Mathematics - 2016-12-23 (page 5 of 5)

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7:00 PM

I really need to cook now, bye

= (√3 cos 20° - sin 20°) / (sin 20° cos 20°)
= 2 (sin 60° cos 20° - cos 60° sin 20°) / (sin 20° cos 20°)
can these steps be expanded?

@ArmaGeddON What are $\sin 60^\circ$ and $\cos 60^\circ$?

(√3 cos 20° - sin 20°) = 2$\cos(\theta\plus\phi $

√3 / 2 and 1/2 respectively

omg finally got it

@Semiclassical?

ya?

Were you teaching me the easier way to do this?

7:07 PM

Depends on what you mean by 'easier.'

i think you were teaching me something of my wildest dream :p

Hah. Do you mean the complex-number approach?

well i got the answer now and solution too

thanks anyways

all

next question

The value of $\sin 10\circ + \sin 20\circ + ... + \sin 360\circ$ is?

isn't my question clear?

It's clear. I'm just not sure what kind of help to give. The simple solution is via Euler's formula.

The answer is 0

7:14 PM

Oh, right. Yeah, there's a really simple solution.

Namely, sin(180+x) = -sin(180-x)

is it $\sin(a+b) = -\sin(a-b)$ ?

That'd imply $\sin(b+a)=\sin(b-a)$ which is generally false.

okay

Maybe think of it like this. It's easy to compute $\sin 360^{\circ}$: it's the same as $\sin0^\circ = 0$.

How about $\sin 350^\circ$?

sin 10?

or - sin 10?

7:18 PM

the latter.

seemed so

It has to be the same as $\sin(-10^\circ)$ by periodicity, and the minus sign can be pulled out of sine.

ok

so now?

What does that suggest about sin(340), sin(330), etc.?

alright, everything got subtracted and nothing left in hand

trigonometry is interestin

7:21 PM

right. one needs to make sure that everything actually cancels, of course.

next

for instance, you can cancel sin(170) off against sin(190).

yeah makes sense

but there's nothing to cancel off sin(180). that's not a problem, though, since sin(180)=0.

$\sin(180) = \sin(90 + 90) = \sin(90) cos(90) + cos(90) sin(90) = 0$

wanna change chapter?

7:24 PM

well, sure. or: sin(x)=sin(180-x), so therefore sin(180)=sin(0)=0.

not really.

i have problems in Trigonometry, Similar Triangles and Heights and Distances

wanna stick to T ?

then the value of $\cos(1\alpha)$ is?

$\cos(2\alpha)$

if

i think i'm going to pass on problems for now, actually.

$cos(\theta-\alpha) = 3/5$ and $sin(\theta+\alpha)=12/13$

7:49 PM

well, 3 hours and still no answer, someone could give my question a try ?
http://math.stackexchange.com/questions/2069774/determine-if-the-set-of-all-eventually-zero-sequences-is-g-delta-in-ell-2

Alessandro Codenotti

$G_\delta$ and $F_\sigma$ are countable union of closed sets and countable intersection of open sets, right?

$G_\delta $ is countable intersection of open sets , and $F_\sigma $ is countable union of closed sets.

3 hours later…

LegionMammal978

10:40 PM

Anyone immediately recognize the sequence 9, 32, 75, 144, 245, 384, 567, 800, 1089, 1440, ...?

@LegionMammal978 n*(n+2)^2

LegionMammal978

@ZachHauk Thanks, these are actually denominators that cause weird patterns in base 2, 3, 4, etc.

@LegionMammal978 The OEIS gives the same answer as Zach: link

11:50 PM

@Semiclassical, I don't O_o. But if we take the ceiling, we have f ^{t-2}*n mod f. Versus taking the floir f^{t-1}* n mod (f-1)$

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