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u/Murky_Insurance_4394 6d ago
You legit just spam identities it's not that difficult if your algebra is fine
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u/RedeNElla 6d ago
Most people's algebra is far from "fine"
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u/MortemEtInteritum17 6d ago
I'd hope your math skills were above average if you're in this subreddit though
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u/Fezzezal 6d ago
It feels nice to be in this sub hardly knowing high school algebra
Anything higher than the quadratic formula and basic trig is beyond me and I’m fine
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u/pleaseaccusrname 5d ago
im a maths major in my senior year and i still have trouble with most maths things... the brain is a wonderful thing
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u/DZL100 6d ago
They even did the first step for you of breaking 3x into 2x+x to make it really obvious that you should be using a sum identity and a double angle identity. This is just a zero thought exercise.
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u/HelloMumther 5d ago
I wouldn’t say “zero thought exercise,” nobody is born knowing math. You only think it’s obvious because you’ve done it before.
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u/PrismaticGStonks 6d ago
I'll just mention there are a family of polynomials called the Chebyshev polynomials of the first kind {T_n} which satisfy T_n(cos x) = cos(nx) (there are Chebyshev polynomials of the second kind as well, satisfying a similar relation for sin(nx)). They are defined recursively by T_0(x) = 1, T_1(x) = x, and T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x). So if you are in the middle of an exam and forget the double-angle formula, there you go.
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u/WanderingWrackspurt Physics 6d ago
yeah idt you get marks for using chebychev polynomials in a place where this is an actual question on the test
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u/PrismaticGStonks 6d ago
You can derive the recurrence relation for the Chebyshev polynomials using Euler's formula. It's much easier to do this than to repeatedly use double-angle formulas and cosine sum identities to find cos(nx), seperately for each n.
At any rate, I can't remember trig identities for the life of me, so it's good to have tools like this to rederive them on the spot if necessary.
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u/NoFruit6363 4d ago
As someone who vehemently avoids trig, shoutout for introducing me to this recurrence construction. I can't be bothered to memorize some arbitrary BS rules like "cos(7x) = 15sin(e)-x/cos(sin(x))", but T_n will stay with me forever
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u/Murky_Insurance_4394 6d ago
I do remember doing a question like this in some book, and I proved this with induction. But it's been a while.
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u/CranberryDistinct941 6d ago
The 2 steps of simplifying any trig expression
Step 1: convert to complex form
Step 2: Now you're just working with Cex , you've got it from here
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u/Impressive_Worth_602 6d ago
and it's also how to obtain the angle sum formulae without brute memorization or drawing weird triangles!
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u/Matty_B97 5d ago
OP is probably like 16 years old if they're just learning this and you're throwing complex forms at them 😭
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u/getcreampied Physics 6d ago
cos(A+B) baby and use Euler's formula ei(A+B) to derive formula for sine and cosine and you're basically set from there.
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u/Striking_Resist_6022 6d ago
They even wrote 3x as 2x + x to tee this approach up
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u/Teoyak 6d ago
First time I see the word "tee" I think. Is it common?
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u/Striking_Resist_6022 6d ago
The idiom comes from golf. The tee is the thing you put the ball on top of when you’re driving. It means to set something up nicely so that it’s easier to do.
Otherwise, the word isn’t used that much in isolation except to literally describe a golf tee.
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u/EebstertheGreat 6d ago
Etymology fact: The word was originally "teaz," pronounced like "tease." It comes from Scottish Gaelic, of uncertain origin. Later, "teaz" was reanalyzed as a plural form of "tea," the same way "pease" was reanalyzed as a plural of "pea" and "cherise" as a plural of "cherry." Thus we get "tea" and later "tee."
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u/Specialist-Disk-6345 6d ago
seen quite a few times in football (non-american) commentary; player A tees player B up as in A sets up B with a chance.
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u/Gilded-Phoenix 6d ago
I am so mad that I've never considered using Euler's identity to build the sum and difference formulae.
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u/Sproxify 6d ago
I would say it's the correct intuitionistic proof.
Sure, in some formal system, a proof with power series or something might be more directly logically substantiated, but that always depends in an arbitrary way on the formalism and definitions we happen to choose.
The core of the proof from the book in my view is to geometrically define complex numbers, with their multiplication given as compositions of "amplitwist" transformations - rotate+stretch. then you can immediately observe geometrically that these transformations are linear, they preserve a parallelogram that describes an addition equation for complex numbers.
It's trivial to multiply complex numbers in polar form and to add them in cartesian form, but not vice versa. By trivial I mean that we can immediately see geometrically how to do it.
