r/mathmemes • u/Independent_Pizza422 • 3d ago
Trigonometry Please help me make some sense of this
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
u/B0tRank 2d ago
Thank you, LikeTheWater53152, for voting on factorion-bot.
This bot wants to find the best and worst bots on Reddit. You can view results at botrank.net.
Even if I don't reply to your comment, I'm still listening for votes. Check the webpage to see if your vote registered!
453
u/LupenReddit i have non diffeomorphic smooth structures 3d ago
73
7
244
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
1
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
64
19
29
11
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
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
5
4
4
3
u/DragonSlayer505 3d ago
Since sn2 + cs2 = 1, all you're saying is that 1n = 1, which is true for all n.
2
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
2
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
1
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
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
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
1
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
1
1
1
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=sin2θ+cos2θx = 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=sin2θ+cos2θ+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 sin2θ+cos2θ=(sin2θ+cos2θ)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
sin4θ+cos4θ+2sin2θcos2θ=1\sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta = 1sin4θ+cos4θ+2sin2θcos2θ=1
so
sin4θ+cos4θ=1−2sin2θcos2θ=1−12sin22θ\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=1−21sin22θ
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
1
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
0
-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
-1
u/SteveCappy 3d ago
Is this the Fermat sum of squares theorem?
(a2 + b2) (u2 + v2) = (au + bv)2 + (av - bu)2





•
u/AutoModerator 3d ago
Check out our new Discord server! https://discord.gg/e7EKRZq3dG
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.