#Solve sin(pi-theta)

[vash3r]

my answer

first estimate, something on the order of a few light-seconds. some quick points of comparison from numbers i have memorized:

1 light-second: ~399,000,000 m/s * 1s = ~400,000 km

sun-earth distance: 1 AU (?? km i forgor), a few light-minutes (4~8?)

diameter of earth: ~40,000 km (radius ~10,000 km)

distance to "earth orbit": ~100km? (atmosphere: ~10km)

moon radius approx 1/4 that of earth? def at least 1/10, moon is a very big moon (=> because volume is cube, only 1/64th the mass: seems an appropriate order of magnitude)

from "Eclipse" (vaguely remembered): moon-sun distance ratio (~=radius ratio, due to eclipses) is about 1:400 ?

refined guess after thinking a little: 1-4 light-seconds.

now it's time to get nerdsniped.

recall equation for gravity: F= (G*m1m2/d^2). i forgot G but we can skip it because we know g=9.8 at surface of earth (1e4 km); at moon orbit (estimated 4e5 km) we need the following to hold: that the moon orbits the earth approx 1x every 29 days (sidereal but let's ignore that for now) (29*86400=30*_-_ =2592000-86400s = 2,508,000s).

let's introduce some variables:

Rmo[radius of moon orbit] = (unknown, estimated 4e8m)

Re[radius of earth] = 1e7 (m)

Cmo[circumference of moon orbit] = Rmo*2pi (m)

Tmo[time of moon to orbit once] = approx 2.5 e6 (sec)

Vmo[velocity of moon in orbit] = Cmo/Tmo = 2pi*Rmo/Tmo (m/s)

Ame[acceleration of moon due to Earth gravity] = 9.8 m/s^2 * (Re/Rmo)^2

we want to solve an equation for circular motion for radius given the centrifugal/centripetal acceleration and orbital period. i forgot the equation though and sadly i can't just do trig on a tiny triangle (not only because i don't have arctan tables memorized).

calculus to the rescue? let's solve the integral of acceleration along the circumference for a quarter-turn (because we know the resulting change in velocity): starting at the standard spot (+Rmo, 0) with velocity [0, +v] (in the positive-y direction) and going counterclockwise: integrate S[0 to pi/2] (a [-cos theta, -sin theta] dtheta) = [-v, -v]. except instead of just integrating along angle, we want to integrate along time; so

S{0 to Tmo/4} (a [-cos (2pi*t/Tmo), -sin (2pi*t/Tmo)] dt) = [-v,-v]

a * (Tmo/2pi) * |0 to Tmo/4| [-sin(2pi*t/Tmo), cos(2pi*t/Tmo)] = [-v,-v]

a * (Tmo/2pi) * [-1,-1] = v*[-1,-1]

a * Tmo/2pi = v (in units of seconds). we've successfully rederived the circular motion equation! now substituting in v=Cmo/Tmo and a=Ame, we get

Ame = 2pi * Cmo/Tmo^2

9.8 *(Re/Rmo)^2 = (2pi)^2 * Rmo/Tmo^2

Rmo = cbrt(Tmo^2 * Re^2 * (2pi)^2 * 9.8)

simplifying to order of magnitude:

Rmo = cbrt(2.5e6^2 * 1e7^2 * 2^2 * pi^2 * 1e1)

Rmo = cbrt((5^2)* 1e12 * 1e14 * 1e1 * 1e1)

Rmo = cbrt(25e28) ~ 2.9e9.3 m, or approx 600,000 km, about 1.5 lightseconds. without using books, internet, or a calculator.

EDIT: turns out while my estimates in terms of lightseconds were relatively accurate, many of my other numbers were super wrong: lightspeed (which is 299,xxx,xxx m/s instead of 399,xxx,xxx), saying "earth diameter" when i meant circumference, using earth diameter in place of radius, multiplying by 2pi^2 instead of dividing, and forgetting an order of magnitude in the answer lmao, so the formula should be

Rmo = cbrt(Tmo^2 * Re^2 * 9.8 / (2pi)^2) = cbrt(1.4e12 * 36e12) = cbrt(54e24) and from 54 between 27 and 64, we get approx 3.8e8 which is practically spot on

Guesstimation game: no googling!

How far away is the moon?

No using the web, books, maps, etc. No checking the notes!

Put your guess and reasoning behind a readmore, so your followers also get a chance to play.

The goal isn't to get the RIGHT answer (or even get close), it's to see how you and others come up with a guess. It's also not a competition, don't worry about how you do in relation to others!

[canmom]

(tl;dr - this is a post about a cool thing I did in Blender for my job.)

It turns out you can 100% use scipy in blender to numerically solve a differential equation and use that to drive the animation of an object and it’s not even that hard. There’s a few quirks - I couldn’t just index the array I produced with the ‘frame’ member of the driver namespace, but I had to write a custom function to look up the frame and index the array. But I got it working! And it looks really cool.

In this case, I was rendering an animation to show a magnetic dipole (a loop of wire carrying a current) flipping from its unstable equilibrium position to the stable one and settling there (with damping). This has exactly the same equation of motion as a pendulum in a uniform gravitational field. Solving it exactly requires some weird special functions, and it's much much much easier to treat it numerically...

