In this note I want to start a prove chain. This means if you want to continue this chain you should prove something, but you must use the result of the previous approach. For example if the first person proved that $a\cdot b=2A$ in a right triangle and you want to prove that the area of a square is $a^2$, then this is a good approach:
If we cut the square into two right triangles: $A_s=2\cdot A_t;a\cdot b=2A_t;a=b\Rightarrow A_s=2\cfrac{a\cdot a}{2}=a^2$
This can be just a fun note. If you are boring or you want to rest for a few minute, then this is a chance to prove something what you want(and you can). You can prove basic, easy or hard things.
Rules:
If you want to share something, what isn't continue this chain, then use the reply button under one of the approaches
Don't share messages to your friends
Don't prove something again in this note
If your approach is wrong, then don't delete your comment. Mention somebody to help you or solve alone
To avoid two approaches at the same time you should reply for the previous approach(use the given text) before you continue this chain
Don't prove evil things what others can't continue
$\color{#eeaa00}\text{Temporary rule(s)}$
$\color{#eeaa00}\text{The first ten approaches should not create a chain}$
$\color{#eeaa00}\text{You can share not original, but not famous approches}$
$\color{#D61F06}\text{Current theme(s)}$
$\color{#D61F06}\text{Number theory}$
$\color{#D61F06}\text{Algebra}$
$\color{#3D99F6}\text{Event(s)}$
$\color{#3D99F6}\text{There has been no events yet(after 10 active members will be the first event)}$
I will change the colored sections sometimes. If you don't want to skip them, then you can subscribe above the comments or if you write an approach, then you will get notifications too. When I change the temporary rule(s) I will notificate you.
$\color{#D61F06}\text{You can also write proofs that do not belong to the given theme(s)}$
You can:
Use things that have already been proven before you(you should name which one is that)
Mention people to join
Mention somebody to help you if your approach is wrong
Mention somebody to help you continue this chain if you wrote an approach, but you are bad in $\LaTeX$
You can prove evil things, but then you must continue the chain and prove something easier to help others
If this is possible you can split up your approach into two approaches
How to:
Mention somebody:
@<name>
Example: @Páll Márton
Write an approach:
Reply for the last approach: I want to continue this chain. If somebody already working on this you should wait.
Write your approach. Use this form below:
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# ID: <serial number of this approach>
## <write a short review about your approach or write the formula>
## <serial number of used formulas>
<write here your approach>
> <P.S. or notes>
When you write an approach don't use the reply button. Just write as a comment.
Use pictures:
The code of your picture will be: ![<text under the picture>](<link to the picture>){: .<position>}
Profile->+Note(in the bottom)->Upload an image
Copy the code of the picture
Paste/Insert your code
If your picture is uploaded to the internet, then you should just paste the link.
The [] section can be empty
The <position> can be center,left,right.
Use $\LaTeX$:
Without spaces:
\ ( <LaTeX code> \ ) - left aligned
\ [ <LaTeX code> \ ] - center aligned
If you are bad in $\LaTeX$, then you can use online editors or web pages to learn.
Use other formatting:
Here you can see how to use links, use bold and italistic styles, use quotes etc.
The code of the first comment is(without spaces between \ and ( or ) ):
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# ID: 1
## Graphically prove that \ (a^2-b^2=(a-b)(a+b)\ )
## I won't use other formulas from this note
Assume that \ (a>b\ ). In this picture below \ (T_?\ ) means territory/area:
![](https://ds055uzetaobb.cloudfront.net/uploads/ZmvJbP3yh4-0.PNG){: .center}
In this picture \ (T_1+T_3+T_4=a^2\ ) and \ (T_4=b^2\implies T_1+T_3=a^2-b^2\ ). We know \ (T_1+T_2=(a+b)(a-b)\ ) from the side length of the rectangle. And we know \ (T_2=T_3\ ), because both of them is equal to \ (b(a-b)\ ). Therefore \ (T_1+T_3=T_1+T_2=(a+b)(a-b)\ ). So \ (a^2-b^2=(a-b)(a+b)\ ).
> P.S.: If \ (a=b\ ) then \ (a^2-b^2=(a-b)(a+b)=0\ ). And if \ (a<b\ ) then, then we can multiply by -1. After that we get this: \ (b^2-a^2=(b-a)(a+b)\ ). If we swap a and b in the picture we get the same result.
Good luck!
Active members:
If you post an approach, I will give you 5 points, except if your solution is wrong, not complete or confusing.
If you post an approach about the current theme, then you can get 3 extra points.
If you write something interesting about the approach (for example ID-3 by David Vreken), then you can get 1 extra points.
