Lmao that's a heavy ass British accent.

easier proof: sum-to-product formula, making use of sinA - sinB = 2sin(A-B /2)cos(A+B /2) and sub A = x+h, B = x, making it (sin(h/2)cos(x + h/2)) / (h/2), directly getting cosx

@samueldeandrade8535

14 күн бұрын

Yep.

Complex definition is so elegant if extending the derivative to the complex plane and holding derivative rules true.

Bro why do you sound different?

nice video

underatted

wow so mis leading here was me thinking this would be a deep video

Using eulers formula is much easier to prove and to understand.

Final 2 limits are easily seen from Taylor series / small angle approximations, sin(h) ~ h and cos(h) ~ 1 - 0.5h^2 as h->0

@EK-bn7jz

11 күн бұрын

correct me if I'm wrong, but I don't think you can use the Taylor series or small angle approximations for sin(x) here as they rely on the derivative of sin(x) being cos(x) which is circular reasoning

@adw1z

11 күн бұрын

@@EK-bn7jz u are half right, it depends which definition u start with. For example, if u start with complex exponential definitions for sine and cosine, u can derive the Taylor series without knowing derivatives of sine and cosine. U can even start by defining sine and cosine with their Taylor Series themselves, as they are unique representations for sin and cos

@EK-bn7jz

9 күн бұрын

@@adw1z interesting, thank you

Use Taylor expansion

@thundercraft0496

14 күн бұрын

That's circulatory The taylor series uses the derivative of the function

@kurrennischal235

13 күн бұрын

@@thundercraft0496 I don’t think it’s circular. If we define sin(x) as the Taylor expansion definition then it’s fine. The difficulty then is having to prove that the Taylor definition is equal to the unit circle definition.

@adw1z

13 күн бұрын

@@thundercraft0496it is not circular. Use the Taylor series for e^x and then compute (exp(ix) - exp(-ix))/(2i)

@thundercraft0496

13 күн бұрын

@@kurrennischal235 well by that definition we wouldn't even need to prove that the derivative of sinx is cosx we could just define cosx as the derivative of sinx

@thundercraft0496

13 күн бұрын

@@adw1z well that also uses the taylor series of sinx When you have the taylor series of e^x substituting ix to x well result to the taylor series of cosx and sinx to appear