Math prove marathon

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 ab=2Aa\cdot b=2A in a right triangle and you want to prove that the area of a square is a2a^2, then this is a good approach:

If we cut the square into two right triangles: As=2At;ab=2At;a=bAs=2aa2=a2A_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

Temporary rule(s)\color{#eeaa00}\text{Temporary rule(s)}

  • The first ten approaches should not create a chain\color{#eeaa00}\text{The first ten approaches should not create a chain}
  • You can share not original, but not famous approches\color{#eeaa00}\text{You can share not original, but not famous approches}

Current theme(s)\color{#D61F06}\text{Current theme(s)}

  • Number theory\color{#D61F06}\text{Number theory}
  • Algebra\color{#D61F06}\text{Algebra}

Event(s)\color{#3D99F6}\text{Event(s)}

  • There has been no events yet(after 10 active members will be the first event)\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.

You can also write proofs that do not belong to the given theme(s)\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\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: <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>}

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

Without spaces:

  • \ ( <LaTeX code> \ ) - left aligned
  • \ [ <LaTeX code> \ ] - center aligned

If you are bad in LaTeX\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<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.
NamePosted approachesPoints
David Vreken16
Páll Márton5
Vinayak Srivastava15

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

No vote yet
1 vote

  Easy Math Editor

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.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Log in to reply

Log in to reply

ID: 1

Graphically prove that a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b)

I won't use other formulas from this note

Assume that a>ba>b. In this picture below T?T_? means territory/area: In this picture T1+T3+T4=a2T_1+T_3+T_4=a^2 and T4=b2    T1+T3=a2b2T_4=b^2\implies T_1+T_3=a^2-b^2. We know T1+T2=(a+b)(ab)T_1+T_2=(a+b)(a-b) from the side length of the rectangle. And we know T2=T3T_2=T_3, because both of them is equal to b(ab)b(a-b). Therefore T1+T3=T1+T2=(a+b)(ab)T_1+T_3=T_1+T_2=(a+b)(a-b). So a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b).

P.S.: If a=ba=b then a2b2=(ab)(a+b)=0a^2-b^2=(a-b)(a+b)=0. And if a<ba<b then, then we can multiply by -1. After that we get this: b2a2=(ba)(a+b)b^2-a^2=(b-a)(a+b). If we swap a and b in the picture we get the same result.

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

Log in to reply

@Vinayak Srivastava What about this note?

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

Log in to reply

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

Vinayak Srivastava - 2 months, 1 week ago

Log in to reply

@Vinayak Srivastava Thank you!

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

Log in to reply

@Páll Márton (no activity) I have proof of something else, can I write it in a separate comment?

Vinayak Srivastava - 2 months ago

Log in to reply

@Vinayak Srivastava Yeah. I will see that :)

Log in to reply

Log in to reply

ID:2

Proof of sin2θ+cos2θ=1{\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 sin2θ+cos2θ=1{\sin}^2{\theta}+{\cos}^2{\theta} =1 is this:

In any right triangle, O2+A2=H2 (Where H is the hypotenuse, O is opposite, and A is adjacent.)O^2+A^2=H^2 \text{ (Where H is the hypotenuse, O is opposite, and A is adjacent.)}     O2H2+A2H2=1\implies \dfrac{O^2}{H^2}+ \dfrac{A^2}{H^2} =1     sin2θ+cos2θ=1\implies {\sin}^2{\theta}+{\cos}^2{\theta} =1

However this proof is not complete! What about an angle θ>90{\theta} >90^{\circ}? It won't form any right triangle!


Here is where our unit circle helps:

In our unit circle, for any angle 360θ0,360^{\circ}\geq {\theta}\geq 0^{\circ}, we have the particular point on its circumference representing cosθ,sinθ\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 x2+y2=1x^2+y^2=1

Hence, for any angle 360θ0,360^{\circ}\geq {\theta}\geq 0^{\circ}, we get sin2θ+cos2θ=1\Huge{{\sin}^2{\theta}+{\cos}^2{\theta} =1 \blacksquare}

Vinayak Srivastava - 2 months ago

Log in to reply

Or: (sin(a))2+(cos(a))2=a2+b2c2=1(sin(a))^2+(cos(a))^2=\cfrac{a^2+b^2}{c^2}=1, where a2+b2=c2a^2+b^2=c^2 :)

Log in to reply

This is the common notation there

Log in to reply

@Páll Márton (no activity) Yes, but I wanted to show that sin=o/h and cos=a/h.

