Would be really funny if one of your student actually wrote down "you do it" as an answer, but then add "just kidding, here's the actual solution:" with the proof below.
Excellent coverage of all important steps and gotchas! I do have one criticism.. When you deal with the "negative r" value, don't describe it as "you have to go back One". Instead of sliding your pen by One, you should 'flip' the tip of the pen over (the base stays where it is) to implement the effect of a "negative radius value" at the subject theta angle. 😉
Regarding you do it, when I took Cal II, most of the test over polar integration was just setting them up. Basically the teacher already knew that we could integrate, but the setup on these problems is the tricky part. Also my teacher favored polar curves with lots of tiny loops, so he knew a fair bit of the test would be taken up just finding the intersections and creating the graphs.
Is setting the absolute values of each equation equal to each other a valid way to find the intersections without the graph? Then you could just find θ at those r values for the limits of integration. (And if the actual values are negatives of each other, you could check if the angles are off by pi, right?)
I am wondering on how can be the radius negative in polar coordinates. I though it should be positive by definition. Isn't the radius a length as a matter of fact? So what is the sense of r=-1?
I must ask about all the explanations about setting the limits. When I last taught this (1995), I was told to expect that students would already know and understand about the relationship between theta and r when r is negative. Moreover, they should have already known about the fact that points where polar curves intersect would not necessarily correspond to the same (r, theta) combination. Is this not true any more?
I think that the first problem might be easier with basic geometry if you graph it. There are points at (0,0) (1,0) (0,-1) (-1,-1). Each of these points have the tan line slope of 1 or -1 so that proves that you can find the center by making normal lines at these points. Of course that could be geometric or algebraic. Then those same lines give you the radius/diameter. To find the segment of the circle in quadrant 1, use the triangles and circle sector formulas. Specifically the triangle would be from (0,0) to (1,0) to (.5,-.5)[the center of the circle]. The lines are perpendicular at the center so that angle is 90°. The area of that portion in the first quadrant can now be defined of a quarter circle minus the triangle. Which is pi/8 minus 1/4. Finally, the area of a quarter of the larger circle minus the section is pi/8 + 1/4. That seems like a lot in typing but actually isn't much. The second problem can be approached similarly but with a little bit more geometry if anyone is interested I can type it out in the comments. However, the important part is the number of problems that can't be solved geometrically and that's why calculus is my favorite subject and this video is entertaining and fun. Thanks BlackPenRedPen!
Putting a negative sign for the radius is a mistake, because it is defined to be positive and the angle makes the direction. Although the integration is still the same you should integrate it from 3pi/2 to 2pi.
(Cos-sin)^2=1(cos-sin)^2=1-2cossin so we got a lot of cancellation to get first one is just plus integral of cossin and 2nd is negative of that need pen and paper to do the rest via di method
Hi BPRP, We stumbled upon a question that is the indefinite integral of tan(cos(x) and we can't seem to figure out how to do it. Do you have any ideas of how to do this?
Couldn't this be done geometrically? line y = x - 1 is a blue diameter, and for red you can do the quarter circle plus the right isosceles on the fourth quadrant...
of course, we guess. and again, you do it! please observe that red quarter circle Minus the right isosceles triangle, if we understood well. how was exactly the question..
r=cosΘ-sinΘ substitute Θ for arctan(y/x) converts to r=cos(arctan(y/x))-sin(arctan(y/x) then from the right triangle you find cos(arctan(y/x)) is x/r and sin(arctan(y/x)) is y/r substituting you get r=x/r-y/r multiply through by r you get r²=x-y substitute r² for x²+y² and you have x²+y²=x-y move everything to left side you get x²-x+y²+y=0 complete the squares you get x²-x+1/4 + y²+y+1/4 = 1/2 factoring you get (x-1/2)² + (y+1/2)² = 1/2 So the other figure is indeed a circle, centered at (1/2,-1/2) with a radius of 1/√2
Can’t you solve algebraically using area of segments and subtracting the segment area from big circle? I tried to do it and I might have made a mistake but I got quantity PI + 2 all over 8
I solved it by ordinary geometry and algebra and got it right. First one should be (pi+2)/8 and the second one should be 1/2. edit: i saw now that you answered this a long time ago, but maybe someone else who searches the comments would like to know.