To get it working in Blender, I wrote a Python script, which generates the angle \(\theta\) that the dipole turns to over 6000 time steps using SciPy’s numerical integration. Then, it takes every 40th time step to get the angle for each frame of animation, and creates a custom function with the data in scope which looks up the current frame and uses that to index the array of angles.

import bpy import scipy.integrate as integrate import numpy as np pi = np.pi sin = np.sin damping = 0.15 spring = 0.04 def derivative(theta_vector, t): theta, thetadot = theta_vector return [thetadot, - damping * thetadot - spring * sin(theta)] t = np.linspace(0, 120, 6000) theta_vector_init = [pi-0.001, 0] theta_vector = integrate.odeint(derivative, vinit, t) theta, thetadot = theta_vector.T theta_frames = theta[0::40] def th(): return thetaframes[bpy.context.scene.frame_current] bpy.app.driver_namespace['th'] = th

After that, I set a driver to call the function ‘th()’, and suddenly my dipole was beautifully flipping around. The arrows etc. are set to follow the rotating loop of wire using various combinations of constraints, so that they’re always facing the orthographic camera. Tweaking the parameters requires me to run the script again, so in that sense it’s not ideal - if I wanted to do this frequently, it might be worth creating a Blender addon which lets you type in a differential equation, then tweak parameters with sliders.

I rendered the frames (150 frames), then used gifski to turn it into an animation with nice quality. (Unfortunately downscaling so that it can go on tumblr reduces that quality quite a bit and introduces a whole lot of aliasing but eh).

The artists on the site have a particular style that's very sharp and vector based, but I have been more or less able to approximate it in Blender using the cycles Toon shader node, not using Filmic colour so I could exactly match the colours they chose, and using Freestyle to draw outlines. For lighting, I used an HDRI of a studio with some bright lights from HDRI Haven, which (with some tweaking using vector nodes) got some decent specular highlights that kind of looked like the drawing.

It was kind of a weird process really - when I was originally writing the questions, I did a render with a realistic metal shader lit by a HDR, then the artist drew over it with vectors and drew in specular highlights manually, and then to match her style for my animations, I had to like... try and figure out a combination of shaders that have more or less the same look as the fake specular highlights she drew.

There’s some glitchiness with some of the Freestyle outlines that I could probably fix by putting things onto separate renderlayers, but my boss is already really happy with this, which is a nice change from ‘hmm how long did you spend on this Blender stuff? please could you just focus on writing’ lol

[matekon]

The “best” definition for the sinus and cosinus functions

I use the following personal conventions:

● - Definitions - Propositions I assume are true

○ - Theorems – Propositions I deduce from the definitions

I also prefer \(\tau\) which equals \(2\pi\) as the circle ratio

_____

In mathematics, there is a common phenomenon: there can be multiple ways of defining the same mathematical object.

For example, here are 2 definitions for an isosceles triangle:

● 1) A triangle is isosceles if it has 2 sides of the same length.

● 2) A triangle is isosceles if it has 2 angles that are equal in measure

These 2 definitions are “equivalent” in the sense that a triangle would be isosceles according to the first definition if and only if it is isosceles according to the second definition. (If you are into analytical philosophy, specifically Frege, you might say these definitions express different “senses” but have the same “reference”.)

The case of the isosceles triangle is pretty simple, but in mathematics, there can be definitions for objects which are equivalent but where it isn’t trivial is the slightest.

Even though it is totally frequent for one mathematical object to have multiple definitions available, the way modern mathematics work (by axiomatisation), we have to choose one definition as a “starting point” and then deduce its equivalence with other definitions later on.

So, with our example of the isosceles triangle, we could either choose the first proposition as our definition and then, the second proposition would follow as a theorem, or we could just as well do the reverse.

But, is there a “starting definition” that is “better” than the others? From experience, I would say that, in the point of view of a “pure mathematician”, this is totally irrelevant and doesn’t matter. But, I do think that we can say some definitions are “better” than others if we allow ourselves to use didactic criteria to evaluate them.

In this article, I will be interested with the functions sinus and cosinus, for which I encountered many different definitions in my school years. In Section 1,  I will present these definitions and say what I like and don’t like about them using a didactic approach. Then, in Section 2, I will introduce a definition of these functions that I personally think is the best one and I will show that it is “equivalent” with some of the definitions of Section 1.

__________

Section 1 - The usual definitions for sin and cos

At secondary school, I learned the following definition:

Definition 1 (by the triangle): 

\(\bullet\) Let \(\triangle\ ABC\) be a right triangle where \(\angle ABC\) is the right angle

Then we define \(sin(\theta)=a/c\) and \(cos(\theta)=b/c  \).

The pros for this definition are the simplicity of the language and the fact that it is directly applicable to problems of geometry.

Among the cons, we have that this definition only makes senses for \(\theta \in ]0,\tau/4[\) (let’s immediately work in radians). Also, I don’t think that this way of presenting sin and cos makes it obvious how to visualize the graphs of these functions. (It is possible to see the graphs but it requires us to be kind of clever.)

You might also say that this definition doesn’t immediately allows us to evaluate the functions for a given input. But I don’t think this is such a big problem and I will explain it soon later.

In CEGEP, I learned the definition involving power series:

Definition 2 (by the power series):

\(\bullet \: sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}t^{2n+1} = t-\frac{t^3}{3!}+\frac{t^5}{5!}-...\)  

\( \bullet \:cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}t^{2n} = 1-\frac{t^2}{2!}+\frac{t^4}{4!}-...\) 

This definition has the main advantage of allowing us to directly calculate the values of the functions. But the inconvenience that immediately comes with these types of definitions is that they make us say: “Where is this coming from? What is its utility?”.