Name
Posted approaches
Points
David Vreken
1
6
Páll Márton
5
Vinayak Srivastava
1
5
If we reach the 10 active members, then I will mark the 3 most active people. And I will mark the winner of the last event too.
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Assume that $a>b$. In this picture below $T_?$ means territory/area:
In this picture $T_1+T_3+T_4=a^2$ and $T_4=b^2\implies T_1+T_3=a^2-b^2$. We know $T_1+T_2=(a+b)(a-b)$ from the side length of the rectangle. And we know $T_2=T_3$, because both of them is equal to $b(a-b)$. Therefore $T_1+T_3=T_1+T_2=(a+b)(a-b)$. So $a^2-b^2=(a-b)(a+b)$.
P.S.: If $a=b$ then $a^2-b^2=(a-b)(a+b)=0$. And if $a<b$ then, then we can multiply by -1. After that we get this: $b^2-a^2=(b-a)(a+b)$. If we swap a and b in the picture we get the same result.
We are usually taught that the proof of the identity ${\sin}^2{\theta}+{\cos}^2{\theta} =1$ is this:
In any right triangle,
$O^2+A^2=H^2 \text{ (Where H is the hypotenuse, O is opposite, and A is adjacent.)}$$\implies \dfrac{O^2}{H^2}+ \dfrac{A^2}{H^2} =1$$\implies {\sin}^2{\theta}+{\cos}^2{\theta} =1$
However this proof is not complete! What about an angle ${\theta} >90^{\circ}$? It won't form any right triangle!
Here is where our unit circle helps:
In our unit circle, for any angle $360^{\circ}\geq {\theta}\geq 0^{\circ},$ we have the particular point on its circumference representing $\cos{\theta},\sin{\theta}$, as its co-ordinates, I have shown in my diagram:
But we know that any point on a unit circle satisfies the equation $x^2+y^2=1$
Hence, for any angle $360^{\circ}\geq {\theta}\geq 0^{\circ},$ we get
$\Huge{{\sin}^2{\theta}+{\cos}^2{\theta} =1 \blacksquare}$
Now I can't post any approaches with pictures. Can you post an approach about $\sin(90^{\circ}-\alpha)=\cos(\alpha)$? Same thing with cos and tangent, and with 180. I think you can make amazing pictures with unit circles :) I will use them in my approach, but I think we can't skip them.
James A. Garfield's Proof of the Pythagorean Theorem (reproduced from here)
I won't use other formulas from this note
Start with a right triangle with legs $a$ and $b$ and hypotenuse $c$. Extend leg $a$ by $b$ units and construct a duplicate right triangle along this extension. Upper leg $a$ is parallel to the original leg $b$ since, in a plane, if a line is perpendicular to each of two lines, then the two lines are parallel.
Draw segment $XY$ to close the figure. The resulting quadrilateral is a trapezoid with bases $a$ and $b$ and altitude $a+b$.
The trapezoid is composed of two congruent right triangles and right $\triangle XYZ$. $\triangle XYZ$ is isosceles since two of its sides have length $c$. $\angle XZY$ is a right angle since $\angle 1 + \angle XZY + \angle 2 = 180°$ and $\angle 1 + \angle 2 = 90°$. (The acute angles of a right triangle are complementary.) Therefore, the area of the trapezoid equals the sum of the areas of the three right triangles of which it is composed, two of which are congruent, so that:
Yeah. In America the presidents are smarter than in Europe. Ferenc Gyurcsány said 11 is the biggest 2-digit prime number :) But the basic approach should be $c^2=a^2+b^2-2ab cos(c);cos(90^{\circ})=0$
Typically the approach to proving the law of cosines is to use the Pythagorean Theorem, so if you use the law of cosines to prove the Pythagorean Theorem you would have circular reasoning. (Now this almost becomes a philosophical conversation, because it all depends on which premises you want to assume are true.)
Wait, Hungary is in Europe! Ugh..............I need to work on my Geography, I thought Hungary was in China or something. I hate my Geo teacher, I think she's a dracanae, trying to kill me.
@Páll Márton (no activity)
–
Most of my teacher's(Gym coach is not, coz he likes how much I can swim and my stamina) are monsters, that's why I'm not good at any subject, and that isn't an excuse for my horrible math and science skills(all right it is an excuse, I don't like studying in english letters, they get jumbled up coz of my dyslexia)
@Percy Jackson
–
Hmmm. My teacher who "teached" me sports was released from sports :) BTW he teached me Ukranian language three years long, but he don't have degree to teach them
@Páll Márton, @David Vreken, @Vinayak Srivastava - Garfield(the mingy, but hilarious cat) was a president of the United States, am I missing something here? And how do you know that Garfield's first name is James?
