This thread is dedicated to solving trig identities. If any of ya'll have any problems which are stumping you I can probably help. (I haven't done calc in a very long time, however I am still extremely good at trig, except for polar coordinates). If any of ya'll need some help with trig identities, just post your problems here and let's see if I can solve them.
>>16793489please explain to me the intuition behind radian, arc and degree. thank you.
>>16793510No problem! I'll try my best (but just in case, do review what I've said).> Intuition behind radiansSo, the idea behind the radian is that one rad is equal to the radius of the circle in question. This means that when you have one rad of a circle (which is approximately 1/6 of a circles circumference). The arc length of that section should be exactly equal to the radius of the circle.The whole idea behind radians vs degrees is that, although degrees are a system that was historically used for trigonometry and geometry, radians are more closely related to the shape itself. Hence the reason why you'll typically be forced to use them when dealing with calculations. (Like arc length, which is s = rθ)In which you can't use degrees to calculate s (the arc length). You have to use radians instead. (Also, a circle has a total rotation of 2pi, so it's easier to use pi as a reference to angular velocity as well. Which is simple rotations per unit measurement of time).> Intuition of an arc?So, the arc of a circle is simply a section of it. I briefly mentioned this before but the whole idea of an arc is that you can use it to cut sections out of the unit circle.For example, if you wanted to know what the total linear velocity a circular object has traveled in a specific direction (Such as a wheel. Or more specifically a the portion of the wheel being driven on) all you'd have to do is simply divide the arc length (distance traveled) over the time that it took. That's how you'd be able to get that. Funnily enough, this is effectively the same as multiplying the angular velocity (rotations per unit of time) by the radius. So take that as you will.
>>16793522thank you, I'll think about it for a few more hours
>>16793489just rewrite all the trig shit with exponentials and see if everything cancels out>>16793522thank you chatgpt
>>16793489A better image for this thread.
please solve the following identity.[math]\cos \alpha - \cos \beta = -2 \sin \frac{\alpha+\beta}{2} \sin\frac{\alpha-\beta}{2}[/math]
>>16794505wtf
>>16796187i remembered this equation from trig sum to product identities section. i crammed the damn thing and mever thought about it's proof.
>>16794505i made an anki card for this literally the other day.
>>16794505A = exp(\alpha i/2)B = exp(\beta i/2)(A^2+1/A^2)/2 - (B^2+1/B^2)/2 = -2 [(AB-1/AB)/2i][(A/B-B/A)/2i]multiply out rhs and everything cancels
>>16796661>>16796696that's concise, thank you.
>>16793489I cannot explain how much I hate trigonometry.
>>16796661[math] \displaystylee^{ix}= \cos(x)+ i \sin(x) \\e^{ix}e^{iy} = e^{ix+iy} = e^{i(x+y)} = \cos(x+y)+ i \sin(x+y) \\e^{ix}e^{iy} = ( \cos(x)+ i \sin(x))( \cos(y)+ i \sin(y)) \\ ~~~ = \cos(x) \cos(y)+ \cos(x) i \sin(y)+ i \sin(x) \cos(y)+ i \sin(x) i \sin(y) \\~~~ = \cos(x) \cos(y) - \sin(x) \sin(y)+ i (\sin(x) \cos(y)+ \cos(x) \sin(y)) \\Re: \cos(x+y) = \cos(x) \cos(y) - \sin(x) \sin(y) \\Im: \sin(x+y) = \sin(x) \cos(y)+ \cos(x) \sin(y)[/math]
[math]\underbrace{\underbrace{\big(2(\frac{(-a + b)}{(a + b + 3)}) +\frac{1}{2}\big)}_{radians}\times90}_{angle} [/math]
>>16799244what the fuck?
>>16799259yeah, galo sengen