The reality is that I think the human mind prefers to start of with definitions that make us directly see why the object in question is interesting and relevant. And then, for a function, we should find a way to “evaluate” it later on.

Let’s also note that this definition doesn’t make the shape of the graphs any more obvious.

In university, I got introduced to the following definition:

Definition 3 (by the differential equation):

\( \bullet \: sin (t) \) and \(cos(t) \) are solutions of the differential equation \( f^{\prime\prime}(t) = -f(t) \)

We first note that, because this equation has infinitely many solutions, we need further specifications to precisely define what \(sin\) and \(cos\) are.

This definition for me just mainly shows us a new interest of the sin and cos functions: there are extremely useful tools for solving differential equations. (This is one of the main motivations behind Fourier Analysis.)

This makes us do an important realization: maybe for didactic reasons, the definition we want to use for different mathematical objects depends on the context: For doing regular problems of geometry,  Definition 1 for sin and cos is this best one to use. But for the theory of differential equations, then  Definition 3 is more relevant.

I do think this argument is very important. But I also have a weak spot for definitions that are kind of more intuitive, more visual and just more “neutral” and “universal” I would say. I think all these criteria apply to the definition I will introduce in Section 2, which I will call “Definition 4 (by the circle)”. In fact, this definition is quite common, but I have never seen it being formalized the way I am about to do.

____________

Section 2 - The “best” definition for sin and cos

Let’s start with a simple question: “How do I describe a circle?”.

Algebraically, the simplest circle is the one of radius 1 centered at the origin. We define it this way:

\( \bullet \: S^1 = \{ (x,y) \in \mathbb{R}^2 | \: ||(x,y)||=1\} \)

We see this way of defining a circle works the following way: We take as points \( (x,y) \) on the circle all the solutions to the equation \( x^2+y^2=1 \).

But what I would like to do instead is to describe the circle with a function, not an equation.

I want a function \(f \) that outputs a point on the unit circle given a real number input.

\(\bullet\:f:\mathbb{R}\to\mathbb{R}^2\) where \(Im(f)=S^1\)

We immediately see that \(f\) is a vector function (has vectors as outputs). Because of that, it can be separated into 2 scalar functions (have real numbers as outputs).

\( f(t) = (x(t),y(t)) \) where \( x: \mathbb{R} \rightarrow \mathbb{R} \) and  \( y: \mathbb{R} \rightarrow \mathbb{R} \)

In case you didn’t guessed it, \(x(t)\) will become \(cos(t)\) and \(y(t)\) will become \(sin(t)\). I will continue to write them as \(x(t)\) and \(y(t)\) mainly because this notation makes their role clearer and because they aren’t fully defined yet.

Now, what I want to do is to fully define the functions \(x(t)\) and \(y(t)\). To do that, I will enumerate a list of properties that I want these 2 functions to have.

Because I said I wanted \(f\) to output a point on the unit circle, that implies:

\( f(t) \in S^1 \iff ||f(t)||=1 \iff ||(x(t),y(t))||=1\)

\(\iff \sqrt{x^2(t)+y^2(t)}=1 \iff  x^2(t)+y^2(t)=1 \)

With this, I will state the first property to these functions, which is the first part of their Definition 4:

\( \bullet \: (1) \: x^2(t)+y^2(t)=1 \)

This property isn’t enough. To illustrate this, let’s remark that the following function \(f^{*} \) does obey the property \( (1) \) but isn’t the “nicest” function we could think of:

Basically, we see that \(f^{*} \) isn’t “continuous”, because it occasionally “jumps”. But, let’s say I want \(f\) to be a function that goes continuously around the circle.

In fact, I want \(f \) to be something more specific: “parameterized by arc length”.

This means the following: 

●  Let \( g(t) \) be a curve in space (2d in this case). Then  \( g(t) \) is parameterized by arc length if the length of the arc between \( g(t_0) \) and \(g(t_1) \) (where \(t_1 > t_0 \) ) is precisely \( t_1 – t_0\).

(In the case of the unit circle, how is this idea related to “radians”?).

I won’t go behind all the theory behind it. It is easy to google anyway. For our purposes, I need to know the theorem that says:

\( \circ \: f \)  is  “parametrized by arc length” \( \iff ||f’(t)||=1 \)

From that I deduce:

\( ||f’(t)||=1 \iff ||(x’(t),y’(t))||=1 \)

\(\iff (x’(t))^2+(y’(t))^2=1 \)

From this, we get the second part of Definition 4:

\( \bullet \: (2) \:  (x’(t))^2+(y’(t))^2=1 \)

\( f \) being “parameterized by arc length” implies that  \( f \) is countinous, as we wanted. 

The properties (1) and (2) largely define \( x(t)\) and \(y(t)\). In fact, with just these 2 properties, we can show that \( x(t)\) and \(y(t)\) must obey Definition 3 (by the differential equation). To prove this is a very fun mathematical exercise. Anyway, here’s my demonstration:

We know:  

\( \bullet \: (1) \: x^2+y^2=1 \)  \( \bullet \: (2) \:  x’^2+y’^2=1 \) 

We want to show that \(  y ^{\prime\prime} =-y \) (The proof for  \(  x ^{\prime\prime} =-x \) is analogous) 

\( (1) \Rightarrow \frac{d}{dt}(x^2+y^2)= \frac{d}{dt}(1) \)

\(\Rightarrow 2xx’+2yy’=0 \)

\(\Rightarrow xx’=-yy’\)

