Trigonometry proof doubt

[ionic niche]

guys i found this on the internet, a similar proof is given in my textbook, I'm here feeling dumb for a while now where I'm trying to generalize this for reflex angles and angles more than 2pi but I don't know how, I made an example diagram with reflex values of a and b (and instead of a-b i did a+b like in my textbook) the angle opposite to the side 'd' (should be equal for d to be equal) to be a+b-2pi for both triangles, but how do I generalize it I do not understand

[queen jasper]

So you want to prove this for all angles?

[ionic niche]

yes

[queen jasper]

The proof seems to work for every angle. So I'm a little lost on which case exactly you are struggling with

[ionic niche]

it ofc works for all cases but I cannot imagine infinite angles on the unit circle

different orientations

[queen jasper]

Infinite angles?

[ionic niche]

I mean to say there are infinite possibilities for a and b

and hence you cant make a diagram for each

[queen jasper]

You don't have to. But the process is exactly the same. No matter your angles

I guess I'm a little lost on your concern

[drowsy osprey]

For example 30° is the same as 390°

[ionic niche]

I know adding pi wont change the geometery

but like an angle between pi and 3pi/2

or

between pi/2 and pi

or 3pi/2 and 2pi

and we have 4*4 possibilities to be proved (4 quadrants in which each a and b angles can lie in)

so if I prove it for these 16 possibilities I can say its always true

but that is tedious

[queen jasper]

You don't have to prove the 16 cases

That's what the unit circle proof here is taking care of for you

The process is exactly the same

[ionic niche]

basically this problem will be simplified if I can show that the angle between the vectors (1,0) ,(cos(x+y) , sin(x+y)) AND (cosx ,sinx ) , (cosy, -siny) is always same for all values of x and y

(the angles opposite to the side d in both diagrams)

[ionic niche]

if we do this, then we can find the cosine of the angle between the vectors using dot product and equate them to get the identity

ig im overthinking

I just wasnt getting that feeling os satisfaction on seeing that proof and feeling that this applies to all angles intuitively, although rationally I know I can show it for any angle I want

[queen jasper]

I disagree that the problem will be simplified at all. The proof already covers all cases, and you are trying to show that some cases are degenerate, which is not necessary. You are adding work to the proof, not reducing it.

[queen jasper]

It sounds like you're still thinking of sine and cosine as defined by ratios on a right triangle, whereas here, there'd defined as coordinates on the unit circle

[ionic niche]

no I dont think of them that way

i know they are coordinates thats why I used this vector idea

[queen jasper]

This tells me that you are.

[ionic niche]

i am what

[queen jasper]

trying to use vector dot product will only work when your angles are not greater than 180 degrees.

You're making it harder to solve the problem

And this is a restriction unique to the right triangle definitions of sine and cosine

[ionic niche]

not really since they still lie on the same unit circle no matter how big the angle

and so the angle between them geometrically will always be less than or equal to pi

[queen jasper]

This is the restriction I am talking about

you are making the problem harder this way

[ionic niche]

hmm

so basically there doesn't seem to be a more fundamental way to derive these identities

[queen jasper]

It'll be correct, yes. But it's now going to take extra work because you need to consider the (pi, 2pi) case separately now.

[queen jasper]

For all angles, no

[ionic niche]

okk

.close