Yeah. I hate it. My math teacher always use this. Once she allowed me to prove one formula on her lesson, so I started to write a long chain(hence the idea) and I proved this formula
I will go to my grandmother, so I won't post approaches in the next couple of days with pictures. Now I will upload an image, to my next approach :) And I will insert it.
$\sin(\alpha\pm360^{\circ}\cdot k)=\sin(\alpha)$ and $\cos(\alpha\pm360^{\circ}\cdot k)=\cos(\alpha)$, where $k$ is an integer
$\sin(180^{\circ}-\alpha)=\sin(\alpha)$ and $\cos(180^{\circ}-\alpha)=-\cos(\alpha)$
$\sin(90^{\circ}-\alpha)=\cos(\alpha)$ and $\cos(90^{\circ}-\alpha)=\sin(\alpha)$
$\tan(\alpha)=\cfrac{\sin(\alpha)}{\cos(\alpha)}$
I won't use other formulas from there
In this picture we can see the $\sin(x)$ and $\cos(x)$ functions. We should calculate in radian. This means:
$2\pi=360^{\circ}$, $\pi=180^{\circ}$, $\cfrac{\pi}{2}=90^{\circ}$
We can see if we shift the functions by $2\pi$, then we get the same functions. Same way if we shift these functions by $4\pi,6\pi,8\pi,10\pi$, then we get the same functions.
From the unit circle:
$\sin(180^{\circ}-\alpha)=\sin(\alpha)$ and $\cos(180^{\circ}-\alpha)=-\cos(\alpha)$
If $\alpha>180^{\circ}$, then you should mirror the image, and you will get the same result.
From interpretation of $\sin,\cos$: $\sin(\alpha)=\cfrac{a}{c},\cos(\beta)=\cfrac{a}{c},\beta=90^{\circ}-\alpha\implies\cos(90^{\circ}-\alpha)=\sin(\alpha)$. Same way $\sin(90^{\circ}-\alpha)=\cos(\alpha)$ .
From similarity: $\cfrac{\sin(x)}{\cos(x)}=\cfrac{\tan(x)}{1}$
a) In this picture AC=BC=x and BD=z
From ID-5:
$\begin{aligned}
AD^2&=x^2+(x-z)^2-2x(x-z)\cdot\cos(\alpha)\\
AD^2&=z^2+AB^2-2z\cdot AB\cdot\cos(\beta)\\
AB^2&=x^2+x^2-2x\cdot x \cdot \cos(\alpha)=x^2(2-2\cos(\alpha))
\end{aligned}$
The left side of the first and the second equations are the same, and we can substitute the third equation to the second, so we can get this long equation:
$x^2+(x-z)^2-2x(x-z)\cos(\alpha)=x^2+x^2-2x^2\cdot\cos(\alpha)+z^2-2zx\sqrt{2-2\cos(\alpha)}\cdot\cos(\beta)$$\blue{x^2+x^2+z^2}-2xz+\blue{2x^2\cdot\cos(\alpha)}+2xz\cdot\cos(\alpha)=\blue{x^2+x^2-2x^2\cdot\cos(\alpha)+z^2}-2zx\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta)$
We can simplify that. We can substract from both sides the blue section and after that we can divide both sides by $-2xz$:
$1-\cos(\alpha)=\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta)$$\begin{aligned}
\cos(\beta)&=\cfrac{1-\cos(\alpha)}{\sqrt{2-2\cos(\alpha)}}\\
&=\cfrac{1-\cos(\alpha)}{\sqrt{2}\sqrt{1-1\cos(\alpha)}}\\
&=\cfrac{\sqrt{1-\cos(\alpha)}}{\sqrt{2}}\\
&= \sqrt{\cfrac{1-\cos(\alpha)}{2}}
\end{aligned}$
We know these things:
b) From ID-2:
$\sin^2(\alpha)+\cos^2(\alpha)=1$
From ID-7a:
$\begin{aligned}
\cfrac{1-\cos(\cfrac{\alpha}{2})}{2}+cos^2(\alpha)&=1\\
1- \cos(\cfrac{\alpha}{2})+2\cos^2(\alpha)&=2\\
\cos^2(\alpha)&=\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}\\
\cos(\alpha)&=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}}
\end{aligned}$
c) From ID-7a and ID-7b:
$\begin{aligned}
\tan(\cfrac{\alpha}{2})&=\cfrac{\sin(\cfrac{\alpha}{2})}{\cos(\cfrac{\alpha}{2})}\\
&=\cfrac{\sqrt{\cfrac{1-\cos(\alpha)}{\cancel{2}}}}{\sqrt{\cfrac{1+\cos(\alpha)}{\cancel{2}}}}\\
&=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}}
\end{aligned}$
d) $\cfrac{1}{\tan(\alpha)}=\cot(\alpha)$. So from ID-7c this is true.