So, we have: \( \circ \: (A)\: xx’=-yy’\\ \)

\( (2) \Rightarrow x^2( x’^2+y’^2)=x^2(1)\)

\( \Rightarrow (xx’)^2+(xy’)^2=x^2 \)

\(\Rightarrow^{(A)} (-yy’)^2+(xy’)^2=x^2 \)

\(\Rightarrow y’^2(y^2+x^2)=x^2 \)

\(\Rightarrow^{(1)} y’^2(1)=x^2 \Rightarrow  y’^2=x^2 \)

Consequently, \( \circ \: (B)\: y’^2=x^2\\ \)

\( (B)\Rightarrow  \frac{d}{dt}(y’^2)= \frac{d}{dt}(x^2) \)

\(\Rightarrow 2y’y ^{\prime\prime} =2xx’\)

\( \Rightarrow^{(A)} y’y ^{\prime\prime} =-yy’ \)

\( \Rightarrow_{*} y ^{\prime\prime} =-y \) 

(I will call this final equation \( (E_y) \) and use it later)

\(QED \)

(The last implication with an asterisk below really needs a bit of justification. Because \( y’ \) can equal 0 for some inputs, we can’t just divide by it. But, there is a way to clean up the mess and make the deduction valid.)

As I said before, Definition 3 (by the differential equation)  is incomplete in its formulation. So, it’s not because I was able to deduce it from \((1)\) and \((2)\) that these 2 properties are enough to define \( x(t) \) and \( y(t) \).  

What I will do now is that I will try do deduce Definition 2 (by the power series). Trying to do so, it will show me what I have to add to \((1)\) and \((2)\) to make the Definition 4 (by the circle) complete.

What I know:

\( (1) \: x^2+y^2=1 \)  \( (2) \:  x’^2+y’^2=1 \)

\( (A)\: xx’=-yy’\\ \) \( (B)\: y’^2=x^2\\ \)

\( (E_x) \: x ^{\prime\prime} =-x \)  \( (E_y) \: y ^{\prime\prime} =-y \)

To deduce Definition 2, I would like to find the Maclaurin Series for \( x(t) \) and \( y(t) \):

\( x(t) = \sum_{n=0}^{\infty} \frac{x^{(n)}(0)}{n!}t^n \)

\(  y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}t^n \)

So, I need to find \( x(0)\), \(x’(0)\), \(x ^{\prime\prime} (0)\), ... and \(y(0)\), \(y’(0)\), \(y ^{\prime\prime} (0)\),...

\((1)\) and \((2)\) aren’t enough to find the Maclaurin Series. So, I will add the following defining property for \(sin\) and \(cos\) that is honestly only justifiable as a convention:

\( \bullet \: (3) \: f(0) = (x(0), y(0)) = (1,0) \)

So we’ve essentially just chosen the starting point for our curve \( f\). It could just as easily have been \( (0,1)\) or \( (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} ) \), which also are on the unit circle. This choice for \( f(0) \) is, I think, mainly justifiable as a way to make Definition 4 (by the circle) equivalent to Definition 1 (by the triangle).

Now, let’s use this equation  in the quest of finding the Maclaurin Series of  \(x(t) \) and \( y(t)\) :

From \((3) \: x(0)=1 \) and \( (E_x) \: x ^{\prime\prime} =-x \), I deduce:

 \( x^{(4k)}(0)=1 \)  and \( x^{(4k+2)}(0)=-1 \) where \(k \in \mathbb{N}\)

From  \(  (3) \: y(0)=0 \) and \( (E_y) \: y ^{\prime\prime} =-y \), I deduce:

 \( y^{(2k)}(0)=0 \) where \(k \in \mathbb{N}\)

We are halfway done. But, for the next step, we need to be kind of clever and use an old equation we proved earlier:

\( (B) \: (y’(t))^2= (x(t))^2\)

\( \Rightarrow (y’(0)^2=(x(0))^2\)

\( \Rightarrow^{(3)} (y’(0))^2=(1)^2 \)

\( \Rightarrow y’(0) = \pm 1 \)

Again, there is a choice to be made, and again, it is a matter of convention.

It can be shown that choosing \( y’(0)= 1\) will make the vector function \(f\) go counterclockwise around the unit circle and that choosing \(y’(0)= -1\) will make it go clockwise instead. Yes, we will choose the first option because of the convention of how we measure angles.

\( \bullet \: (4) \:  y’(0) = 1 \)

This will be the last property we add to Definition 4. Let’s see what we can do with it:

From \(  (4) \: y’(0)=1 \) and \( (E_y) \: y ^{\prime\prime} =-y \), I deduce:

 \( y^{(4k+1)}(0)=1 \)  and \( y^{(4k+3)}(0)=-1 \) where \(k \in \mathbb{N}\)

From \( (4) \: y’(0)=1 \) and \( (2) \: x’^2+y’^2=1 \), I deduce:

\( (*) \: x’(0) = 0 \)

Finally,  from  \( (*) \: x’(0)=0 \) and \( (E_x) \: x ^{\prime\prime} =-x \), I deduce:

\( x^{(2k+1)}(0)=0 \) where \(k \in \mathbb{N}\)

Putting all of this together, we finally get the MacLaurin Series:

\( x(t) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k)!}t^{2k} \)

\(  y(t) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)!}t^{2k+1} \)

And so we’ve just proven that Definition 4 (we’ve just completed) is equivalent to Definition 2!