@@justabeardedguythatisahero9848, well let's say, please don't confuse (x, y(x)) coordinates, orthonormal, with (thêta, r(thêta)) ones, polar. the ‘place’ bprp pointed to, is expressed in terms of x,y (+1, 0). now in polar terms of th,r both (0, 1) and (π, -1). further of course (2π, 1) and (3π, -1) &c. explanation needed? with thêta = 0 or 2π, 4π,… the radius axis points along the positive x-axis, then r=+1 coming up on x=+1. with thêta = π or 3π, 5π,… the radius axis points along the negative x-axis, that is, r>0 lies along *x
@@blackpenredpen I think I have to disagree. I'm trying them myself, went through them, and double-checked my answers with Wolfram Alpha, and I didn't get pi/8 for the first answer. Instead, I got (pi+2)/8. Here's my math as best as I can enter it into here. So obviously the first integral is pi/4, and we can check this geometrically by just seeing that it's a quarter of a circle with radius 1. The area of such a circle is pi, so a quarter of it is pi/4, and this is a straightforward integral anyway. Now for the second integral, the integral from 0 to pi/4 of 1/2(cos(theta)-sin(theta))^2 d(theta). First, let's work out that square, to get the integral from 0 to pi/4 of 1/2(cos^2(theta)+sin^2(theta)-2*cos(theta)*sin(theta)) d(theta). We simplify this with one of our favorite identities and get the integral from 0 to pi/4 of 1/2(1 - 2*cos(theta)*sin(theta)) d(theta). We can break this apart into two integrals and get the integral from 0 to pi/4 of 1/2*(1) d(theta) - the integral from 0 to pi/4 of 1/2*2*cos(theta)*sin(theta) d(theta). Focusing on the first of those, it's simply the integral from 0 to pi/4 of 1/2 d(theta). This is a straightforward integral and yields pi/8. For the second integral, we simplify it to the integral from 0 to pi/4 of cos(theta)*sin(theta) d(theta). This seems tricky, but it's easy enough to do a u-substitution with u = sin(theta), du = cos(theta) d(theta), and change the integral bounds from 0 to pi/4 into 0 to sqrt(2)/2. We now have the integral from 0 to sqrt(2)/2 of u du. This gives us 1/2 u^2 evaluated from 0 to sqrt(2)/2, which becomes 1/2(1/2-0) = 1/4. Now we take our three results and add or subtract as is appropriate. We should have pi/4 - pi/8 + 1/4, which first simplifies to pi/8 + 1/4. If we want, we can rewrite this as (pi+2)/8. For the second problem, I followed a very similar method (seeing as the integrals have only changed in their orders and boundaries, this isn't too complicated) and got 1/2, the same as the poster above.
There is a serious misconception of polar coordinates in this video which URGENTLY needs to be corrected: In polar coordinates r is ONLY defined in the domain [0,infinity), and is therefore always positive. The blue circle is displaced to a centre at (1/2,-1/2) with radius 1/2. You are right in that for this case r= cos (theta)-sin (theta), but then the domain of theta is [-3*pi/4,pi/4], which keeps r positive. THE WHOLE DOMAIN (pi/4, 5*pi/4) IS EXCLUDED. As usual, x= r*cos(theta) and y= r*sin(theta). Therefore, the circle is drawn clockwise in the allowed domain for x=[cos(theta)-sin(theta)]cos(theta), and y=[cos(theta)-sin(theta)]sin(theta). If you check it, you'll find the circle goes clockwise from (0,0) to (0,0) within this domain. The limits in your integral would then be 0 and pi/4 in your first example, and -pi/2 and 0 in the second one. The geometrical interpretation is straightforward as you can see. I love your videos, but I REALLY URGE you to correct this one, since it's misleading.
We rewrite the function as r = a*sin(theta+b), then use the property of right triangle in a circle, we can prove that the blue one is circle. The functions like a*sin(theta+b) are all circle.
The real challenge . . . solve them WITHOUT using calculus AT ALL. Just use plane geometry. You crack a nut every time with a sledge hammer, that's a sure win but far less fun. Crack it with a usual nutcracker and you'll get a lot more fun.
I think it's easier without calculus. I took courses for learning this a long time ago and had forgotten how to do it with calculus. With plane geometry it was straightforward with just adding and subtracting parts.
