I feel like some checking of the final results would be useful, since there really aren't 3 solution, let alone 7 (as the video suggests): - All three solutions in the case of x=0,+/- pi give us a degenerate triangle with side lengths (0,0,0), which some might not consider as a proof for the existence of a triangle with the given form. - Both of the solution in the case x=+/- pi give a triangle with one side of length zero, and another side with a negative length, the possibilities being either (1,0,-1) or (-1,0,1). again, both of these sets of side length uphold the Pythagorean theorem, but they do not in any way prove the existence of a triangle with the wanted form (if we allow for side lengths o zero, this case be corrected by taking the absolute value of each side). - Only in the case of x=+/- pi/6, +/- 5*pi/6 do we get an actual solution. The given values of x correspond to side lengths of (1/2.sqrt(3)/2,1), (-1/2,-sqrt(3)/2,-1),(1/2,-sqrt(3)/2,1),(-1/2,sqrt(3)/2,-1). All solutions except one have negative side length (which again, might make sense if placed on some grid, but has absolutely no manning in the strictly geometric problem). I would say that a correct solution to the question in this video would be: Yes, a solution does exist, which is a triangle with side lengths (1/2.sqrt(3)/2,1), given only by x=pi/6 (assuming x on the interval (-pi, pi)). I get that the point of the video is mostly to show the beautiful connection between complex numbers and the trigonometric multi-angle formulas for sin and cos. But still, it is important to remember what the initial question was.
Thank you for finishing the video by resolving the part that makes it interesting. Failing to answer the actual question would give a low score on an exam.
As it was about right triangles on a plane which must have internal angles adding up to 180 degrees or pi radians then 5 pi /6 + pi /2 = 8 pi /6 > pi can be dismissed leaving the answer x= pi/6 as you suggested.
@@tedr.5978 or two other farmore likely scenarios... 1) it's a legitimate mistake, which happens. 2) it's deliberate in order to get people thinking for themselves. (akin to the deliberate mistakles in ye olde meccano instruction booklets) If you give up watching these videos on the basis of this thumbnail, maybe maths isn't for you?
Triangle sides must obviously be positive, so sin(x), sin(2x) and sin(3x) cannot be negative or zero. If x = pi/2 then sin(2x) = 0, which doesn't make a triangle. The only possible case is when x=pi/6. The triangle sides are sin(x) = 1/2, sin(2x) = sqrt(3)/2, sin(3x) = 1.
@@Apollorion Yes, of course, since sin is 2pi periodic you trivially have an infinite family of solutions, but upon restricting to a reasonable domain, say -pi to pi, you don't have any redundant ones.
This was not a good place to stop. When talking about triangle side lengths, we should only consider the positive values of sin(x). So only sin(x) = 1/2 and sin(x) = 1. From there we get x = pi/6, 5pi/6, or pi/2, a.k.a. 30, 90, 150 degrees. BUT! When x = 90 degrees, 2x = 180 degrees => sin (2x) = 0, not a side length. When x = 150 degrees, 2x = 300 degress => sin (2x) = -SQRT(3)/2 < 0, not a side length. Finally, when x = 30 degrees, 2x = 60 degrees, 3x = 90 degrees, so we get sin(x) = 1/2, sin(2x) = sqrt(3)/2, sin(3x) = 1, and 1/2^2 + (sqrt(3)/2) = 1/4 + 3/4 = 1 = 1^2. So there is only 1 valid x = pi/6 a.k.a. 30 degrees, and not the 3+4+2 = 9 solutions shown on the board.
This is epic I used the sine sum and difference formula by moving sin^2x to the right. This is ingenious and you already know which angle is the right one
x = pi/6 seems to be the only solution that would really make sense as the others would give you either side lengths of 0 or negative side lengths which wouldn't make an actual triangle.
If we only accept positive numbers as the sides, excluding 0 and negative values, then sinx = 1/2 and x = pi/6 + 2npi is the only solution, the length of the three sides will be 1/2, sqrt(3)/2 and 1 respectively
I don't think x = 0 or +/- pi counts as a right triangle - side lengths would be (0,0,0). With x = +/- pi/2 you get side lengths (1,0,1) and (-1,0,-1) respectively - again, hard to call those "real" right triangles. Restricting to positive real values for the sides, pi/6 (giving sides 1/2, \sqrt{3}/2, 1) is the only one.