Vinayak Srivastava - 2 months ago

Log in to reply

@Páll Márton (no activity) Please invite more people, it will be fun to prove many things!

Vinayak Srivastava - 2 months ago

Log in to reply

@Vinayak Srivastava Can you help me in it?

Log in to reply

@Páll Márton (no activity) I have some work of school. Sorry! But, mention few people at a time, it will work.

Vinayak Srivastava - 2 months ago

Log in to reply

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

Log in to reply

@Páll Márton (no activity) Same with me! :) But I have to write also! :(

Vinayak Srivastava - 2 months ago

Log in to reply

@Vinayak Srivastava Write a book? :)

Log in to reply

@Páll Márton (no activity) No! Write answers(at least 20 in one chapter)!

Vinayak Srivastava - 2 months ago

Log in to reply

@Vinayak Srivastava I should write a journal about all of them

Log in to reply

@Páll Márton (no activity) LOL 18 books :) I won't read them. I want to learn english :)

Log in to reply

Now I can't post any approaches with pictures. Can you post an approach about sin(90α)=cos(α)\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.

Vinayak Srivastava - 2 months ago

Log in to reply

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 aa and bb and hypotenuse cc. Extend leg aa by bb units and construct a duplicate right triangle along this extension. Upper leg aa is parallel to the original leg bb since, in a plane, if a line is perpendicular to each of two lines, then the two lines are parallel. Draw segment XYXY to close the figure. The resulting quadrilateral is a trapezoid with bases aa and bb and altitude a+ba+b.

The trapezoid is composed of two congruent right triangles and right XYZ\triangle XYZ. XYZ\triangle XYZ is isosceles since two of its sides have length cc. XZY\angle XZY is a right angle since 1+XZY+2=180°\angle 1 + \angle XZY + \angle 2 = 180° and 1+2=90°\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:

AVWXY=AZVY+AXWZ+AXYZ12(a+b)(a+b)=12ab+12ab+12c2(a+b)(a+b)=ab+ab+c2a2+2ab+b2=2ab+c2a2+b2=c2\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.

David Vreken - 2 months ago

Log in to reply

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 c2=a2+b22abcos(c);cos(90)=0c^2=a^2+b^2-2ab cos(c);cos(90^{\circ})=0

Log in to reply

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.)

David Vreken - 2 months ago

Log in to reply

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

David Vreken - 2 months ago

Log in to reply

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.

Percy Jackson - 2 months ago

Log in to reply

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

Log in to reply

@Páll Márton (no activity) Wow, that must be hard, it probably is with a dracanae as your teacher.................

Percy Jackson - 2 months ago

Log in to reply

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

Log in to reply

@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 - 2 months ago

Log in to reply

@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

Log in to reply

@Páll Márton (no activity) Released from sports??? @Páll Márton

Percy Jackson - 2 months ago

Log in to reply

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

Log in to reply

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

Vinayak Srivastava - 2 months ago

Log in to reply

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.

Log in to reply

@Páll Márton (no activity) I haven't heard of anything like this. Can you give a link, or explain yourself?

Vinayak Srivastava - 2 months ago

Log in to reply

@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?

Percy Jackson - 2 months ago

Log in to reply

ID: 4

Prove that 2a2+2b2=e2+f22a^2+2b^2=e^2+f^2, where aa and bb is the two sides of the parallelogram and ee and ff is the diagonals of it

I will use this formula from there: ID-3

From ID-3:

  • In ΔBFC\Delta BFC: x2+m2=b2x^2+m^2=b^2
  • In ΔAFC\Delta AFC: (a+x)2+m2=f2(a+x)^2+m^2=f^2
  • In ΔEBD\Delta EBD: (ax)2+m2=e2(a-x)^2+m^2=e^2

The sum of the two last statement: e2+f2=2a2+2x2+2m2e^2+f^2=2a^2+\blue{2x^2+2m^2}. From the first statement: e2+f2=2a2+2b2e^2+f^2=2a^2+2b^2

You can use this formula too: c2=a2+b22abcos(γ)c^2=a^2+b^2-2ab\cdot cos(\gamma)

Log in to reply

Where is point FF?