First I did it without any integral. I guess its way easier ;) But than I doublechecked it with your integrals :) Im happy that I got the same results xD
Writing "You do it" is the easiest way to do well in a test. You just have to hope the examiner is good at maths or will cheat by using the mark scheme.
@@obinnanwakwue5735 On the second one, yes. On the first one you (probably) have a sign wrong on the way to the final answer (I did too). Your self-checking hint for these kind of questions is that you're looking for area, answers must be positive.
@@Ni999 oh I get it I jacked up with the sign integrating one of the functions in the first problem, that should be + 1/2 and in the second one that should be 1/2 as well. Let me edit that.
@@obinnanwakwue5735 π/4 is the area of a full quarter red circle, the first problem is less than that, less than 0.785. (π/8) + ½ ≈ 0.893 Double check your terms, you're close. Second one is correct.
That feeling when you click because you know the area of a circle and then he starts using integrals
Oh Boy! It's calculus time!
That feeling when you think you know integrals and then he starts talking about polar
HAHAHA
Would be really funny if one of your student actually wrote down "you do it" as an answer, but then add "just kidding, here's the actual solution:" with the proof below.
I did it once with an exam bc i didn't know how to solve the problem except I didn't wrote the actual proof lol
You uploaded this less than one hour after the AP Calculus tests.
Gabriel Shapiro AB Calc BC 🤧
a = (pi + 2) / 8
b = 1 / 2
"now we have find the second angle"
Hey, that's easy, it's pi/2
"it's not pi/2"
Why do I even bother, honestly
You see a hard question on your test that your professor gave you to solve and you are like: The answer is = "You do it".
Excellent coverage of all important steps and gotchas!
I do have one criticism..
When you deal with the "negative r" value, don't describe it as "you have to go back One". Instead of sliding your pen by One, you should 'flip' the tip of the pen over (the base stays where it is) to implement the effect of a "negative radius value" at the subject theta angle.
😉
Regarding you do it, when I took Cal II, most of the test over polar integration was just setting them up. Basically the teacher already knew that we could integrate, but the setup on these problems is the tricky part. Also my teacher favored polar curves with lots of tiny loops, so he knew a fair bit of the test would be taken up just finding the intersections and creating the graphs.
5:43 this math problem out of context is funny
I LOLd at 4:40. YOU DO IT!
LOVE IT!
Is setting the absolute values of each equation equal to each other a valid way to find the intersections without the graph? Then you could just find θ at those r values for the limits of integration. (And if the actual values are negatives of each other, you could check if the angles are off by pi, right?)
Thanks for your efforts.
-pi/2
Thank you for this video! I've been having a hard time with the angles when the circles aren't centered at the origin. Your explanation was great!
Pretty 🥰
I think limits for both the integrals giving second area can be -π/2 to 0.
Please upload videos on flux, surface integrals. It would be really helpful.
I got the same result in the second one by using 3pi/2 and 2pi as lower and upper limits of integration, respectively.
and i got the same thing integrating using (-pi/2 to 0). :)
I am wondering on how can be the radius negative in polar coordinates. I though it should be positive by definition. Isn't the radius a length as a matter of fact? So what is the sense of r=-1?
For a sec I thought this notification was a reply from bprp and I was like wait I do it 😂
Do It! Just do it! Make your dreams come true!
upload ap calc frqs when they release!!
Would it be easier if you equalled the two r functions together to find the bounds?
Can we calculate the volume of a 3d polar function using some type of formula like this? For a function in the form:
f(theta, alpha) = r
You can use spherical coordinates
Thank u man
Hey, bprp, could you please do a video about why the sum of 1/n! equals e?
I must ask about all the explanations about setting the limits. When I last taught this (1995), I was told to expect that students would already know and understand about the relationship between theta and r when r is negative. Moreover, they should have already known about the fact that points where polar curves intersect would not necessarily correspond to the same (r, theta) combination. Is this not true any more?
try i^i^i^i^i^i^i^i^i^..... I tried it myself and got -1 and e^(pi/2) as answers. Thanks
Which univ are you from?
any student would be lucky to have u as their teacher, so easy to understand
yes we already know u is loved in integrals
I think that the first problem might be easier with basic geometry if you graph it.
There are points at (0,0) (1,0) (0,-1) (-1,-1). Each of these points have the tan line slope of 1 or -1 so that proves that you can find the center by making normal lines at these points. Of course that could be geometric or algebraic. Then those same lines give you the radius/diameter.