e) From ID-7b:
$\begin{aligned}
\cos(2\alpha)&=2\cos^2(\alpha)-\blue{1}\\
&=2\cos^2(\alpha)-(\sin^2(\alpha)+\cos^2(\alpha))\\
&=\cos^2(\alpha)-\sin^2(\alpha)
\end{aligned}$
This is a proof by contradiction, so let's assume that $\sqrt{2}$ is rational.
If so, then it can be expressed in an equation as $\dfrac{p}{q}$ where $p$ and $q$ are co-primes - $\begin{aligned} \sqrt{2} &= \dfrac{p}{q} \\ 2 &= \dfrac{p^{2}}{q^{2}} \\ 2p^{2} &= q^{2} \end{aligned}$
By this simplification, we know that $q^{2}$ even, because it is $2$ multiplied by some other number. So $q$ must also be an even number. That means $q$ can be represented as $(2m)$.
Back to the equation, if we use the new value of $q$ and simplify further - $\begin{aligned} 2p^{2} &= q^{2} \\ 2p^{2} &= (2m)^{2} \\ 2p^{2} &= 4m^{2} \\ p^{2} &= 2m^{2} \end{aligned}$
As $p$ can also be represented like this, $p$ must also be even.
This creates a contradiction, as we had earlier said that $p$ and $q$ were co-primes, but now we say they are both even, which means they have a common divisor of $2$. Since this creates a contradiction, our original assumption must have been wrong...
$\text{Q.E.D - } \sqrt{2} \text{ is irrational!}$
This was originally proved by the Greek Mathematician and Father of Geometry, Euclid of Alexandria
@Páll Márton (no activity) - Thoughts? Also, do I have to wait till ID:10 to start chains?
Graphical Proof of Circle Area formula - $\pi r^{2}$
I will not use any formulas from other proofs
Let's cut a circle into $8$ equal sectors and arrange them as shown in the image below -
Image credits - https://medium.com
As you can see, it forms a figure that looks very close to a parallelogram that is going towards becoming a rectangle. The next two images show what happens when we take even smaller arcs.
In the last image, you see an figure that is almost a rectangle. TO get a perfect rectangle, we would need infinitesimally small sectors, but we can evaluate the area of the circle using this figure, as we know the formula for the area of a rectangle - $A = bh$
The height is easy to find, as it is just the radius. The base, if you look closely at the next image, is actually half of the circumference or half of $2 \pi r$ or $\pi r$
Image credits - https://medium.com
Thus, the formula for the area of a circle $\pi r \times r = \boxed{\pi r^{2}}$
@Nscs 747
–
K that's good! I have recently learned Aussie slang such as dead horse! Even though I am a migrant Australia who has stayed in the land of Oz for 7 years!!!!!
@Bithiah Koshy
–
for me i always get credit lol and for some reason i always get 81% for maths like always even two part exams part a and part b i get 81%
Proof that $2\cos^{-1} n+2 \sin^{-1} n=\pi$ in radians
Not using other identities
Construct an isosceles triangle $\triangle ABC$.
In isosceles$\triangle ABC$, $BC$ is the base. Construct the median $AM$ where $M$ is the midpoint of $BC$.
Since $\triangle ABC$ is isosceles, $\angle AMB$ is a right angle. $\Rightarrow \triangle AMB$ is a right angled triangle.
This part uses basic definitions of sin & cos.
Call the length of the hypotenuse of $Rt\triangle AMB$$k$.
Call the length of median $AM$$l$. $\Rightarrow \angle BAM=\cos^{-1} \frac{l}{k}.$ $\Rightarrow \angle BAC=2\angle BAM=2\cos^{-1} \frac{l}{k}.$
Since $\triangle AMB$ is right angled, and $\triangle ABC$ is isosceles, $\sin \angle B=\sin \angle C=\frac{l}{k}.$ $\Rightarrow \angle B=\angle C=\sin^{-1} \frac{l}{k}.$
Let$\frac{l}{k}=n.$.
Finally, since $\angle A,\angle B,\angle C$ are the interior angles of a triangle, i.e. they add up to $\pi (rad)$,
$2\cos^{-1} n+2 \sin^{-1} n=\pi$ in radians.