Vinayak Srivastava - 2 months ago

Log in to reply

Ohhh. Wait a minute :)

Log in to reply

@Páll Márton (no activity) See now. Thank you!

Log in to reply

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

Vinayak Srivastava - 2 months ago

Log in to reply

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

Log in to reply

I will post the next approach :)

Log in to reply

ID: 5

Prove that: a2=b2+c22bccos(α)a^2=b^2+c^2-2bc\cdot cos(\alpha)

I will use these formulas from there: ID-2, ID-3

Assume that A>B>C\angle A>\angle B>\angle C There are three cases:

1.case A=90\angle A=90^{\circ}:

From ID-3, this is true, because cos(90)=0cos(90^{\circ})=0

2.case A<90\angle A<90^{\circ}:

Swap A and B\huge\text{Swap A and B} AD=AB  cos(A)AD=AB\cdot\;cos(\angle A) and BD=AB  sin(A)BD=AB\cdot\;sin(\angle A)
From the 1. case: BC2=BD2+DC2=BD2+(ACAD)2=(ABsin(A))2+AC2+(ABcos(A))22ACABcos(A)=AB2(cos2(A)+sin2(A))+AC22ACABcos(A)\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: BC2=AB2+AC22ACABcos(A)BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)

3.case A>90\angle A>90^{\circ}:

Same way: AD=AB  cos(BAD)AD=AB\cdot\;cos(\angle BAD) and BD=AB  sin(BAD)BD=AB\cdot\;sin(\angle BAD)
From the 1. case: BC2=BD2+DC2=BD2+(AC+AD)2=(ABsin(BAD))2+AC2+(ABcos(BAD))2+2ACABcos(BAD)=AB2(cos2(BAD)+sin2(BAD))+AC2+2ACABcos(BAD)\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: BC2=AB2+AC2+2ACABcos(BAD)BC^2=AB^2+AC^2+2AC\cdot AB\cdot cos(\angle BAD), but cos(180α)=cos(α)cos(180^{\circ}-\alpha)=-cos(\alpha). So BC2=AB2+AC22ACABcos(A)BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)

Log in to reply

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.

Log in to reply

And see the actual theme :)

Log in to reply

@Páll Márton (no activity)

I will prove these things:

  • sin(α2)=±1cos(α)2sin(\cfrac{\alpha}{2})=\pm\sqrt{\cfrac{1-cos(\alpha)}{2}}
  • same thing with cos, tan, cot
  • 1tan2(α)=cos(2α)cos2(α)1-tan^2(\alpha)=\cfrac{cos(2\alpha)}{cos^2(\alpha)}
  • same thing with cot

Sorry if there are typos!

Log in to reply

@Páll Márton (no activity) And cos(α)=cos2(α/2)sin2(α/2)cos(\alpha)=cos^2(\alpha /2)-sin^2(\alpha /2)

Log in to reply

For case 2 I don't think your letters match with your picture (specifically BC2=BD2+DC2BC^2 = BD^2 + DC^2)

David Vreken - 2 months ago

Log in to reply

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

Log in to reply

@Páll Márton (no activity) that should do it!

David Vreken - 2 months ago

Log in to reply

ID:6

Prove these formulas:

  • sin(α±360k)=sin(α)\sin(\alpha\pm360^{\circ}\cdot k)=\sin(\alpha) and cos(α±360k)=cos(α)\cos(\alpha\pm360^{\circ}\cdot k)=\cos(\alpha), where kk is an integer
  • sin(180α)=sin(α)\sin(180^{\circ}-\alpha)=\sin(\alpha) and cos(180α)=cos(α)\cos(180^{\circ}-\alpha)=-\cos(\alpha)
  • sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) and cos(90α)=sin(α)\cos(90^{\circ}-\alpha)=\sin(\alpha)
  • tan(α)=sin(α)cos(α)\tan(\alpha)=\cfrac{\sin(\alpha)}{\cos(\alpha)}

I won't use other formulas from there

In this picture we can see the sin(x)\sin(x) and cos(x)\cos(x) functions. We should calculate in radian. This means:

2π=3602\pi=360^{\circ}, π=180\pi=180^{\circ}, π2=90\cfrac{\pi}{2}=90^{\circ} We can see if we shift the functions by 2π2\pi, then we get the same functions. Same way if we shift these functions by 4π,6π,8π,10π4\pi,6\pi,8\pi,10\pi, then we get the same functions.


