Most often they are used for simplifying expressions. For that purpose it's not a matter of deducing what they are, but recognizing them in the expression, and spotting the opportunity to use the identity on the path to eliminating terms.
For that purpose one needs to have them memorized in order to spot the opportunity -on the carefully derived homework problem- in the wild.
If I'm stuck with a trig function I don't know what to do with, I just look on Wikipedia for a trig identity and don't worry about derivations or memorization.
This trick is good for exam settings though! My memory is terrible so I always relied on tricks like this in school
Yes, most high schools spend a few months teaching “trigonometry” as a system for simplifying formulas that was primarily useful historically to save time for human calculators working with pen/paper and a printed lookup table. In the modern world this is entirely anachronistic and the content could be profitably compressed down to a couple weeks, with the saved time spent on something else. The main remaining utility is as rather cumbersome, unpleasant, and unenlightening algebra practice.
In the same sense that doing this work by hand will build depth of understanding, I know that remembering the trig identities came before really understanding their relationships. And that also came far before I could truly understand Euler's formula, radians, etc.
"i know two things:
[cos(x)]^2 + [sin(x)]^2 = 1
i*i = -1
but can re-derive all the rest"
He said it relates the most important numbers and constants of the universe. He had it embroidered and framed on his office wall
The part that makes the expression flashy is that the rotation is written via its logarithm, which can be represented as the length of a circular arc (“angle measure”).
It’s a useful fact to know, but is vastly overrated.
So you need at least that one, since it is by convention, and not really derivable[1]:
exp(ix) = cos(x) + i*sin(x)
[cos(x)]^2 + [sin(x)]^2 = 1
[1] OK, you can look at the taylor series expansion, but you need to remember the derivatives for this.
Because x^2 + y^2 = 1 defines a unit circle, if you perform x^2 + i * y^2 = 1, you get a unit circle over the real/imaginary planes.
Or to put it in a picture: https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Eu...
Remember the definition of sin / cos: https://upload.wikimedia.org/wikipedia/commons/b/bd/Sine_and...
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Once you realize that "the imaginary axis" is just an arbitrary 2-dimension extension field (and that "imaginary" is a very bad name for it), it becomes way easier to see.
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Now if you had a 3-dimension graph (real, imaginary, and time), and then you project the unit-circle over the real + time axis, you get a sin (or cos) respectively.
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Or to put it another way: there's a periodic nature of i:
* i = i
* i^2 = -1
* i^3 = -i
* i^4 = 1 -- Cycle-length 4
* i^5 = i
* i ^ x = i^(x mod 4) (In general)
This period itself forms a circle, moving from real-to-imaginary and back. (e^i^x) = (e^(ix)) therefore is periodic as well. And the most natural periodic cycle is a circle (which is of course, just sin/cos).
To make this "period 4" into a proper circle, multiply by pi/2.
i^(pi/2)^x == period 3.141592.... or the "circumference" of the unit circle.
e^i^(pi/2)^x == e^(i * pi/2 * x). Done.
(Though, yes, there's a bit of a connection with group theory, but not enough to help you.)
From there, you get extension fields from real vs imaginary already. (Ex: you can form a new extension field from x + y*j, where x and y are complex numbers), which forms a new periodic cycle.
I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts.
But that's just a way to embed some of group theory. It doesn't actually help you much.
(It's similar to how you can use eg set theory to construct the integers. It's possible, but doesn't actually help you prove anything about interesting about integers that you wouldn't have been able to prove without embedding them like this.)
> I mean, deriving it all is hard because group theory is hard. I'm not sure if its because the tools "aren't there". Some super-AI or super-human probably can derive it all from those given facts.
Yes, but that task wouldn't be made easier by this approach compared to starting from just the group theory axioms instead.
e^(iθ) times e^(-iθ) = e^(iθ-iθ) = e^0 = 1 = (cosθ+isinθ)(cosθ-isinθ) = cos²θ + sin²θ
Why do people even have to memorize stuff like this? I simply don't understand what's wrong with referencing stuff you don't remember during a test. That's what we do in real life!
Incidentally, we can derive [cos(x)]^2 + [sin(x)]^2 = 1 from the fact that cos and i * sin are even and odd components of exponentiation with some base, like so: 1 = exp(ix) * exp(-ix) = (cos(x) + i * sin(x)) * (cos(x) - i * sin(x)) = [cos(x)]^2 + [sin(x)]^2.
Any complex number can be written as an exponent like rexp(ix), where x is the angle and r is the distance from the origin in the complex plane. Using this understanding, we can interpret complex number multiplication as a rotation in the complex plane:
Ok, the above was a digression, let's get to the trig identities. Let's derive the double angle formula. We can start with the point on the unit circle representing an angle of 2x, and use Euler's formula to simplify it one way: Ok, we can also use the fact that a^(bc) = (a^b)^c to write the same expression a different way, then FOIL it: Now, comparing the two expressions, we see that Taking real and imaginary parts of both sides, we find that: This works for most identities you want to derive, and there is a variation for things like cosh and sinh. This trick came in handy on calc 3 exams whenever I couldn't remember the silly trig identities that were required to perform integrals.