Prove cos2(x)+sin2(x) = 1

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Reply 220

Reminisce

Has this thread been resurrected AGAIN? :biggrin:

Reply 221

TheMagicMan

Original post by Jodin

Showing my ignorance here once again, isn't the usual way to just make use of the derivative of lnx? (which you determine by the definition of e)

(ln(x+h)-lnx)/h etc.

Then of course y = ln(x) so dy/dx = 1/x = e^-y so dx/dy = e^y and hence
d/dy e^y = e^y ?

This really depends on how you define ln tbh. The most common way to prove the derivative of lnx is 1/x is to use the result we are trying to prove.

If you use the very first definition of e (

\displaystyle e= \lim_{n \to \infty} (1+1/n)^n

), then the way of showing it is to use this to demonstrate that

\displaystyle e^x= \sum_{n=0}^\infty \frac{x^n}{n!}

. Thence

\displaystyle\frac{d}{dx}e^x= \lim_{h \to 0}(\frac{e^{x+h}-e^x}{h})=e^x \lim_{h \to 0} \frac{e^h-1}{h}=e^x

. Obviously this isn't totally rigorous in it's current form, but I find it easier than working from first principles with ln

(edited 12 years ago)

Reply 222

gff

Beaten to it. Just checked the OP.

Spoiler

(edited 12 years ago)

Reply 223

hippogriff

I have a proof which revolves around Pythagoras' theorem and right-angled triangle trigonometry. I attached in as a photo... sorry about it not being very clearly layed out, I had to put sin^2 and cos^2 because I could find a notation for power functions!