Derivative of trigonometric functions

1) lim as theata approaches 0 --> 1-cos theata/theata^2

2) lim as theaata approaches 0 --> tan theata/2 theata


Please explain how each is done please. THank yoU! I am supposed to determine the limits for each.

first of all, it's "theta"

it's a lot better than some people though :D

i don't mind mispelling as much as some people though, but now you know, so you can spell it right next time

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Originally Posted by superweirdo

1) lim as theata approaches 0 --> 1-cos theata/theata^2

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one way that everyone will hate is to apply L'Hopital's rule:

thus, $\displaystyle \lim_{\theta \to 0} \frac {1 - \cos \theta}{\theta ^2} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta}$

i think you can take it from there. Hint: think "special limit"


if you want to do it without L'Hopital's, power series is another method. too much work here though, i think

(do you know what L'Hopital's rule is? do you know how to do it using power series?)

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2) lim as theaata approaches 0 --> tan theata/2 theata

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$\displaystyle \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta}$ is also a special limt (without the 2 in the denominator that is, but it doesn't matter, we can factor constants out).

if you want to do it the hard way, you can note that: $\displaystyle \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta \cos \theta}$

Hello, superweirdo!

We need the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}\;=\;1$

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$\displaystyle 1)\;\;\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}$

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Multiply by $\displaystyle \frac{1+\cos\theta}{1+\cos\theta}$: . $\displaystyle \frac{1-\cos\theta}{\theta^2}\cdot\frac{1+\cos\theta}{1+\c os\theta} \;=\;\frac{1-\cos^2\!\theta}{\theta^2(1+\cos\theta)} \;=\;\frac{\sin^2\!\theta}{\theta^2(1+\cos\theta)}$

then: . $\displaystyle \lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac {1}{1+\cos\theta}\right] \;=\;1^2\cdot\frac{1}{1+1} \;=\;\frac{1}{2}$

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$\displaystyle 2)\;\;\lim_{\theta\to0}\frac{\tan\theta}{2\theta}$

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We have: .$\displaystyle \frac{\frac{\sin\theta}{\cos\theta}}{2\theta}\;=\; \frac{\sin\theta}{2\theta\cos\theta}
$

Then: . $\displaystyle \lim_{\theta\to0}\left[\frac{1}{2}\cdot\frac{\sin\theta}{\theta}\cdot\fra c{1}{\cos\theta}\right] \;=\;\frac{1}{2}\cdot1\cdot\frac{1}{1}\;=\;\frac{1 }{2}$

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Originally Posted by Soroban

Hello, superweirdo!

We need the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}\;=\;1$

Multiply by $\displaystyle \frac{1+\cos\theta}{1+\cos\theta}$: . $\displaystyle \frac{1-\cos\theta}{\theta^2}\cdot\frac{1+\cos\theta}{1+\c os\theta} \;=\;\frac{1-\cos^2\!\theta}{\theta^2(1+\cos\theta)} \;=\;\frac{\sin^2\!\theta}{\theta^2(1+\cos\theta)}$

then: . $\displaystyle \lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac {1}{1+\cos\theta}\right] \;=\;1^2\cdot\frac{1}{1+1} \;=\;\frac{1}{2}$



We have: .$\displaystyle \frac{\frac{\sin\theta}{\cos\theta}}{2\theta}\;=\; \frac{\sin\theta}{2\theta\cos\theta}
$

Then: . $\displaystyle \lim_{\theta\to0}\left[\frac{1}{2}\cdot\frac{\sin\theta}{\theta}\cdot\fra c{1}{\cos\theta}\right] \;=\;\frac{1}{2}\cdot1\cdot\frac{1}{1}\;=\;\frac{1 }{2}$

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