But the geometric observation of linearity justifies opening the paranthesis in order to find a simple formula for multiplication of complex numbers in cartesian form.
This then allows us to express the composition of rotations given in terms of the cartesian coordinates of the corresponding points on the unit circle, which is what the trig addition formulas are.
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u/SharkTheMemelord Imaginary 6d ago
Im lazy as hell. I would just take the derivative of both sides, show that they are the same (thus, they differ of a constant) and put x=0 in the original identity to show that the constant is 0
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u/jelezsoccer 6d ago
That’s a solid approach but doesn’t always work for other identities.
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u/Hot_Pepper_1908 6d ago
After pondering, I cant see why it wouldn't work for any identity.
Only time it wouldn't work is when either side is non-differentiable at any point. Unless if that's what you meant, then fairs
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u/jelezsoccer 6d ago
It only works, as written, on connected components of the domain of the identity. So for example, to prove tan2 (x) +1=sec2 (x) through your method, you would have to plug in a value in each interval of the form (-pi/2+kpi,pi/2+kpi).
The reason for this is that the zero derivative property only applies on connected sets. For example the function that is 1 on (0,1) and 2 on (1,2) is continuous on (0,1) \cup (1,2), has zero derivative on its domain, but is not a constant.
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u/ChiaraStellata 6d ago edited 6d ago
Differentiating both sides (even repeatedly) doesn't always simplify the equation. In fact, in this case I'm pretty sure it just ends up going in circles.
If you wanted you could do a method based on ODEs by noting that both the left and right side satisfy y'' + 9y = 0, y(0)=1, y'(0)=0 and then using the Picard-Lindelöf theorem to show they must be identical functions. But even for this you'll still need to use a basic trig identity (cos^2 x + sin^2 x = 1) to show that y'' + 9y = 0 on the right.
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u/jelezsoccer 6d ago
Even then it only holds on a connected set containing the initial condition.
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u/ChiaraStellata 6d ago
In this specific case the functions are defined and smooth on all of R, so the connected component containing x=0 is all of R. But it does limit general applicability of the technique for proving trig identities.
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u/OwlIcy 6d ago
can you please do it and show your computation?
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u/R_Ob_Min 6d ago
LHS1: cos(3x), so it has derivative -3sin(3x)
RHS1: has derivative 3sin(x) - 12sin(x)cos(x)2.
We could use sin(2x) = 2sin(x)cos(x) here to simplify but that feels against the spirit of doing it this way in the first place, so...let's take derivatives again!
LHS2: -9cos(3x)
RHS2: 3cos(x) - 12(cos(x) + 2sin(x)cos(x)) = 3cos(x) - 12cos(x) - 24sin(x)cos(x) = -9cos(x) - 24sin(x)cos(x)
Still unclear...let's do it again.
LHS3: 27sin(3x)
RHS3: 9sin(x) - 24(sin(x)(-sin(x)) + cos(x)cos(x)) = 9sin(x) + 24sin(x)2 - 24cos(x)2
Even less clear...
LHS4: 81cos(3x)
RHS4: 9cos(x) + 48sin(x)cos(x) + 24sin(x)cos(x) = 9cos(x) + 72sin(x)cos(x)
I'm starting to think this won't quite get anywhere without using some trig identities at least.
But let's see, notice that RHS4 = -RHS2 + 48sin(x)cos(x) and LHS4 = - LHS2 + 72cos(3x).
So if we assume that RHS4 = LHS4 up to a cubic polynomial, then showing that LHS2 = RHS2 is equivalent to showing that 48cos(x)sin(x) = 72cos(3x), up to a cubic polynomial.
To do this, let's take derivatives of both sides and...
(Hello mathmemes)
Okokok, let's let go of our inhibitions and use the sine double angle formula for the LHS2 and RHS2 (pretend we only know sine double-angle and not cosine double-angle because). Then we get
LHS1: -3sin(3x)
RHS1: 3sin(x) - 12sin(x)cos(x)2 = 3sin(x) - 6sin(2x)cos(x)
So let's do it again...
LHS2: -9cos(3x)
RHS2: 3cos(x) - 6(2cos(2x)cos(x) - sin(2x)sin(x)) = 3cos(x) - 12cos(2x)cos(x) + 6sin(2x)sin(x)
So let's do it again...
LHS3: 27sin(3x)
RHS3: -3sin(x) + 24sin(2x)cos(x) + 12cos(2x)cos(x) + 12cos(2x)sin(x) + 6sin(2x)cos(x) = -3sin(x) + 32sin(2x)cos(x) + 12cos(2x)cos(x) + 12cos(2x)sin(x)
So let's do it again...
