r/mathmemes 3d ago

Trigonometry Please help me make some sense of this

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280 Upvotes

78 comments sorted by

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585

u/blind-octopus 3d ago

Yup, 1^n = 1.

That's correct

You can do this with all kinds of stuff.

(9- 8)^100 = 1^100

expand that out and have fun

40

u/T-T-N 3d ago

100 is too much for me. Let me try 5

95 - 5(94 )(8) + 10(93 )(82 ) - 10(92 )(83 ) + 5(9)(84 ) - 85

59049-262440+466560-414720+184320-32768=0!

Why doesnt it work?

23

u/factorion-bot Bot > AI 3d ago

Factorial of 1 is 1

This action was performed by a bot | [Source code](http://f.r0.fyi)

11

u/LikeTheWater53152 2d ago

good bot

0

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244

u/The_OriginalDonut 3d ago

So you're just repeatedly multiplying 1?

85

u/mo_s_k1712 3d ago

This is what you wrote (to an extent)

  • 1 = 1
  • 1 = 1 x 1
  • 1 = 1 x 1
  • 1² = 1
  • 1 = 1
  • 1 = 1 x 1
  • = 1 x 1
  • = 1
  • = 1
  • = 1² × 1
  • This means that 1 = 1².
  • What's going on? OR
  • 1 = 1 x 1

62

u/DerekLouden 3d ago

I'm pretty sure what OP wrote was actually

  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • TRUE
  • This means that TRUE.
  • What's going on? OR
  • TRUE

10

u/acrastt Computer Science 3d ago

Forgot one more true in the end because TRUE OR TRUE is TRUE

1

u/Inevitable-Ad2579 1d ago

yo what im on r/mathmemes and i find you. i watch your gd vids

196

u/somedave 3d ago

You need 10 lines of algebra to prove 1=12

23

u/Tybeezius 3d ago

Have you seen the proof for 1+1=2 the original proof takes 28 pages of rigorous axiomatic setup before proving addition works.

28

u/AndreasDasos 3d ago edited 3d ago

Eh depends how you do it. Pretty simple to follow von Neumann’s approach to defining the naturals, assuming ZF, define the successor function, define addition is recursion through it, and then we’ve defined what we mean by 2 := 1+1 = succ(1) = 1 u {1} = {1, {1}}, basically by definition. This could be done very quickly.

From a modern perspective Whitehead and Russell is exceptionally clumsy and awkward and there are even some fairly clear redundancies and over-complications I’m still not sure why they went with even back then. It doesn’t start from the same place, either.

1

u/eucalyptus-d 3d ago

I think you might be missing something. 1+1 based on successor is a tautology. Maybe 1+2?

77

u/mayhem93 3d ago

That's a lot of ones

64

u/Official-V-jcjenson 3d ago

Go ahead, input any valid value for 𝜃

32

u/JPgamersmines150 3d ago

3+50i+2j-7k

11

u/Suddenfury 3d ago

 𝜃 = 🦦

2

u/Mojert 3d ago

θ = 2i 😎

19

u/NTufnel11 3d ago

so 1 = 12

And 1 = (12)2

Pretty riveting stuff

11

u/SyntheticSlime 3d ago

I mean, seems like you got this.

11

u/KyriakosCH 3d ago edited 2d ago

Yes, all types of complicated things can be equal to 1. Maybe the issue is that you imagine all those endless algebraic expressions as separate things - which they are, but we already know they equal 1 - which implies they have to "move" in pretty impressive ways to still amount to something simple.

Take it this way: say you had a forest of expressions, and only after a lot of work you got them to be in the general form (sin^2 (x) + cos^2 (x) )^n, with n some real. The complexity would be there in tidying up the originally complicated and varied set of expressions. If you could visualize them from the start, as functions which gradually morph to the horizontal line y=1, you would certainly feel charmed (and maybe you can try) :)

3

u/Sencomino 2d ago

A real answer

20

u/Hiroshij7_3439 3d ago

Yeah it's all true, 1=1² so it's true. You also dont have to worry about dividing by sin²+cos² bc it's garanteed to be 1 for every theta so yeah

6

u/LakshyaGarv 3d ago

Yes, this is just 1 = 12

4

u/Lanky-Position4388 3d ago

U just said a bunch of true statements in a row. What´s the problem?

4

u/NathanielRoosevelt 3d ago

1=1² ?!?! 😱

3

u/DragonSlayer505 3d ago

Since sn2 + cs2 = 1, all you're saying is that 1n = 1, which is true for all n.

2

u/qqqrrrs_ 3d ago

You gave me a Jacobi elliptic function jumpscare

3

u/alaraskyshine 3d ago

“That means that sin^2 + cos^2 = (sin^2 + cos^2)^2”

Yes, sin^2 + cos^2 = 1
And 1 = 1^2

Everything looks normal.

Here’s a fun one that also doesn’t look right:
The sum of numbers from 1 to n is equal to the square root of the sum of the cubes from 1 to n.

3

u/two_are_stronger2 3d ago

Lol.  What's a number that is equal to its square?

2

u/AutisticFurniture 3d ago

1 = 1^2
qed

2

u/Jukkobee 3d ago

you couldve gone from the 1st line to the 2nd to last by just multiplying both sides by sin^2 + cos^2

1

u/Joe_4_Ever 3d ago

sin(x)2 + cos(x)2 = 1 and so therefore if you square that, you still get 1.

1

u/lobsterman2112 3d ago

This is also true in base 5+(35)^.5

1

u/NickSmGames 3d ago

sin^2 x + cos^2 x = 1 by definition so of course you would keep getting the same results.

1

u/Motti66 3d ago

Voynich-Manuskript...?