For clarity, let’s put all the parts of Definition 4 together and we will posture \(x(t) = cos(t) \) and \(y(t) = sin(t)\).

Definition 4 (by the circle):

\( \bullet \: sin: \mathbb{R} \rightarrow \mathbb{R} \) and \( cos: \mathbb{R} \rightarrow \mathbb{R} \), where

\( (1)\: cos^2(t)+sin^2(t)=1 \)

\( (2)\: (\frac{d}{dt}cos(t))^2+ (\frac{d}{dt}sin(t))^2 =1 \)

\( (3)\: sin(0) = 0 \) (which implies \( cos(0) =1) \) 

\( (4) \: ( \frac{d}{dt}sin(t))|_{t=0} = 1 \)

We can also put it on words like that:

Definition 4 (by the circle):

\( \bullet \: f(t) = (cos(t),sin(t)) \) is a function from \(\mathbb{R}\) to  \(\mathbb{R}^2\) where \( Im(f) = S^1\) (unit circle). Also, \( f\) is parameterized by arc length, starts on \( (1,0) \) and goes counterclockwise. 

The language can really be seeing as harsh, but once this definition is really understood, it allows us to directly visualize what \(sin(t)\) and \(cos(t)\) mean. It is also not hard to see the shape of their graphs, especially with the help to the following gif: http://i.imgur.com/jvzRYnC.gif

This is the reason why I think this is the best definition in a didactic point of view.

If we want a more accessible language for people who aren’t specialized in math, this formulation would also be valid:

Definition 4:  ● If I start on the unit circle at \( (1,0)\) and I walk t units of distance counterclockwise while staying on the unit circle, my x position will be \(cos(t) \) and my y position will be \( sin(t)\)

I let to you the proof that Definition 4 is equivalent to Definition 1 (for \( t \in ]0,\tau/4[) \). It is definitely the most straightforward proof on the bunch.

[mathematicianadda]

A circle is divided into $5$ parts as shown in the diagram and parts are colored either red or green. Find which area is bigger. https://ift.tt/2PeDy6n

In the given diagram, there are $5$ points $A, B, C, D$ and $E$ on the circumference of the circle such that $\angle ABC = \angle BCD = \angle CDE = 45^{\circ}$ and $O$ is the center of the circle.

Sectors made by $AB$ and $DE$, and area of the circle between $BC$ and $CD$ are highlighted in green. Area of the circle between $AB$ and $BC$, and between $CD$ and $DE$ are highlighted in red.

Which area is bigger, the area highlighted in red or the area highlighted in green?

This was sent to me by someone. While I solved the problem (given below), the sender said that the source solution arrived at the conclusion that points $A$, $O$ and $E$ are collinear and $OC \perp AE$, so $\displaystyle \angle OCB = \angle OCD = \frac{45^{\circ}}{2}=22.5^{\circ}$. While I agree with points being collinear and $OC \perp AE$ but that cannot obviously be the reason for the angles being equal. In fact the solution does not depend on them being equal as we can see. I am seeking help in establishing $\angle OCB = \angle OCD$ if that is indeed true, which I cannot see how one can conclude based on what is given.

My solution: Say, $\angle OCB = \theta$. Then, $\angle ACB = \angle OCD = (45^{\circ}-\theta)$ and $\angle DCE = \theta$.

Segment $AB= \displaystyle r^2 \left[\frac{\pi}{4}-\theta-\sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)\right]$ Segment $DE= \displaystyle r^2 \left[\theta-\sin \theta \cos \theta\right]$

$\triangle OBC = r^2 \sin \theta \cos \theta$ $\triangle ODC = r^2 \sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)$

Section $BOD = \dfrac{\pi}{4} r^2$

Adding all of the above, total area in green $= \dfrac{\pi}{2} r^2$. So the red area has to be the same too.

In addition to my question on $OC$ being bisector of $\angle BCD$, let me also know if any of you have a simpler solution.

from Hot Weekly Questions - Mathematics Stack Exchange Math Lover from Blogger https://ift.tt/3fjqLu1

[rowanstree-blog]