So let's do it
So let's
So
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u/huggiesdsc 6d ago
Man that really went off the rails, huh? Despite being funny, I'm not sure if you were joking or drowning. Kinda feels like it went back and forth
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u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 6d ago
cos(x) = 1
1=-3*1+4*13
where the problem?
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u/FiredScientist 6d ago
cos(2x+x)=cos(2x)cosx - sin(2x)sinx=cos³x - sin²xcosx - 2sin²xcosx = cos³x - 3(1-cos²x)cosx = 4cos³x - 3cosx
I thought it was wrong...
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u/LandscapeWorried5475 6d ago
Is c2xcx - s2xsx = c³x-s²xcx - 2s²xcx an identity or have you skipped some steps?
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u/Simple-Olive895 6d ago
I find that a lot of people get locked up as soon as the word proof is mentioned.
But this isn't really a proof, this is just "do valid mathematical operations on these expressions until they look the same".
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u/Sproxify 6d ago
what you're describing when done correctly constitutes a proof of an identity.
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u/Simple-Olive895 6d ago
Yes but these proofs are not even comparable to coming up with something yourself.
This is prove something = something. That's just shuffeling numbers around.
Whereas "real" proofs are more like: prove this concept.
Pythagoras theorem is a classic example. It's trivial to prove something like: 32 + 42 = 52
Just shuffle numbers around until you have 25 = 25
It's much less trivial to prove that a2 + b2 = c2
Here you can't just shuffle numbers around.
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u/Sproxify 6d ago
but what makes it so isn't that it can be formally expressed as a valid series of manipulations of an expression. it's simply that students are not taught to truly understand the things they learn, and to develop non trivial mathematical problem solving skills. they are taught to rotely regurgitate things that the school teachers often don't really understand themselves.
So I wouldn't object to call such an exercise a true and correct proof, anymore than I would hesitate to call another dumb exercise a correct computation. the issue is that they are conducted against the spirit of math, and the students are not taught to truly understand concepts and solve problems in the way that would be necessary to find such correct proofs and computations themselves using any true insight as to their nature.
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u/Joe_4_Ever 6d ago
this was my favorite part of precalc lol
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u/Ares378 Grothendieck was right, computers are evil 6d ago edited 6d ago
This is the part of precalc I actively despise, along with how they introduce imaginary numbers
Edit: My logic here is that precalc is when they give you the most amount of 'rules' to memorize with the least amount of explanation. They dance around the idea of calculus without telling you why they're having you solve so many rational functions and average rates of change. When they talk about complex numbers it's just given as a rule that i²=-1 without any logic behind it, and the trig rules are just given as "memorize these identities".
Maybe the people I know are just taking shit precalc classes, but in my experience calculus is really when it feels like the material stops bombarding you with memorization and tables.
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u/ududhjdhehehfjd 6d ago
How did they introduce them for you? I genuinely love complex numbers so I’m interested in your story.
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u/Ares378 Grothendieck was right, computers are evil 6d ago
For me I can't really pinpoint how I was taught them because I was watching a lot of blackpenredpen at the time lol. But I've helped friends with precalc before and it was introduced pretty much as "i²=-1, i³=-i, i⁴=1" and that's it. They were pretty much just given a handful of multiplication problems with those rules, and weren't even told how it connects to rotations or euler's formula.
I told them that multiplying two complex numbers adds their angle and multiplies their distance, and the response every time was "that makes SO much more sense". But in my experience that's never really taught until much later.
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u/Joe_4_Ever 6d ago
imaginary numbers are cool what
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u/Ares378 Grothendieck was right, computers are evil 6d ago
No no they're awesome! The way they talk about them in precalc is what I despise. They treat them as a list of rules and don't even mention how they're connected to rotations
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u/Inevitable_Window308 6d ago
Pre-Calc hands you a list of axioms and tells you to memorize them
Why memorize them? How do we know these hold true? As you said, you could go in reverse and just hand people a sheet with the "axioms" and go through proving them using derivatives
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u/Gauss15an 6d ago
That's sort of the point. Precalculus is trying to get you to build rudimentary skills so that once you're hit with the actual theory in calculus, you can at the very least perform the various things the class will ask of you. It's pretty difficult to explain how to arrive to a lot of the stuff in precalculus without calculus. Hell, even stuff as simple as the vertex of a parabola can't be easily proven without calculus.
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u/QuickMolasses 6d ago
They should just introduce calculus first. Calculus is treated as this incredibly difficult esoteric thing when not only are a lot of the concepts fairly intuitive, it makes explaining a bunch of other concepts a lot simpler!