To find the segment of the circle in quadrant 1, use the triangles and circle sector formulas.
Specifically the triangle would be from (0,0) to (1,0) to (.5,-.5)[the center of the circle]. The lines are perpendicular at the center so that angle is 90°. The area of that portion in the first quadrant can now be defined of a quarter circle minus the triangle. Which is pi/8 minus 1/4.
Finally, the area of a quarter of the larger circle minus the section is pi/8 + 1/4.
That seems like a lot in typing but actually isn't much. The second problem can be approached similarly but with a little bit more geometry if anyone is interested I can type it out in the comments.
However, the important part is the number of problems that can't be solved geometrically and that's why calculus is my favorite subject and this video is entertaining and fun. Thanks BlackPenRedPen!
nicee
Putting a negative sign for the radius is a mistake, because it is defined to be positive and the angle makes the direction.
Although the integration is still the same you should integrate it from 3pi/2 to 2pi.
Nice
(Cos-sin)^2=1(cos-sin)^2=1-2cossin so we got a lot of cancellation to get first one is just plus integral of cossin and 2nd is negative of that need pen and paper to do the rest via di method
I messed up somewhere cause now I get both answers being zero
@@robertcotton8481 use the double angle identity after expanding.
2sin(ø)cos(ø) = sin(2ø)
cos(ø)^2 + sin(ø)^2 = 1
1/2 ∫ (cos(ø) - sin(ø))^2 dø
1/2 ∫ cos(ø)^2 - 2sin(ø)cos(ø) + sin(ø)^2 dø
1/2 ∫ 1 - sin(2ø) dø
1/2 ø + 1/4 cos(2ø) + C
hope this helps!
Hi BPRP,
We stumbled upon a question that is the indefinite integral of tan(cos(x) and we can't seem to figure out how to do it. Do you have any ideas of how to do this?
it doesn’t have an antiderivative
Pls load a video showing formula for sum 1+1/2+...+1/n.
a. π/4-(π/8-1/4)=π/8+1/4
b π/2-2(π/8-1/4)-π/4=1/2
No integrals, just (quarter) circle and triangle areas
Couldn't this be done geometrically? line y = x - 1 is a blue diameter, and for red you can do the quarter circle plus the right isosceles on the fourth quadrant...
of course, we guess.
and again, you do it!
please observe that red quarter circle Minus the right isosceles triangle, if we understood well.
how was exactly the question..
TRANSFORMATION OF GRAPHS PLEASE REQUEST!!!!!
r=cosΘ-sinΘ
substitute Θ for arctan(y/x)
converts to r=cos(arctan(y/x))-sin(arctan(y/x)
then from the right triangle you find cos(arctan(y/x)) is x/r and sin(arctan(y/x)) is y/r
substituting you get r=x/r-y/r
multiply through by r you get r²=x-y
substitute r² for x²+y² and you have x²+y²=x-y
move everything to left side you get x²-x+y²+y=0
complete the squares you get x²-x+1/4 + y²+y+1/4 = 1/2
factoring you get (x-1/2)² + (y+1/2)² = 1/2
So the other figure is indeed a circle, centered at (1/2,-1/2) with a radius of 1/√2
Things I learned from this video: it's been way too long since I did anything with polar coordinates and blue pens have somehow become acceptable.
Can’t you solve algebraically using area of segments and subtracting the segment area from big circle? I tried to do it and I might have made a mistake but I got quantity PI + 2 all over 8
I solved it by ordinary geometry and algebra and got it right. First one should be (pi+2)/8 and the second one should be 1/2. edit: i saw now that you answered this a long time ago, but maybe someone else who searches the comments would like to know.
Thanks it will help in my exam
You could also use 3pi/2 and 2pi as boundaries, right?
Sensei yes
@@blackpenredpen too confusing to be honest why did you assume the cooridnate ( pi , -1 ) for a point on the positive theta axis ?