Easy Math Editor
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ID: 1
Graphically prove that $a^2-b^2=(a-b)(a+b)$
I won't use other formulas from this note
Assume that $a>b$. In this picture below $T_?$ means territory/area: In this picture $T_1+T_3+T_4=a^2$ and $T_4=b^2\implies T_1+T_3=a^2-b^2$. We know $T_1+T_2=(a+b)(a-b)$ from the side length of the rectangle. And we know $T_2=T_3$, because both of them is equal to $b(a-b)$. Therefore $T_1+T_3=T_1+T_2=(a+b)(a-b)$. So $a^2-b^2=(a-b)(a+b)$.Log in to reply
@Vinayak Srivastava What about this note?
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Good, I will try to find something good! Nice note!
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ID:2
Proof of ${\sin}^2{\theta}+{\cos}^2{\theta} =1$
I won't use other formulas from this note
We are usually taught that the proof of the identity ${\sin}^2{\theta}+{\cos}^2{\theta} =1$ is this:
However this proof is not complete! What about an angle ${\theta} >90^{\circ}$? It won't form any right triangle!
Here is where our unit circle helps:
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Or: $(sin(a))^2+(cos(a))^2=\cfrac{a^2+b^2}{c^2}=1$, where $a^2+b^2=c^2$ :)
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This is the common notation there
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Now I can't post any approaches with pictures. Can you post an approach about $\sin(90^{\circ}-\alpha)=\cos(\alpha)$? Same thing with cos and tangent, and with 180. I think you can make amazing pictures with unit circles :) I will use them in my approach, but I think we can't skip them.
Log in to reply
Actually, I don't know this formula! I'll search it, and if I understand, I will try.
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ID: 3
James A. Garfield's Proof of the Pythagorean Theorem (reproduced from here)
I won't use other formulas from this note
Start with a right triangle with legs $a$ and $b$ and hypotenuse $c$. Extend leg $a$ by $b$ units and construct a duplicate right triangle along this extension. Upper leg $a$ is parallel to the original leg $b$ since, in a plane, if a line is perpendicular to each of two lines, then the two lines are parallel. Draw segment $XY$ to close the figure. The resulting quadrilateral is a trapezoid with bases $a$ and $b$ and altitude $a+b$.
The trapezoid is composed of two congruent right triangles and right $\triangle XYZ$. $\triangle XYZ$ is isosceles since two of its sides have length $c$. $\angle XZY$ is a right angle since $\angle 1 + \angle XZY + \angle 2 = 180°$ and $\angle 1 + \angle 2 = 90°$. (The acute angles of a right triangle are complementary.) Therefore, the area of the trapezoid equals the sum of the areas of the three right triangles of which it is composed, two of which are congruent, so that:
$\begin{aligned} A_{VWXY} &= A_{\triangle ZVY} + A_{\triangle XWZ} + A_{\triangle XYZ} \\ \frac{1}{2}(a + b)(a + b) &= \frac{1}{2}ab + \frac{1}{2}ab + \frac{1}{2}c^2 \\ (a + b)(a + b) &= ab + ab + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2 \end{aligned}$
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Yeah. In America the presidents are smarter than in Europe. Ferenc Gyurcsány said 11 is the biggest 2-digit prime number :) But the basic approach should be $c^2=a^2+b^2-2ab cos(c);cos(90^{\circ})=0$
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Typically the approach to proving the law of cosines is to use the Pythagorean Theorem, so if you use the law of cosines to prove the Pythagorean Theorem you would have circular reasoning. (Now this almost becomes a philosophical conversation, because it all depends on which premises you want to assume are true.)
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By the way, if you put a slash in front of cos in latex, it won't be italicized (like this: "\cos")
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Wait, Hungary is in Europe! Ugh..............I need to work on my Geography, I thought Hungary was in China or something. I hate my Geo teacher, I think she's a dracanae, trying to kill me.
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@Páll Márton
Released from sports???Log in to reply
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Good proof Sir! I remember that I once saw a proof in which the triangle is copied 4 times to form a square.
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Yeah. And there are this formula: the length of the two sides of the parallelogram on the square is equal to the sum of the diagonals on the square.
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@Páll Márton, @David Vreken, @Vinayak Srivastava - Garfield(the mingy, but hilarious cat) was a president of the United States, am I missing something here? And how do you know that Garfield's first name is James?
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ID: 4
Prove that $2a^2+2b^2=e^2+f^2$, where $a$ and $b$ is the two sides of the parallelogram and $e$ and $f$ is the diagonals of it
I will use this formula from there: ID-3
From ID-3:The sum of the two last statement: $e^2+f^2=2a^2+\blue{2x^2+2m^2}$. From the first statement: $e^2+f^2=2a^2+2b^2$
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Where is point $F$?