From the unit circle:

sin(180α)=sin(α)\sin(180^{\circ}-\alpha)=\sin(\alpha) and cos(180α)=cos(α)\cos(180^{\circ}-\alpha)=-\cos(\alpha)

If α>180\alpha>180^{\circ}, then you should mirror the image, and you will get the same result.


From interpretation of sin,cos\sin,\cos: sin(α)=ac,cos(β)=ac,β=90α    cos(90α)=sin(α)\sin(\alpha)=\cfrac{a}{c},\cos(\beta)=\cfrac{a}{c},\beta=90^{\circ}-\alpha\implies\cos(90^{\circ}-\alpha)=\sin(\alpha). Same way sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) .


From similarity: sin(x)cos(x)=tan(x)1\cfrac{\sin(x)}{\cos(x)}=\cfrac{\tan(x)}{1}

Log in to reply

ID:7

Prove these formulas:

  • sin(α2)=1cos(α)2\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}
  • cos(α)=1+cos(α2)2\cos(\alpha)=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}}
  • tan(α2)=1sin(α)1+cos(α)\tan(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}}
  • cot(α2)=1+cos(α)1sin(α)\cot(\cfrac{\alpha}{2})=\sqrt{\cfrac{1+\cos(\alpha)}{1-\sin(\alpha)}}
  • cos(2α)=cos2(α)sin2(α)\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)
  • 1tan2(α)=cos(2α)cos2(α)1-\tan^2(\alpha)=\cfrac{\cos(2\alpha)}{\cos^2(\alpha)}
  • 1cot2α=cos(2α)sin2(α)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: AD2=x2+(xz)22x(xz)cos(α)AD2=z2+AB22zABcos(β)AB2=x2+x22xxcos(α)=x2(22cos(α))\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: x2+(xz)22x(xz)cos(α)=x2+x22x2cos(α)+z22zx22cos(α)cos(β)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) x2+x2+z22xz+2x2cos(α)+2xzcos(α)=x2+x22x2cos(α)+z22zx22cos(α)cos(β)\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-2xz: 1cos(α)=22cos(α)cos(β)1-\cos(\alpha)=\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta) cos(β)=1cos(α)22cos(α)=1cos(α)211cos(α)=1cos(α)2=1cos(α)2\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:

  • β=180α2=90α2\beta=\cfrac{180^{\circ}-\alpha}{2}=90^{\circ}-\cfrac{\alpha}{2}
  • sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) sin(α2)=1cos(α)2\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}

b) From ID-2: sin2(α)+cos2(α)=1\sin^2(\alpha)+\cos^2(\alpha)=1 From ID-7a: 1cos(α2)2+cos2(α)=11cos(α2)+2cos2(α)=2cos2(α)=1+cos(α2)2cos(α)=1+cos(α2)2\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: tan(α2)=sin(α2)cos(α2)=1cos(α)21+cos(α)2=1sin(α)1+cos(α)\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) 1tan(α)=cot(α)\cfrac{1}{\tan(\alpha)}=\cot(\alpha). So from ID-7c this is true.


e) From ID-7b: cos(2α)=2cos2(α)1=2cos2(α)(sin2(α)+cos2(α))=cos2(α)sin2(α)\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) 1tan2(α)=1sin2(α)cos2(α)=11cos(2α)2cos2(α)=cos2(α)1cos(2α)2cos2(α)=12cos2(α)cos(2α)2cos2(α)=24cos2(α)2cos2(α)=2cos2(α)1cos2(α)=cos(2α)cos2(α)\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) 1cot2α=sin2(α)1+cos(2α)2sin2(α)=1cos(2α)21+cos(2α)2sin2(α)=cos(2α)cos(2α)2sin2(α)=cos(2α)sin2(α)\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}

Log in to reply

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

Log in to reply

see the solution

Jeff Giff - 1 month, 1 week ago

Log in to reply

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

Jeff Giff - 1 month, 2 weeks ago

Log in to reply

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

Jeff Giff - 1 month, 2 weeks ago

Log in to reply

Whup!

Jeff Giff - 1 month, 2 weeks ago

Log in to reply

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

Percy Jackson - 1 month, 2 weeks ago

Log in to reply

You can use buttons only in notes :)

Páll Márton (no activity) - 1 month, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...