Hashrima Senju because the negative 1 makes u move back into the first quadrant so its the same thing
@@isaacaguilar5642 explain more intutively pls
@@justabeardedguythatisahero9848, well let's say, please don't confuse (x, y(x)) coordinates, orthonormal, with (thêta, r(thêta)) ones, polar. the ‘place’ bprp pointed to, is expressed in terms of x,y (+1, 0). now in polar terms of th,r both (0, 1) and (π, -1). further of course (2π, 1) and (3π, -1) &c.
explanation needed? with thêta = 0 or 2π, 4π,… the radius axis points along the positive x-axis, then r=+1 coming up on x=+1.
with thêta = π or 3π, 5π,… the radius axis points along the negative x-axis, that is, r>0 lies along *x
But you get the same answer regardless of whether you pick "theta = 3*pi/2 to 2*pi" OR "theta = pi/2 to pi" ?? what's the difference.
Either way works. Thinking parametrically, an interval from 0
2:43 with subtitles "for the Virgin right" lmaaaaaaaaao
Lol
yooh, the same at 8:30 "…only what i just besides the Virgin, of course?"
funny, should be about points on the y-axis..
Homework from bprp XD
Mak Vinci yup!
It’s actually a video from a few weeks ago.
@@blackpenredpen It's ok, the world will know the area of circle soon 😁
The first curve in Cartesian is equal to:
2xy= (x² +y²)(1-(x²+y²))
Imagine integrating that!
What is the thing called after d?
theta
Aww I only know how to use Cartesian coordinates.
Bprp: r=cos()-sin()
Me: What is that? Is it y=cosx-sinx?
Me: wait... ()=π/4? What is going on?
I like the "you do it", then I did it : pi/8 for the first and 1/2 for the second. Right ?
Egill Andersson yes! : )))
Couldn't we define in the second example that -pi/2
@@blackpenredpen I think I have to disagree. I'm trying them myself, went through them, and double-checked my answers with Wolfram Alpha, and I didn't get pi/8 for the first answer. Instead, I got (pi+2)/8. Here's my math as best as I can enter it into here.
So obviously the first integral is pi/4, and we can check this geometrically by just seeing that it's a quarter of a circle with radius 1. The area of such a circle is pi, so a quarter of it is pi/4, and this is a straightforward integral anyway.
Now for the second integral, the integral from 0 to pi/4 of 1/2(cos(theta)-sin(theta))^2 d(theta).
First, let's work out that square, to get the integral from 0 to pi/4 of 1/2(cos^2(theta)+sin^2(theta)-2*cos(theta)*sin(theta)) d(theta).
We simplify this with one of our favorite identities and get the integral from 0 to pi/4 of 1/2(1 - 2*cos(theta)*sin(theta)) d(theta).
We can break this apart into two integrals and get the integral from 0 to pi/4 of 1/2*(1) d(theta) - the integral from 0 to pi/4 of 1/2*2*cos(theta)*sin(theta) d(theta).
Focusing on the first of those, it's simply the integral from 0 to pi/4 of 1/2 d(theta). This is a straightforward integral and yields pi/8.
For the second integral, we simplify it to the integral from 0 to pi/4 of cos(theta)*sin(theta) d(theta). This seems tricky, but it's easy enough to do a u-substitution with u = sin(theta), du = cos(theta) d(theta), and change the integral bounds from 0 to pi/4 into 0 to sqrt(2)/2. We now have the integral from 0 to sqrt(2)/2 of u du. This gives us 1/2 u^2 evaluated from 0 to sqrt(2)/2, which becomes 1/2(1/2-0) = 1/4.
Now we take our three results and add or subtract as is appropriate. We should have pi/4 - pi/8 + 1/4, which first simplifies to pi/8 + 1/4. If we want, we can rewrite this as (pi+2)/8.
For the second problem, I followed a very similar method (seeing as the integrals have only changed in their orders and boundaries, this isn't too complicated) and got 1/2, the same as the poster above.
I got (pi + 1)/4 and 1/2.
you do it
There is a serious misconception of polar coordinates in this video which URGENTLY needs to be corrected:
In polar coordinates r is ONLY defined in the domain [0,infinity), and is therefore always positive. The blue circle is displaced to a centre at (1/2,-1/2) with radius 1/2. You are right in that for this case r= cos (theta)-sin (theta), but then the domain of theta is [-3*pi/4,pi/4], which keeps r positive. THE WHOLE DOMAIN (pi/4, 5*pi/4) IS EXCLUDED. As usual, x= r*cos(theta) and y= r*sin(theta). Therefore, the circle is drawn clockwise in the allowed domain for x=[cos(theta)-sin(theta)]cos(theta), and y=[cos(theta)-sin(theta)]sin(theta). If you check it, you'll find the circle goes clockwise from (0,0) to (0,0) within this domain. The limits in your integral would then be 0 and pi/4 in your first example, and -pi/2 and 0 in the second one. The geometrical interpretation is straightforward as you can see.