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Ohhh. Wait a minute :)
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Nice proof! I had not heard of this theorem, but seems interesting!
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Yeah. I hate it. My math teacher always use this. Once she allowed me to prove one formula on her lesson, so I started to write a long chain(hence the idea) and I proved this formula
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I will post the next approach :)
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ID: 5
Prove that: $a^2=b^2+c^2-2bc\cdot cos(\alpha)$
I will use these formulas from there: ID-2, ID-3
Assume that $\angle A>\angle B>\angle C$ There are three cases:1.case $\angle A=90^{\circ}$:
From ID-3, this is true, because $cos(90^{\circ})=0$
2.case $\angle A<90^{\circ}$:
$\huge\text{Swap A and B}$ $AD=AB\cdot\;cos(\angle A)$ and $BD=AB\cdot\;sin(\angle A)$
From the 1. case: $\begin{aligned} BC^2&=BD^2+DC^2\\ &=BD^2+(AC-AD)^2\\ &=(AB\cdot sin(\angle A))^2+AC^2+(AB\cdot cos(\angle A))^2-2AC\cdot AB\cdot cos(\angle A)\\ &=AB^2(cos^2(\angle A)+sin^2(\angle A))+AC^2-2AC\cdot AB\cdot cos(\angle A) \end{aligned}$ From ID-2: $BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)$
3.case $\angle A>90^{\circ}$:
Same way: $AD=AB\cdot\;cos(\angle BAD)$ and $BD=AB\cdot\;sin(\angle BAD)$
From the 1. case: $\begin{aligned} BC^2&=BD^2+DC^2\\ &=BD^2+(AC+AD)^2\\ &=(AB\cdot sin(\angle BAD))^2+AC^2+(AB\cdot cos(\angle BAD))^2+2AC\cdot AB\cdot cos(\angle BAD)\\ &=AB^2(cos^2(\angle BAD)+sin^2(\angle BAD))+AC^2+2AC\cdot AB\cdot cos(\angle BAD) \end{aligned}$ From ID-2: $BC^2=AB^2+AC^2+2AC\cdot AB\cdot cos(\angle BAD)$, but $cos(180^{\circ}-\alpha)=-cos(\alpha)$. So $BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)$
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I will go to my grandmother, so I won't post approaches in the next couple of days with pictures. Now I will upload an image, to my next approach :) And I will insert it.
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And see the actual theme :)
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I will prove these things:
Sorry if there are typos!
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$cos(\alpha)=cos^2(\alpha /2)-sin^2(\alpha /2)$
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For case 2 I don't think your letters match with your picture (specifically $BC^2 = BD^2 + DC^2$)
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Sorry. Swap A and B :) I will edit it, when I can
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ID:6
Prove these formulas:
I won't use other formulas from there
In this picture we can see the $\sin(x)$ and $\cos(x)$ functions. We should calculate in radian. This means:From the unit circle:
If $\alpha>180^{\circ}$, then you should mirror the image, and you will get the same result.
From interpretation of $\sin,\cos$: $\sin(\alpha)=\cfrac{a}{c},\cos(\beta)=\cfrac{a}{c},\beta=90^{\circ}-\alpha\implies\cos(90^{\circ}-\alpha)=\sin(\alpha)$. Same way $\sin(90^{\circ}-\alpha)=\cos(\alpha)$ .
