# Fast review:

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
• 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 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:

1. Reply for the last approach: I want to continue this chain. If somebody already working on this you should wait.
2. Write your approach. Use this form below:
 1 2 3 4 5 6 # ID: ## ## > 

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>}

1. Profile->+Note(in the bottom)->Upload an image
2. Copy the code of the picture
• 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 ) ):

 1 2 3 4 5 6 7 8 # 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

# 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.

Note by Páll Márton (no activity)
2 months, 1 week ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
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# up as a code block.

print "hello world"
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\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

- 1 month, 4 weeks ago

# ID: 1

## 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 $a^2-b^2=(a-b)(a+b)$.

P.S.: If $a=b$ then $a^2-b^2=(a-b)(a+b)=0$. And if $a 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.

- 2 months, 1 week ago

- 2 months, 1 week ago

Good, I will try to find something good! Nice note!

- 2 months, 1 week ago

Thank you!

- 2 months, 1 week ago

I have proof of something else, can I write it in a separate comment?

- 2 months ago

Yeah. I will see that :)

Done!

- 2 months ago

# 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:

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 $\Huge{{\sin}^2{\theta}+{\cos}^2{\theta} =1 \blacksquare}$

- 2 months ago

Or: $(sin(a))^2+(cos(a))^2=\cfrac{a^2+b^2}{c^2}=1$, where $a^2+b^2=c^2$ :)

This is the common notation there

Yes, but I wanted to show that sin=o/h and cos=a/h.

- 2 months ago

Please invite more people, it will be fun to prove many things!

- 2 months ago

Can you help me in it?

I have some work of school. Sorry! But, mention few people at a time, it will work.

- 2 months ago

Yeah. I have to read 10+ books, but I hate reading :) I will try mention people

Same with me! :) But I have to write also! :(

- 2 months ago

Write a book? :)

No! Write answers(at least 20 in one chapter)!

- 2 months ago

I should write a journal about all of them

LOL 18 books :) I won't read them. I want to learn english :)

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.

Actually, I don't know this formula! I'll search it, and if I understand, I will try.

- 2 months ago

# ID: 3

## 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}

P.S. Written in 1876, James A. Garfield was the only U.S. president to contribute a proof to a mathematical theorem.

- 2 months ago

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.)

- 2 months ago

By the way, if you put a slash in front of cos in latex, it won't be italicized (like this: "\cos")

- 2 months ago

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.

- 2 months ago

LOL I learned the location of all the countries. And also the states of the United States But I don't remember too much

Wow, that must be hard, it probably is with a dracanae as your teacher.................

- 2 months ago

No, I went to a competition. But I failed on evil geometry. I proved twice this is impossible to solve the problems.

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)

- 2 months ago

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

Released from sports??? @Páll Márton

- 2 months ago

I can't translate it :( When he was young he couldn't play sports.

Good proof Sir! I remember that I once saw a proof in which the triangle is copied 4 times to form a square.

- 2 months ago

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.

I haven't heard of anything like this. Can you give a link, or explain yourself?

- 2 months ago

@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?

- 2 months ago

# ID: 4

## I will use this formula from there: ID-3

From ID-3:

• In $\Delta BFC$: $x^2+m^2=b^2$
• In $\Delta AFC$: $(a+x)^2+m^2=f^2$
• In $\Delta EBD$: $(a-x)^2+m^2=e^2$

The sum of the two last statement: $e^2+f^2=2a^2+\blue{2x^2+2m^2}$. From the first statement: $e^2+f^2=2a^2+2b^2$

You can use this formula too: $c^2=a^2+b^2-2ab\cdot cos(\gamma)$

Where is point $F$?

- 2 months ago

Ohhh. Wait a minute :)

See now. Thank you!

Nice proof! I had not heard of this theorem, but seems interesting!

- 2 months ago

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 post the next approach :)

# ID: 5

## 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 $BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)$

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.

And see the actual theme :)

I will prove these things:

• $sin(\cfrac{\alpha}{2})=\pm\sqrt{\cfrac{1-cos(\alpha)}{2}}$
• same thing with cos, tan, cot
• $1-tan^2(\alpha)=\cfrac{cos(2\alpha)}{cos^2(\alpha)}$
• same thing with cot

Sorry if there are typos!

And $cos(\alpha)=cos^2(\alpha /2)-sin^2(\alpha /2)$

For case 2 I don't think your letters match with your picture (specifically $BC^2 = BD^2 + DC^2$)

- 2 months ago

Sorry. Swap A and B :) I will edit it, when I can

that should do it!

- 2 months ago

# ID:6

## Prove these formulas:

• $\sin(\alpha\pm360^{\circ}\cdot k)=\sin(\alpha)$ 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: $\cfrac{\sin(x)}{\cos(x)}=\cfrac{\tan(x)}{1}$

# ID:7

## Prove these formulas:

• $\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}$
• $\cos(\alpha)=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}}$
• $\tan(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}}$
• $\cot(\cfrac{\alpha}{2})=\sqrt{\cfrac{1+\cos(\alpha)}{1-\sin(\alpha)}}$
• $\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)$
• $1-\tan^2(\alpha)=\cfrac{\cos(2\alpha)}{\cos^2(\alpha)}$
• $1-\cot^2{\alpha}=-\cfrac{\cos(2\alpha)}{\sin^2(\alpha)}$

## 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:

• $\beta=\cfrac{180^{\circ}-\alpha}{2}=90^{\circ}-\cfrac{\alpha}{2}$
• $\sin(90^{\circ}-\alpha)=\cos(\alpha)$ $\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}$

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}

I'm working on this yet :) I should prove that, when $\alpha>180^{\circ}$.

- 1 month, 1 week ago

You can add one more rule: use buttons to decrease length. :)

- 1 month, 2 weeks ago

[[show-hide-button|H]]
[[start-hidden]]
my proof
[[end-hidden]]

- 1 month, 2 weeks ago

Whup!

- 1 month, 2 weeks ago

@Jeff Giff - Buttons don't work in chat, only in notes and wikis, not even problems :)

- 1 month, 2 weeks ago

You can use buttons only in notes :)

- 1 month, 2 weeks ago