Wikipedia:Reference desk/Archives/Mathematics/2006 October 20

=October 20=

How do I prove this?
How do I prove that d/dx (cosx)=-sinx?

Thank you!

-EggYolk


 * If you're working from first principles, you need to use the formula $$\cos(x+y) = \cos x\cos y - \sin x\sin y$$, as well as the small angle approximation (that $$\lim_{x \rightarrow 0}{(\sin x)/x} = 1$$), and then the only tricky part is a term where you have to prove that $$lim_{x\rightarrow 0}\frac{\cos x - 1}{x} = 0$$, and I can't recall the proof for that offhand. Of course, if you already have your proof that (sin x)' = cos x, you can use the chain rule: $$\frac{d\cos x}{dx} = \frac{d\sin(\pi/2 - x)}{dx} = -\cos(\pi/2 - x) = -\sin x$$. Confusing Manifestation 05:15, 20 October 2006 (UTC)

$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{x} \frac{\cos x + 1}{\cos x + 1} = \lim_{x\rightarrow 0}\frac{\cos^2 x - 1}{x(\cos x + 1)} = \lim_{x\rightarrow 0}\frac{-\sin^2 x}{x(\cos x + 1)}$$

I'll let you finish the proof that $$lim_{x\rightarrow 0}\frac{\cos x - 1}{x} = 0$$ from there. Yes, it does need the small angle approximation. -- Ķĩřβȳ ♥  Ťįɱé  Ø  05:57, 20 October 2006 (UTC)

Also, you can check out this newly created article, lol. -- Ķĩřβȳ ♥  Ťįɱé  Ø  08:53, 20 October 2006 (UTC)


 * A more geometric argument begins with a unit circle. For any parametric curve, the instantaneous movement at any point is a vector tangent to the curve. If the curve is parameterized by arclength, the tangent vector will have unit length. On the circle, that tangent is the radial vector to the point, rotated counterclockwise 90°. For the radial vector (cos &thetasym;,sin &thetasym;), the tangent is (−sin &thetasym;,cos &thetasym;). This simultaneously gives the derivative of cosine and of sine.
 * This argument uses concepts and methods not available in first-year calculus, but the insights can still be of use. --KSmrqT 09:00, 20 October 2006 (UTC)
 * Not that your point isn't valid, but the idea that geometry is inherently more advanced than formal calculus seems to be a property of the Western system of elementary math education, rather than a reflection of some purely mathematical stratification. Tesseran 15:24, 20 October 2006 (UTC)
 * Perhaps I did not make myself clear. It is not the geometry that is learned later, but the use of vectors and differential geometry. --KSmrqT 19:38, 20 October 2006 (UTC)