From similarity: $\cfrac{\sin(x)}{\cos(x)}=\cfrac{\tan(x)}{1}$
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ID:7
Prove these formulas:
I will use these formulas: ID-5, ID-6
a) In this picture AC=BC=x and BD=z From ID-5: $\begin{aligned} AD^2&=x^2+(x-z)^2-2x(x-z)\cdot\cos(\alpha)\\ AD^2&=z^2+AB^2-2z\cdot AB\cdot\cos(\beta)\\ AB^2&=x^2+x^2-2x\cdot x \cdot \cos(\alpha)=x^2(2-2\cos(\alpha)) \end{aligned}$ The left side of the first and the second equations are the same, and we can substitute the third equation to the second, so we can get this long equation: $x^2+(x-z)^2-2x(x-z)\cos(\alpha)=x^2+x^2-2x^2\cdot\cos(\alpha)+z^2-2zx\sqrt{2-2\cos(\alpha)}\cdot\cos(\beta)$ $\blue{x^2+x^2+z^2}-2xz+\blue{2x^2\cdot\cos(\alpha)}+2xz\cdot\cos(\alpha)=\blue{x^2+x^2-2x^2\cdot\cos(\alpha)+z^2}-2zx\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta)$ We can simplify that. We can substract from both sides the blue section and after that we can divide both sides by $-2xz$: $1-\cos(\alpha)=\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta)$ $\begin{aligned} \cos(\beta)&=\cfrac{1-\cos(\alpha)}{\sqrt{2-2\cos(\alpha)}}\\ &=\cfrac{1-\cos(\alpha)}{\sqrt{2}\sqrt{1-1\cos(\alpha)}}\\ &=\cfrac{\sqrt{1-\cos(\alpha)}}{\sqrt{2}}\\ &= \sqrt{\cfrac{1-\cos(\alpha)}{2}} \end{aligned}$ We know these things:
b) From ID-2: $\sin^2(\alpha)+\cos^2(\alpha)=1$ From ID-7a: $\begin{aligned} \cfrac{1-\cos(\cfrac{\alpha}{2})}{2}+cos^2(\alpha)&=1\\ 1- \cos(\cfrac{\alpha}{2})+2\cos^2(\alpha)&=2\\ \cos^2(\alpha)&=\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}\\ \cos(\alpha)&=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}} \end{aligned}$
c) From ID-7a and ID-7b: $\begin{aligned} \tan(\cfrac{\alpha}{2})&=\cfrac{\sin(\cfrac{\alpha}{2})}{\cos(\cfrac{\alpha}{2})}\\ &=\cfrac{\sqrt{\cfrac{1-\cos(\alpha)}{\cancel{2}}}}{\sqrt{\cfrac{1+\cos(\alpha)}{\cancel{2}}}}\\ &=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}} \end{aligned}$
d) $\cfrac{1}{\tan(\alpha)}=\cot(\alpha)$. So from ID-7c this is true.
e) From ID-7b: $\begin{aligned} \cos(2\alpha)&=2\cos^2(\alpha)-\blue{1}\\ &=2\cos^2(\alpha)-(\sin^2(\alpha)+\cos^2(\alpha))\\ &=\cos^2(\alpha)-\sin^2(\alpha) \end{aligned}$
f) $\begin{aligned} 1-\tan^2(\alpha)&=1-\cfrac{\sin^2(\alpha)}{\cos^2(\alpha)}\\ &=1- \cfrac{\cfrac{1-\cos(2\alpha)}{2}}{\cos^2(\alpha)}\\ &=\cfrac{ \cos^2(\alpha)-\cfrac{1-\cos(2\alpha)}{2}}{\cos^2(\alpha)}\\ &=-\cfrac{1-2\cos^2(\alpha)-\cos(2\alpha)}{2\cos^2(\alpha)}\\ &=-\cfrac{2-4\cos^2(\alpha)}{2\cos^2(\alpha)}\\ &=\cfrac{2\cos^2(\alpha)-1}{\cos^2(\alpha)}\\ &=\cfrac{\cos(2\alpha)}{\cos^2(\alpha)} \end{aligned}$
g) $\begin{aligned} 1-\cot^2{\alpha}&=\cfrac{\sin^2(\alpha)-\cfrac{1+\cos(2\alpha)}{2}}{\sin^2(\alpha)}\\ &=\cfrac{ \cfrac{1-\cos(2\alpha)}{2}- \cfrac{1+\cos(2\alpha)}{2}}{\sin^2(\alpha)}\\ &=\cfrac{ -\cos(2\alpha)-\cos(2\alpha)}{2\sin^2(\alpha)}\\ &=-\cfrac{\cos(2\alpha)}{\sin^2(\alpha)} \end{aligned}$
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I'm working on this yet :) I should prove that, when $\alpha>180^{\circ}$.
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ID:8
Proof that $\sqrt{2}$ is irrational
I will not use any formulas from other proofs
This is a proof by contradiction, so let's assume that $\sqrt{2}$ is rational.If so, then it can be expressed in an equation as $\dfrac{p}{q}$ where $p$ and $q$ are co-primes - $\begin{aligned} \sqrt{2} &= \dfrac{p}{q} \\ 2 &= \dfrac{p^{2}}{q^{2}} \\ 2p^{2} &= q^{2} \end{aligned}$
By this simplification, we know that $q^{2}$ even, because it is $2$ multiplied by some other number. So $q$ must also be an even number. That means $q$ can be represented as $(2m)$.
Back to the equation, if we use the new value of $q$ and simplify further - $\begin{aligned} 2p^{2} &= q^{2} \\ 2p^{2} &= (2m)^{2} \\ 2p^{2} &= 4m^{2} \\ p^{2} &= 2m^{2} \end{aligned}$
As $p$ can also be represented like this, $p$ must also be even.
