Osvaldo Apodaca
2021-12-30
Answered

How can I evaluate $\underset{v\to \frac{\pi}{3}}{lim}\frac{1-2\mathrm{cos}v}{\mathrm{sin}(v-\frac{\pi}{3})}$ without using LHospitals rule?

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Medicim6

Answered 2021-12-31
Author has **33** answers

You can use the fact that

$2\mathrm{cos}v=2\mathrm{cos}((v-\frac{\pi}{3})+\frac{\pi}{3})$

$=\mathrm{cos}(v-\frac{\pi}{3})-\sqrt{3}\mathrm{sin}(v-\frac{\pi}{3})$

It follows from this that

$\frac{1-2\mathrm{cos}v}{\mathrm{sin}(v-\frac{\pi}{3})}=\frac{1-\mathrm{cos}(v-\frac{\pi}{3})}{\mathrm{sin}(v-\frac{\pi}{3})}-\sqrt{3}$

and therefore all that remains to be done is to compute

$\underset{t\to 0}{lim}\frac{1-\mathrm{cos}t}{\mathrm{sin}t}$

and for this you can use the fact that

$1-\mathrm{cos}t=2{\mathrm{sin}}^{2}\left(\frac{t}{2}\right)\text{}\text{}\text{and that}\text{}\mathrm{sin}t=2\mathrm{sin}\left(\frac{t}{2}\right)\mathrm{cos}\left(\frac{t}{2}\right)$

It follows from this that

and therefore all that remains to be done is to compute

and for this you can use the fact that

temnimam2

Answered 2022-01-01
Author has **36** answers

just use Taylor expansions:

$\mathrm{cos}x=\frac{12}{-}\frac{\sqrt{3}\cdot (x-\frac{\pi}{3})}{2}+o(x-\frac{\pi}{3}),\text{}x\to \frac{\pi}{3}$

$\mathrm{sin}x=x-\frac{\pi}{3},\text{}o(x-\frac{\pi}{3}),\text{}x\to \frac{\pi}{3}$

$\underset{x\to \frac{\pi}{3}}{lim}\frac{1-2\mathrm{cos}x}{\mathrm{sin}(x-\frac{\pi}{3})}=\underset{x\to \frac{\pi}{3}}{lim}\frac{1-1+\sqrt{3}(x-\frac{\pi}{3})+o(x-\frac{\pi}{3})}{x-\frac{\pi}{3}+o(x-\frac{\pi}{3})}=\underset{x\to \frac{\pi}{3}}{lim}\sqrt{3}+o\left(1\right)=\sqrt{3}$

Vasquez

Answered 2022-01-08
Author has **460** answers

Use Prosthaphaeresis fromula and

Now cancel out

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P(t) models the distance of a swinging pendulum (In CM) from the place it has travelled t seconds after it starts to swing. Here, t is entered in radians.

$P\left(t\right)=-5\mathrm{cos}\left(2\pi t\right)+5$

What is the first time the pendulum reaches 3.5 CM from the place it was released?

Round your final answer to the hundredth of a second.

Ok, so I did this to get the solution:

$3.5=-5\mathrm{cos}\left(2\pi t\right)+5$

$-1.5=-5\mathrm{cos}\left(2\pi t\right)$

$0.3=\mathrm{cos}\left(2\pi t\right)$

$\mathrm{cos}-1\left(0.3\right)=2\pi t$

$\frac{{\mathrm{cos}}^{-1}\left(0.3\right)}{2\pi}=t$

The fact is this gives me 11.55 seconds to get to 3.5 CM, which does not sound right. Where did I go wrong and how can I fix it?

What is the first time the pendulum reaches 3.5 CM from the place it was released?

Round your final answer to the hundredth of a second.

Ok, so I did this to get the solution:

The fact is this gives me 11.55 seconds to get to 3.5 CM, which does not sound right. Where did I go wrong and how can I fix it?

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I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into$\mathrm{sin}2x$ but then I have ${\mathrm{sin}}^{2}x$ and $\mathrm{sin}2x$ together.

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Could you solve it from opposite?

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Could you solve it from opposite?

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