Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0. I can sort of see from the graph of y=sinx that it (looks like it) tends to y=x but is there a better reason?
Could you use eix=cosx+isinx to give sinx=2ieix−e−ix and cosx=2eix+e−ix and note that differentiating this expression for sin x gives the expression for cos x, or does this formula presuppose Taylor's expansion (the only way I can think of proving it does)?
Edit: the wierd s-like thingies in the exponents in the formulae are actually x's which have been mutilated by being rendered at too low a resolution.
Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0. I can sort of see from the graph of y=sinx that it (looks like it) tends to y=x but is there a better reason?
You can prove it geometrically by drawing the unit circle and considering a sector with angle 0<x<pi/2, then comparing the areas of the inscribed triangle, the sector and the circumscribed triangle. That would give you the inequality: cos(x)sin(x)/2 <= x/2 <= tan(x)/2
Which can be simplified to: cos(x) <= sin(x)/x <= 1/cos(x)
Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.
It's a bit complicated proof, using geometry as I studied before. But maybe there's another way (I don't mention about Taylor series or L'Hopital as it might not be taught in your degree)
It's a bit complicated proof, using geometry as I studied before. But maybe there's another way (I don't mention about Taylor series or L'Hopital as it might not be taught in your degree)
yar thats L'hospital's territory, if you have a limit with indeterminate form 0/0 or inf./inf., and both sin x and x go to zero, but dx/dx = 1 so there is no circular argument in this path as the denominator vanishes! we win!
Because sin(x) is defined as Σn=0∞(−1)n(2n+1)!x2n+1. Theorems about power series tell us that this has infinite radius of convergence, and thus its derivative has infinite radius of convergence. Its derivative is Σn=0∞(−1)n(2n)!x2n, which is the definition of cos(x).
Because sin(x) is defined as Σn=0∞(−1)n(2n+1)!x2n+1. Theorems about power series tell us that this has infinite radius of convergence, and thus its derivative has infinite radius of convergence. Its derivative is Σn=0∞(−1)n(2n)!x2n, which is the definition of cos(x).