Day #167

          Today I started with a math class on the graphs of the trigonometric functions, sine, cosine and tangent. I started the lesson of with a video on the graph of the sinusoidal function, y=sin(x). I started with a question asking what the domain of the sinusoidal function was. The domain is all sets of valuable inputs for the function. I used the unit circle and a y, theta graph to wrote and plot the sinusoidal equation. Once I did all of that, I concluded that the domain of this function is (-1,1) including -1 and 1. I then watched a video on how to construct the graph of y=tan(x). I concluded that the graph would have a vertical asymptote every five radians and approach positive snd negative infinity while doing so. Considering the tangent of theta is the slope of the terminal ray, I just had to take the sin of theta over the cosine of theta to find out the tangent of several pi radians. I then watched a video on intersection points of y=sin(x) and y=cos(x). In this video, I wrote terms of the table in terms of radians for both the cosine of theta and the sine of theta on the unit circle and graphed accordingly. After that, I watched a video on basic trigonometric entities involving symmetry. In the lesson, I drew several terminal rays in all four quadrants of a graph in a unit circle. I then wrote down which sine and cosine functions were equivalent and then followed it up with a video on tangent identities which is the proportion of a sin function over its equivalent cosine function. Next, I watched a video on sine and cosine identities involving symmetry in which I took random sine and cosine functions and determined if they were equal to each other. I then watched a video on tangent identities periodicity. In this video, I did pretty much the same thing except I determined if its slope, or tangent was equal to 1/2. After all of that, I took an Algebra 1 class on the Quadratic equation sample problems as a refresher. I then took a French lesson on Duolingo followed by a Chemistry class on how to fill in an ICE table according to the equilibrium constant expression of a reaction. I stands for initial concentration which should be given in order to calculate E which stands for equilibrium concentration which can be found after adding or subtracting I and C which stands for change in concentration. First, you must write all of your initial concentrations If your equilibrium constant is larger than or equal to 10^4, then your products are favored. If Kc is less than or equal to 10^-2, then your reactants are favored. This information can help you figure out which side is equal to zero which will later determine how you fill in your ICE table. Depending on which side is favored, you must add or subtract the molecules stoichiometric coefficient of x. Add, subtract and solve for x to find out your equilibrium constant. After that, I took a space science class on a simulation video of Earth’s tilt and rotation which determines its season. This information can contribute to the reasoning behind why the equator doesn't have seasons. It is because it is never turned away from the sun, like us hence our four seasons as we turn towards and away from the sun. I then took an environmental science class on various studies launched and questions asked on the transgenic contamination of Mexican maize. Even though there have been both false positives and false negatives in order to protect and expose the firms causing this transgenic contamination. It is true that transgenic contamination is becoming more and more abundant and the effects are still unknown. I then did a quick exercise followed by an hour long lunch. After that, I took a women’s history class on Mary Harris “Mother” Jones who was a labor organizer who played an outspoken role in the world that was mislead by her appearance. She was a passionate and resilient woman that fought for what she believed in and due to that, became known as “the most dangerous woman in America.” I then took an economics class on Allocatice efficiency which is when the marginal cost is equal to the marginal benefit and I did an example that showed this both literally and graphically. Next, I took a grammar class on commas and introductory elements which are followed by a comma and are placed in the beginning of a sentence. After that, I read some of “The Beauty Myth” on how markets rely so much on feminine appearance more that the actual product to “appeal” to the reader while they are actually making them more self conscious.

[1dxdydz]

The ring w/Spherical Coordinates

We are going to solve the ring problem. Imagine we have a sphere with radius 2. Inside of our sphere, we have a cylinder with radius 1. We take out the cylindrical piece, and we are left with an object like a ring. We can imagine it like we have a drill on top of our sphere, and we drill out through the center. We need to find the volume of this object.

Let’s use the spherical coordinates for this problem because we have a sphere in the end! Before writing our triple integral in spherical coordinates, let’s examine the object a bit. We know that we have two distinct surfaces for this object. We have a cylindrical surface as a result of our drilling out operation. We have also a surface from our sphere with radius 2. We have a full object, meaning we don’t have a half or quarter of the object.

We have three integrals, so we need to find limits for each one. Let’s start with our inner most integral that’s respect to rho. When we deal with rho, imagining a light bulb in the origin can help us. Let’s imagine that our object, ring, is transparent. If we turn the light bulb on in the origin, what surface does it hit first? It would hit the cylindrical surface first because we have nothing before the cylindrical surface but empty space that’s coming from our drill. Alright, now our light rays penetrate through our solid. It hit the cylindrical surface first as we know it. Now, light rays from the light bulb in the origin traveled through the solid, and they are about to leave. What surface does it touch right before leaving the solid? It’s the spherical surface. My light rays hit the cylindrical surface first and left the solid by touching the spherical surface.

We actually identified the limits of the inner most integral that’s respect to rho. We need to write it in spherical coordinates though. Anything we have in spherical coordinates, limits and integrand, must be in terms of spherical coordinates! First, we hit cylindrical surface with radius 1. How can I describe it in terms of spherical coordinates? It’s 1 csc(phi) because our radius is 1 for the cylinder. So, what about the csc(phi) part? If we say rho=csc(phi), I can multiply both sides by sin(phi) to cancel out csc(phi). I got 1 on the right side, and sin(phi)rho on the left side which equal r in polar coordinates. r=1 represent a cylinder with radius 1 in cylindrical coordinates,the 3d version of polar coordinates. Or you can just accept it as acsc(phi) represents a cylindrical surface with radius a in spherical coordinates. My upper limit is quite easy. It is our spherical surface from our sphere with radius 2. So, the sphere with radius 2 is just rho=2. So my lower limit is csc(phi) and my upper limit is just 2 for the inner most integral that’s respect to rho. We are done for our first integral!

For my middle integral that’s respect to phi, I need to know the angle interval that my solid exists. We can think it as radar. It can start from 0 to pi, meaning that I can make 0 degree to z-axis. For this instance, I am actually touching my z axis. Then, I can let my radar cover starting from the top point on z-axis,0, to the bottom point,phi. I’d make a total coverage of phi for a full sphere. However, we drilled out the cylindrical part in our sphere. We don’t have a full sphere. Our coverage starts from an angle,phi, we cover our solid. We stop our coverage where the solid ends. We need to find this angle. It’s not hard.We’ll use trigonometry for finding the angles. Let’s imagine out solid in z-y or z-x plane, meaning we look from the sides. I see a circle with radius 2(my sphere) and I see two straight lines(from my cylinder with radius 1).I think one of the most challenging/confusing concepts in spherical coordinates is the presence of phi. I heard that spherical coordinates are confusing or hard from my friends but it’s because we are so used to work in cartesian coordinates. We spend a couple weeks to adapt into a total different coordinates system so, it’s very normal to be confused at first. 