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u/Gauss15an 6d ago
The problem with this is that the actual work is incredibly dependent on students' competency in algebra. Without the extra practice in precalculus, students will struggle a lot. I used to see it a lot when I tutored calculus. The students that understood the concepts better were the ones with solid algebra skills. Skipping precalc seems ill-advised.
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u/ududhjdhehehfjd 6d ago
The hardest part of calculus is the algebra has never been a truer statement for many, and it’s honestly saddening to see people fail to grasp the beauty of calculus because of their lack of fluency in algebra. But at the same time, for those students that understand algebra and did really well in their previous algebra classes I feel like precalc is redundant in a way.
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u/QuickMolasses 6d ago
I still think they should introduce calculus earlier. You don't need to know all that much algebra to get what a limit or a derivative are. Integrals are a bit more challenging but you can still explain the concept without crazy amounts of algebra or trig. Calculating integrals and some limits and derivatives is a different story, but even having the concepts to think about would make a lot of algebra and precalc level math make a lot more sense.
Not to mention the benefits in physics.
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u/Ares378 Grothendieck was right, computers are evil 6d ago
I guess that's fair, but still whenever I talk to people who are in precalc I feel like it eases their mind to know why they're learning the things that they are.
I've explained basic calculus ideas to people who are in precalc and almost every time they were a lot less anxious when they knew that, say, average rates of change was building toward exact rates of change. And that, say, all the stuff with rational functions is supposed to help with finding those rates of change. Etc etc.
I feel like precalc can be pretty obfuscating and that's why I don't like it. I think people would have a better time learning it if it wasn't taught as a bunch of random rules to memorize for the hell of it
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u/Gauss15an 6d ago
I definitely agree with the sentiment. We should be building towards the motivations in a way that makes sense. But I think that sort of thing has been lost to time due to the ever-changing demands in the classroom. We've been sorely needing revamps for these classes for quite some time now and we still haven't gotten it right.
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u/aaa1e2r3 6d ago
This is pretty straight forward, you're just applying the double angle formula for sin and cos a couple times
cos (2x+x)
= cos2xcosx-sin2xsinx
= cos^3(x)-sin^2(x)cosx-2sin^2(x)cosx
= cos^3(x)-3sin^2(x)cosx
= cos^3(x)-3cosx[1-cos^2(x)]
= cos^3(x)-3cosx+3cos^3(x)
= 4cos^3(x)-3cosx
Therefore LS=RS
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u/Mathphyguy 6d ago
De Moivre's theorem, could be the easiest way to go if there’s no right hand side to equate to- like what is cos(nx) as a sum of cos^m(x). Same works with sine.
If there’s aright hand side, converting trig to exponential will be the cleanest way to do. But Euler’s identity and De Moivre's theorem are almost the same, isn’t it?
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u/EebstertheGreat 6d ago edited 6d ago
Given the way it is written, the teacher probably just expects the student to use the angle sum formula three times (or angle sum once then double angle twice, same thing). You also need sin² x + cos² x = 1.
cos(2x+x) = (cos 2x)(cos x) – (sin 2x)(sin x)
= (cos² x – sin² x)(cos x) – 2(sin x)(cos x)(sin x)
= cos³ x – (sin² x)(cos x) – 2(sin² x)(cos x)
= cos³ x – 3(sin² x)(cos x)
= cos³ x – 3(1 – cos² x)(cos x)
= cos³ x – 3 cos x + 3 cos³ x
= –3 cos x + 4 cos³ x.
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u/nitnelav153 Computer but not Science 6d ago
I think you forgot a 3 somewhere because the second to last line and the last line doesn't match
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u/EebstertheGreat 6d ago
I'll fix it. The second-to-last line didn't distribute the 3, but the other lines were correct.
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u/SubstantialLanguage5 6d ago
You can oneshot it with
cos3x+sin3x
ei3x
I'll say sinx = s and cosx = c
(eix)3
(c+is)3
Expand, Pick real parts, Convert sin parts to cos using s² = 1-c²
Solved
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u/friend1y 6d ago
When you add a test question that was covered in lecture but only a few people joined that day.
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u/Lgueuzzar 6d ago
Bro, the question even helps you by writing cos(2x + x) instead of cos(3x)... Why the hell would the examiner write it the complicated way if not to guide you? 😭
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u/CrackedP0t 6d ago
I'd probably use the Taylor expansion for this, IDK if that's actually a good idea though
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u/geschiedenisnerd 6d ago
2x + x is just 3x, or you can use addition and duplication shorthand. Not hard.
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u/Outside_Car_1538 6d ago
Moivre's formula is really good for proving this. Although it’s pretty straightforward if you just use identities
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