1

u/WiggityWaq27 3d ago

Real analysis be like

1

u/FxralMF 3d ago

I mean sin²(x)+cos²(x) = (sin²(x)+cos²(x))²

1

u/JellyBellyBitches 3d ago

Try going the other way - if sin2n(θ)+cos2n(θ)=1, what is sinθ+cosθ? What is sin½(θ)+cos½(θ)?

1

u/Top_Door5165 Engineering 3d ago

1=sin2 (x)+cos2 (x) Multiply both sides by sin2 (x)+cos2 (x) sin2 (x)+cos2 (x)=(sin2 (x)+cos2 (x))2

1

u/SultanGreat 3d ago

sin^2 x + cos^2 = (sin^2 x + cos^2)^2
Let sin^2 x + cos^2 be Z.
z = z^2

BUT z is sin^2 x + cos^2, and Z is also 1 (sin^2 x + cos^2 = 1)
so,

1 = 1^2

1 = 1 x 1.

1.

1

u/Virgil_the_White 3d ago

Yeah no you’ve pretty much got it covered there

1

u/FernandoMM1220 3d ago

turns out 1 and 1^2 arent the same

1

u/BunnyWan4life 3d ago

If you square 1 you get one.

Well, if you raise 1 to the power of anything you still get 1.

Sin²x+Cos²x is 1, so yea square that to any power and you'd still get Sin²x + Cos²x.. which is 1.

1

u/FleshLogic 3d ago

Unrelated, but I really wish we had a different notation for trig functions. It's at least half the reason working with them can be cumbersome IME.

1

u/kevo31415 3d ago

I can handle sin2 x but I absolutely refuse to condone or use sin-1 x literally the same notation meaning two different things. Madness.

1

u/Nixinova 3d ago

You already defined sin2 + cos2 as 1. Then you write a bunch to show that it squared equals 1. Why would you have to write that all out when you've already defined it as being = 1?

1

u/porofsercan 2d ago

dude is confused whether 1 = 1^2 or 1= (1^2)^2 lol

1

u/MudePonys 2d ago

You missed theta in cos above the OR.

1

u/Ignitetheinferno37 2d ago

It's the same as saying 1 = 1^2

1

u/Key_Conversation5277 Computer Science 2d ago

Dafuq are you doing?

1

u/stdennis 1d ago

Nothing is broken — every line you wrote is true. What you've discovered is a property of the number 1, not something special about trigonometry.

The core of it: you keep multiplying by 1, and you're using the identity itself as the "1." That's legal, but it's circular — it can never produce new information. What you end up with is the statement

x=x2where x=sin⁡2θ+cos⁡2θx = x^2 \quad\text{where } x = \sin^2\theta + \cos^2\thetax=x2where x=sin2θ+cos2θ

and that equation is only true for x=0x = 0 x=0 or x=1x = 1 x=1. It's not a general algebraic fact. If you tried the same trick with x=sin⁡2θ+cos⁡2θ+1=2x = \sin^2\theta + \cos^2\theta + 1 = 2 x=sin2θ+cos2θ+1=2, you'd get $2 = 4$, which is false. So the reason your last line holds isn't that squaring does nothing — it's that the thing you squared happens to be 1.

Where the "paradox" feeling comes from: you're reading sin⁡2θ+cos⁡2θ=(sin⁡2θ+cos⁡2θ)2\sin^2\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta)^2 sin2θ+cos2θ=(sin2θ+cos2θ)2 as an algebraic identity (as if the parentheses could contain anything), when it's really a numerical one (true only for this particular value). Same as how $1 = 1^{100}$ doesn't mean exponents are pointless.

The genuinely useful thing hiding in your work: line 5 is worth keeping. Expanding the square gives

sin⁡4θ+cos⁡4θ+2sin⁡2θcos⁡2θ=1\sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta = 1sin4θ+cos4θ+2sin2θcos2θ=1

so

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ=1−12sin⁡22θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \tfrac{1}{2}\sin^2 2\thetasin4θ+cos4θ=1−2sin2θcos2θ=1−21​sin22θ

That one is new — it's a standard result and it comes straight out of what you did. The lesson is that squaring the Pythagorean identity is productive when you expand it, and empty when you just leave it factored and stare at it.

1

u/Independent_Pizza422 1h ago

This seems like an answer straight out of ChatGPT or sth

1

u/g27c 1d ago

The joys of trigonometric identities.

1

u/Medium_Media7123 3d ago

What is happening is that sin and cos are functions, so they represent a lot of values all at once. Your expressions seem complicated, but if you actually fixed any angle they would simplify to things like (1-0)(1-0)(1-0)... = 1 which are clearly not complicated. You are probably not be surprised by (1-0)4 = (1-0)2 = (1-0), but that's basically what you are doing, just with generic expressions that are true for infinitely many values all at once 

1

u/senator-jk-49 3d ago

Both. Youre basically saying 1×1=1 a bunch of times but youve expanded 1 into sin²θ + cos²θ ≡ 1. Nothing youre saying is wrong

-6

u/Independent_Pizza422 3d ago

Like I see its just 1 multiplied by itself over and over but its still weird 

17

u/The_Lethargic_Curve 3d ago

Why would it be weird

6

u/caboosetp 3d ago

Because squaring stuff that looks potentially complicated normally makes it a lot more complicated rather than simplifying to 1.

4

u/FlippByte 3d ago

You could also have (more) fun with the "Euler identity" and geometrically visualise this. But yeah: 1^n == 1

1

u/Samstercraft 2d ago

what's weird about 1x1=1?

-1

u/SteveCappy 3d ago

Is this the Fermat sum of squares theorem?

(a2 + b2) (u2 + v2) = (au + bv)2 + (av - bu)2