I love your videos, but I REALLY URGE you to correct this one, since it's misleading.
Plz make a VEDIO on transformation of graphs plzzzzzzzzzzzsssssssssssss👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👌🏽👌🏽👌🏽👌🏽👌🏽👌🏽👌🏽👌🏽👌🏽👌🏽
well i'll let you know that 3pi/2 to 2pi also works
I figured out a way to solve this with geometry alone!
8:20 „Whenever you are on the y axis the theta is either pi/2 or 3pi/2.” - actually the origin is also on the y axis, and the theta is pi/4 there. :-)
Przemyslaw Kwiatkowski
I think I also said “anywhere on the y-axis beside the origin” after that.
Besides*
Indeed... that was 10sec later... :-)
They are all circles, we can do it with geometry. There are two “o’s” in “You do it”, is it a coincidence? I don’t think so.
The second curve is NOT a circle !
It’s a circle.
We rewrite the function as r = a*sin(theta+b), then use the property of right triangle in a circle, we can prove that the blue one is circle. The functions like a*sin(theta+b) are all circle.
@@JianJiaHe Soory ! You're right.
第二題角度要怎麼判斷？用零代入還是在同一個點。
呂永志 因為下面已經是π/2,所以我們必須用π
對，你英文是這樣說的。但我的意思是，如果沒有前一個角度，其實這角度不能確定對嗎？
呂永志
對 “要找鄰居”
我們也可用-π/2 to 0
@@blackpenredpen 這題圖不算複雜所以可行，但我想有些圖不易這樣判斷。
呂永志 是啊 我有算這個算到快瘋掉的時候
Oh no, geometry.
A:0.0535
B:0.678
The answer for the first us (pi-sqrt(8)+2)/4
= you do it
had my ib hl math paper 2 exam today, anyone else?
AddPrada Did you find it harder than paper 1?
nah p1 was harder, did p3 calculus yesterday and was pre easy. wbu
2+2=you do it.
The real challenge . . . solve them WITHOUT using calculus AT ALL. Just use plane geometry.
You crack a nut every time with a sledge hammer, that's a sure win but far less fun. Crack it with a usual nutcracker and you'll get a lot more fun.
I think it's easier without calculus. I took courses for learning this a long time ago and had forgotten how to do it with calculus. With plane geometry it was straightforward with just adding and subtracting parts.
If I wre your students and I have time in the test, I will write 'you do it' and then cross it out, and give the correct answer below it. XD
I solve the peoblem without integral
The asnwer its why!? xD
First I did it without any integral. I guess its way easier ;) But than I doublechecked it with your integrals :) Im happy that I got the same results xD
Writing "You do it" is the easiest way to do well in a test. You just have to hope the examiner is good at maths or will cheat by using the mark scheme.
SlenderCam lilll
If u have guts sir then try to solve JEE MAINS AND ADVANCE PAPER because it is the toughest paper in the world
Kiran Mahabole
Why is it the toughest test?
But the fun part is to solve this without calculus
this guy loves to hold balls!
a) pi/8 + 1/2
b) 1/2
Double check your work. 😉
@@Ni999 wrong signs?
@@obinnanwakwue5735 On the second one, yes. On the first one you (probably) have a sign wrong on the way to the final answer (I did too). Your self-checking hint for these kind of questions is that you're looking for area, answers must be positive.
@@Ni999 oh I get it I jacked up with the sign integrating one of the functions in the first problem, that should be + 1/2 and in the second one that should be 1/2 as well. Let me edit that.
@@obinnanwakwue5735 π/4 is the area of a full quarter red circle, the first problem is less than that, less than 0.785.
(π/8) + ½ ≈ 0.893
Double check your terms, you're close.
Second one is correct.
We havr homework😂
Redpenbluepen, not much of black pen
The first one is (pi-2)/8
Serouj Ghazarian i got (pi + 2)/8
Let me make you both happy by settling for (π±2)/8😁
@@d1o2c3t4o5r d it! I accidently put theta-(cos(2theta))/2 instead of theta+(cos(theta))/2
And the second one gives you....
One! Wow!