This creates a contradiction, as we had earlier said that $p$ and $q$ were co-primes, but now we say they are both even, which means they have a common divisor of $2$. Since this creates a contradiction, our original assumption must have been wrong...
$\text{Q.E.D - } \sqrt{2} \text{ is irrational!}$
@Páll Márton (no activity) - Thoughts? Also, do I have to wait till ID:10 to start chains?
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ID:9
Graphical Proof of Circle Area formula - $\pi r^{2}$
I will not use any formulas from other proofs
Let's cut a circle into $8$ equal sectors and arrange them as shown in the image below -Image credits - https://medium.com
As you can see, it forms a figure that looks very close to a parallelogram that is going towards becoming a rectangle. The next two images show what happens when we take even smaller arcs.
Image credits - https://medium.com Image credits - https://medium.com
In the last image, you see an figure that is almost a rectangle. TO get a perfect rectangle, we would need infinitesimally small sectors, but we can evaluate the area of the circle using this figure, as we know the formula for the area of a rectangle - $A = bh$
The height is easy to find, as it is just the radius. The base, if you look closely at the next image, is actually half of the circumference or half of $2 \pi r$ or $\pi r$
Image credits - https://medium.com
Thus, the formula for the area of a circle $\pi r \times r = \boxed{\pi r^{2}}$
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Excellent animations!
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@David Vreken Quite true. How do they even do that?
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@Percy Jackson1-2 more mins untill daily challenge
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@NSCS 747 yea for Australian timing right?
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@NSCS 747 I live in Victoria. Why do so many people live in NSW?!!
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Yes! I found them on medium.com, so I image credited each of them under the pics.
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ID:10
Algebraic Proof of following Trig Identities
b) $\cos(90-\beta) = \sin(\beta)$
I will not use any formulas from other proofs
a) and b) can be proven using the sine and cosine angle subtraction formulas
$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$
$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)$
a) $\sin(90- \beta) = \cos(\beta)$
Using the above formula, substituting $\alpha$ with $90$,
$\begin{aligned} \sin(90- \beta) &= \sin(90)\cos(\beta) - \cos(90)\sin(\beta) \\ \sin(90- \beta) &= 1 \times\cos(\beta) - 0 \times\sin(\beta) \\ \color{#3D99F6}\sin(90- \beta) \color{#3D99F6} &= \color{#3D99F6}\cos(\beta) \end{aligned}$
b) Using the above formula again, substituting $\alpha$ with $90$,
$\begin{aligned} \cos(90- \beta) &= \cos(90)\cos(\beta) + \sin(90)\sin(\beta) \\ \cos(90- \beta) &= 0 \times \cos(\beta) + 1 \times \sin(\beta) \\ \color{#3D99F6}\cos(90- \beta) \color{#3D99F6} &= \color{#3D99F6}\sin(\beta)\end{aligned}$
@Páll Márton (no activity) - We have reached ID:10!!! Can we start the chain proofs now?
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ID:11
Proof that $2\cos^{-1} n+2 \sin^{-1} n=\pi$ in radians
Not using other identities
Construct an isosceles triangle $\triangle ABC$.In isosceles$\triangle ABC$, $BC$ is the base. Construct the median $AM$ where $M$ is the midpoint of $BC$.
Since $\triangle ABC$ is isosceles, $\angle AMB$ is a right angle.
$\Rightarrow \triangle AMB$ is a right angled triangle.
This part uses basic definitions of sin & cos.
Call the length of the hypotenuse of $Rt\triangle AMB$ $k$. Call the length of median $AM$ $l$.
$\Rightarrow \angle BAM=\cos^{-1} \frac{l}{k}.$
$\Rightarrow \angle BAC=2\angle BAM=2\cos^{-1} \frac{l}{k}.$
Since $\triangle AMB$ is right angled, and $\triangle ABC$ is isosceles,
$\sin \angle B=\sin \angle C=\frac{l}{k}.$
$\Rightarrow \angle B=\angle C=\sin^{-1} \frac{l}{k}.$
Let$\frac{l}{k}=n.$.
Finally, since $\angle A,\angle B,\angle C$ are the interior angles of a triangle, i.e. they add up to $\pi (rad)$, $2\cos^{-1} n+2 \sin^{-1} n=\pi$ in radians.
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I think this is a new identity that has never been discovered ago?
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Maybe, but if so, you're a mathematician officially.
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see the solution
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You can add one more rule: use buttons to decrease length. :)
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[[show-hide-button|H]]
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my proof
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Whup!
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@Jeff Giff - Buttons don't work in chat, only in notes and wikis, not even problems :)
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You can use buttons only in notes :)
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