We can plug in out x value, 1, in the equation. We ended up with square root of 3. I know the vertical and horizontal components of the triangle. Thus, we know that tangent of square root 3 is the angle in my triangle. It’s not phi though!!! Phi is the angle between my rho and z-axis. I got pi/3 for my angle in the triangle. So, phi is pi/6. This is the angle where we begin to cover the solid. It’s our lower limit. This is a symmetric solid. So, the angle is 5pi/6 which is my upper limit. So, in the interval of pi/6 and 5pi/6, I cover my solid. These are my limit for my middle integral that’s respect to phi. So, my lower limit is pi/6 because I first begin covering the solid at this angle and my upper limit is 5pi/6 where I completely cover the solid. We are done with our middle integral, too.

Finally, the last and sweetest one, the most outer integral that’s respect to theta. It’s a full solid, not a half or quarter one. Our limits will be 0 and 2pi for the last integral.

We can finally start solving our triple integral.

It’s always good to check our answers by using other methods. What method can we use for this problem rather than another triple integral? The old and good washer method can be a good fit. We have circles with different radii stacked up on top of each other along the y-axis. I know my range for my y-axis from the previous method which is from negative square root of 3 to positive square root of 3. I’ll subtract the circle inside the bigger circle. I end up with the same answer. It’s always good to check your answers because a small mistake in triple integrals can give you a different result because it’s a long process and can be tricky.

[devphilamaths]

Solve sin(pi/2+theta) | sin(pi/2+ x) | sin pi/2 + x formula, Find value sin pi by 2 + x

[devphilamaths]

Solve sec(pi/2+theta) | sec(pi/2 +x) | sec pi/2 + x formula, Find Exact value sec pi by 2 + x

[devphilamaths]

Solve cosec(pi/2+theta) | cosec pi/2 + x formula, csc (pi/2 +x) | Find value cosec pi by 2 + x

[devphilamaths]

sin(π/2-x) =cosx solve trigonometric identities | sin(pi/2-theta) =sintheta

sin(π/2-x) =cosx

sin(π/2-x) 

sin

12th IIT JEE

[mathematicianadda]

Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution https://ift.tt/eA8V8J

I will explain my approach, help me with the last step please! $$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$

substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$ $$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$

For solving modulus, it was important to determine range of $\theta$ , therefore I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as , $ -1 \le x \le 1 \,$ , from domain ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.

$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$ = dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$

= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$

That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$ Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ . $$ (\theta - \pi/4) +\pi \,$$ for $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and

$$\theta -\pi/4$$ for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$

My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.

So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.

So please help me, very hopefully I signed up in stackexchange!

Found Solution :-

I was confused because I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$ as then we would not simply write like $$ \theta = \sin^{-1} x $$ we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$, now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people. (I write the answer in this edit to help anyone who will reach here after searching web , thanks to everyone for answers)

from Hot Weekly Questions - Mathematics Stack Exchange Aryaman from Blogger https://ift.tt/32uV1PE

[mathematicianadda]

Integrate a weighted Bessel function over the unit disk https://ift.tt/eA8V8J

I would like to evaluate a complex-valued integral of the form

$$ I_e = \int_0^1 x e^{iax} J_0(b \sqrt{1-x^2}) dx $$

where $a$ and $b$ are positive real numbers and $J_0(z)$ is the Bessel function of the first kind. I am particularly interested in the special case of $a=(c+1)b$ for $c \ll 1$.

The task boils down to evaluating two real-valued integrals

$$ I_s = \int_0^1 x \sin(ax) J_0(b \sqrt{1-x^2}) dx $$ $$ I_c = \int_0^1 x \cos(ax) J_0(b \sqrt{1-x^2}) dx $$

The integral with the sine has a simple form given by Gradshteyn and Ryzhik (6.738.1) which, after simplification, becomes

$$ I_s =\sqrt{\frac{a^2}{a^2 + b^2}}j_1(\sqrt{a^2 + b^2}) $$

where $j_1(z)$ is the spherical Bessel function of the first kind.

I am not exactly sure how this expression was derived. Perhaps it holds a clue. I tried substituting the integral form of the Bessel function and integrating analytically but did not get very far.

By symmetry, I naively expected the integral involving the cosine to be proportional to the spherical Bessel function of the second kind $y_1(z)$ (and thus, the complex-valued integral to be proportional to the spherical Hankel function of the second kind), but that does not appear to be the case.

$$ I_c \approx -\sqrt{\frac{a^2}{a^2 + b^2}}y_1(\sqrt{a^2 + b^2}) $$

The agreement for the special case of $a=b$ is not too terrible, but something is going wrong for small values of $b$. If I take the difference between the true value of the integral and my guess, it is proportional to another Bessel function, but the arguments are crazy and it does not make any sense.

Now, if I assume that it is proportional to the spherical Bessel function of the first kind instead, and compute the ratio between the true value of the integral and my guess, I get the classic graph of the cotangent.

For $b \gg 1$, it appears

$$ I_c \approx \cot(\sqrt{a^2 + b^2} + \phi)) \sqrt{\frac{a^2}{a^2 + b^2}}j_1(\sqrt{a^2 + b^2} + \phi) $$

In practice, this is hardly useful, since the period of the cotangent would have to correspond to zeros of the spherical Bessel function, and these are hard to compute. Additionally, $j_1(z)$ and the actual $I_c$ appear to be slightly off-phase.

Of course, ideally, I would like to solve the problem analytically, but I am not sure how (using a series expansion, perhaps?). I would appreciate any tips or guidance.

