Well I'm allowed to use anything, I'm just doing it for my own satisfaction. Tho i have no idea about Taylor series.is it possible to clear the basics in a few hours?.Scott said:
You could always define ##\sin## and ##\cos## using the Taylor expansion for ##e^{ix}## and Eulers formula.haushofer said:
PeroK said:
It makes sense if you want to verify that your definitions using the Taylor series are the same as the precalculus trigonometry definitions of sin() and cos() that the OP was almost certainly referring to.Dragon27 said:
To answer the OP question, it is best to deal with their question instead of a question that you wish had been asked.Dragon27 said:
@pasmith already answered the OP's question as it was likely intended, but it doesn't hurt to point out that a proof depends on how the trig functions are defined and that there is more than one way to define them.FactChecker said:
The limits that a left for your last step can be found without any derivatives.Mr X said:
Well, tbh, I felt encouraged to try and probe more into the assumptions of the question by the fact that the assumptions were more guessed at, than clearly stated, and by this statement:FactChecker said:
Mr X said:
Dragon27 said:
The projection of a unit vector onto the X-axis is easily made rigorous. I think it is just as "rigorous" and more intuitive than the power series. The power series requires a lot of work to make it "rigorous". To start with, limits and convergence need to be defined and established for that series.Dragon27 said:
Well, yes. For this you need some existence and uniqueness theory for ODEs, but that's not too difficult for these equations.wrobel said:
But you still have a lot of work to define angles rigorously. Analytical definition of sine/cosine is, imho, comparatively less effort.FactChecker said:
Not really. In an inner product space, the "angle" is the relative orientation of a unit vector that gives a certain inner product with a unit vector along the X-axis.Dragon27 said:
How do you define angle measure?FactChecker said:
That's the downside!pasmith said:
Thank you, I understood that one.DaveE said:
The derivative of sin(x) is cos(x).
To derive the derivative of sin(x) using the limit definition, we start with the limit definition of the derivative: \[ \frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} \]Using the sum-to-product identities, this becomes:\[ \frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} \]\[ = \lim_{h \to 0} \left[ \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} \right] \]As \( h \to 0 \), \(\frac{\sin(h)}{h} \to 1\) and \(\frac{\cos(h) - 1}{h} \to 0\), so:\[ = \sin(x) \cdot 0 + \cos(x) \cdot 1 \]\[ = \cos(x) \]Thus, the derivative of sin(x) is cos(x).
Yes, the chain rule can be used to derive the derivative of sin(x). Consider the function \( y = \sin(x) \). Let \( u = x \), then \( y = \sin(u) \) and \( \frac{du}{dx} = 1 \). By the chain rule:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]Since \( \frac{dy}{du} = \cos(u) \) and \( u = x \), we get:\[ \frac{dy}{dx} = \cos(x) \cdot 1 = \cos(x) \]Therefore, the derivative of sin(x) is cos(x).
The derivative of sin(x) equals cos(x) because of the way the sine and cosine functions are defined and their properties in trigonometry. The derivative measures the rate of change of a function, and the rate of change of the sine function at any point x is given by the cosine of x. This relationship can be derived from the limit definition of the derivative or understood through the geometric interpretation of the unit circle, where the cosine