Thank you!

Edit: by transforming using $x=\sin{\theta}$, there is an almost perfect match in Gradshteyn and Ryzhik (6.688.2), except that the leading term in my case is $\sin{2\theta}$ rather than $\sin{\theta}$. Additionally, it gives a result in terms of the spherical Bessel function of the first kind rather than the second kind, so it does not explain my empirical approximation.

$$ I_c = \int_0^{\pi/2} \frac{1}{2} \sin{2\theta} \cos(a \cos{\theta}) J_0(b \sin{\theta}) d\theta $$

Using the representation given above, we can expand $\cos(a \cos{\theta})$ in a series, which does give an analytic solution, but it converges rather poorly and does not appear to contain any spherical Bessel terms (just the regular Bessel functions). I am yet to find an expansion in terms of spherical Bessel functions.

from Hot Weekly Questions - Mathematics Stack Exchange zalbard from Blogger https://ift.tt/2AXLvcJ

[mathematicianadda]

The integral $\int_{0}^{\pi/2} \sin 2\theta ~ \mbox{erf}(\sin \theta)~ \mbox{erf}(\cos \theta)~d\theta=e^{-1}$

In a work it was required to find an integral akin to $$I=\int_{0}^{\pi/2} \sin 2 \theta~\tanh(\sin\theta) \tanh(\cos \theta)~d\theta.$$ Since $\tanh x$ and $\operatorname{erf}(x)$ are similar functions so it was a pleasure to see that its variant namely $$J=\int_{0}^{\pi/2} \sin 2\theta ~ \operatorname{erf}(\sin \theta)~ \operatorname{erf}(\cos \theta)~d\theta=e^{-1}$$ could be solved by Mathematica.

So the question is: Can one do it ($J$) by hand?

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2IqV2ty

[mathematicianadda]

A tricky integration over the unit sphere

Please help me solve the following integral, $$ I=\int\limits_{\{x,y,z\}\in \mathbb{S}^2}\!\!\!\max\{0,x,x\cos{\theta}+y\sin{\theta}\}\,\mathrm dx\,\mathrm dy\,\mathrm dz $$ where $\mathbb{S}^2$ is a unit sphere and $\theta$ is some constant such that $0\le\theta\le2\pi$. Numerically (up to arbitrary precision) this is equal to $$ I=\pi(1+\frac{\sqrt{(1-\cos{\theta})^2+\sin^2{\theta}}}{2}). $$ I am unable to solve/prove this analytically.

This is for a research project, any help would be appreciated and acknowledged in the article.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2HUq6Sq

[mathematicianadda]

Elegant way to evaluate $\int_0^\pi\frac{\sin\left(M+\frac{1}{2}\right)\theta\;\sin\left(N+\frac{1}{2}\right)\theta}{\sin^2(\theta/2)}d\theta$?

Solve in terms of $M, N$ $$I(M. N) = \int_0^\pi\frac{\sin\left[\left(M + \frac{1}{2}\right)\theta\right]\sin\left[\left(N + \frac{1}{2}\right)\theta\right]}{\sin^2\left(\frac{\theta}{2}\right)}d\theta$$ where $M, N$ are non-negative integers.

I've tried solving it by using trigonometric identities and brute force, but it gets extremely annoying. Is there a simpler way?

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2XBsxyb

[mathematicianadda]

Proving that $\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\sum_{n=1}^\infty \frac{\sin n}{n}$.

Proving that $$\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\frac{\pi -1}{2}$$

I've known a similar conclusion $$ \sum_{n=1}^\infty \frac{\sin nx}{n}= \begin{cases} \dfrac{\pi - x}{2} & x \in (0, 2\pi),\\ \quad 0 & x = 0, \\ f(x+2\pi) & x \in \Bbb{R}. \end{cases} $$ And one of my classmates found the equation mentioned above by mathematica.

I was amazed by the equation

$$ \sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\sum_{n=1}^\infty \frac{\sin n}{n} $$

My attempt

\begin{align} \sum_{n=1}^\infty \frac{\sin^2 n}{n^2} & = \sum_{n=1}^\infty \frac{1-\cos 2n}{2n^2} \\ & = \sum_{n=1}^\infty \int_0^1 \frac{\sin 2n\theta}{n} d\theta \\ & = \int_0^1 \sum_{n=1}^\infty \frac{\sin 2n\theta}{n} d\theta \\ & = \int_0^1 \frac{\pi}{2}-\theta \,d\theta \\ & = \frac{\pi-1}{2} \end{align}

Oh. Actually I hadn't solved it before I edited this question, but I seemed to have worked it out.

So is there any other method to solve this problem? And deeper insights?

I've heard that it can be worked out via complex analysis and fourier analysis.

Thanks in advance!

Added

Thanks for your comments!

Here is another possible generalization

$$ \sum_{n \in \mathbb{Z} } \left[\frac{\sin (n \alpha + \theta) }{ n \alpha + \theta} \right]^2 = \frac{\pi}{\alpha} \,\, \forall \alpha , \theta \in \mathbb{R} $$

I got stuck on it. For $\theta=0$ we can use the following equation $$ \sum_{n \in \mathbb{Z} } \frac{\cos n\theta}{n^2} = \frac{\pi^2}{6}-\frac{\pi \theta}{2} + \frac{\theta^2}{4} \,\, \theta \in [0,2\pi] $$ But how to deal with the situation that $\theta \ne 0$?

Can you give me some hints? Thanks in advance!

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2GpD0H2