Easy problem. For ANY triangle with angles A,B,C sin(A)=Sin(B+C) , sin(B)=sin(A+C), & sin(C) = sin(A+B), because A+B+C=180 degrees. For Michael's problem 1x+2x+3x= 6x = 180, thus x = 30 degrees => 30-60-90 triangle Euclidean terms: the measure of an exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles.
of these, the only valid (non-degenerate) triangle is the one with x=π/6, and we get the triangle with sides 1/2, √3/2, 1. you might notice that the hypotenuse is 1, so the small angle there is actually x. 2x=π/3=π/2-π/6, so sin(2x) = sin(π/2-x)=cos(x). 3x=π/2. The other two have sidelengths 0,0,0, and 1,0,1
Curious where you're going with this. Per the thumbnail, we have a triangle with sides sin(x), sin(2x) and sin(2x). This is a 36-72-72-degree isosceles triangle in planar geometry. It appears to be right in the picture, but of course that would mean sin(x) = 0 per Pythagoras and we'd have a degenerate triangle with area 0. Edit: And the actual problem is not what's in the thumbnail. Now we clearly have a 30-60-90 triangle and sin(x) = 1/2, sin(2x) = sqrt(3)/2 and sin(3x) = 1. Still curious.... Edit 2: sin(3x) is opposite a right angle per the construction. This means sin(3x) = 1. The only angle we can have in a triangle that has sin 1 is pi/2 (as all usable angles must be strictly between 0 and pi), so 3x = pi/2. This means x = pi/6. If we accept the basics of triangles, this is the only possible solution. I mean, it was fun, but it's kinda like a theme park safari-venture ride. You wind up back where you started except with a sunburn at best.
Only one of these solutions, x = pi/6 is acceptable for a true triangle. The constraint imposed here is to comply with the pythagorean theorem but you would also like to add the constraint of all sides been larger than zero since they are lenghts of a polygon.
The x=π/6=30° solution, with sides sin30°=1/2, sin60°=√3/2 and sin90°=1, also happens to be a 30°-60°-90° right triangle. It's not magic, it's literally one of the definitions of sine
If we are asking is such a right triangle ‘possible’, the one possible solution is to have the largest angle, 3x=90deg… Hence x would be pi/6. Then the sides would be 1/2, root(3)/2, and 1, which works, so it is definitely possible. Is it the only solution? TBH I also solved it first using the double and triple angle formulae, as you did, and apart from x=pi/6+2K*pi, all of the others either are degenerate, or give negative sides… eg x=5*pi/6 gives sides: 1/2, -root(3)/2, and 1. Nice problem though. Thank you
A bit lazy to not check the solutions against some obvious conditions... like non-degenerate triangles, or more importantly, *non-negative side lengths*
@GroundThing hmm, maybe - even so, I think that approach isn't ideal, since now it's assuming the whole viewership will pick up on the joke, and pick up on the fact that not all the cases work
I did a simple graph of sin(3x) compared to sqrt(sin(x)^2+sin(2x)^2), the pythagorean one looks like infinite teeth, but taking the pythagorean sin function to greater lengths ended me up with a whole other rabbit hole, so thanks! also pi/6 is one of the infinite solutions.
Following some comments in different threads, I'm going to offer an alternative solution. Using the sine rule, a/sinA = b/sinB = c/sinC. on the triangle gives us a = sin(x), b = sin(2x) and c = sin(3x). Note than the acute angles A and B are clearly less than 90°, which means that arcsin(sinA) has a single value, namely A, in that range, and the same is true for B. Now we have sin(x)/sinA = sin(2x)/sinB = sin(3x)/sinC, and the only way for that to be true is if B = 2A and C = 3A, by taking arcsin of the top and bottom of each fraction. So A + 2A + 3A = 180°, making A = 30°, hence B = 60° and C = 90°. And that's the right-angled triangle we want. [Edit:] Oops! Taking arcsin of the top and bottom of a fraction does not preserve its value, so there's a huge flaw in my argument. Oddly enough the result turns out to be true in this case, but I'm yet to figure out why.
I've replied to you on our original discussion, but i have a new wonder. If we exclude the last sentence of your comment, didn't you prove somehow that every triangle with side lengths (sin(x),sin(2x),sin(3x)) has to be a 30-60-90 triangle, which isn't correct? I actually think you found a very cool proof, but you got the order wrong at the end., and the correct order should be something like: If we require that C=90deg then we must have A=30deg and B=60deg. Since this happens to uphold the requirement A+B+C=180deg, then this triangle can exist.
@@andyneeman4378 I thought I proved that every triangle with side lengths sin(x), sin(2x), sin(3x) _that is a right triangle_ has to be 30-60-90. but you may be right that I assumed the condition, rather than using it! Thanks for the interesting comments.
@@RexxSchneider You should also know there is an inherent flaw in your argument, which i explained in the original thread. This has been a very fun, interesting, and respectful exchange, so thank you.
ISTM the only valid answer is sin x = 1/2. So the triangle's side lengths are 1/2, √3/2, and 1. The other values of sin x result in nonpositive side lengths.
I dont agree with the, "That's a good place to stop" statement. A good place to stop would be after feeding the answers back in and exploring the significance of the answers.
I am not sure what you call it, but the "teaser" which invites people to view this video has the hypotenuse as equal to the base as equal to 2x. Obviously it would be nice to see that fixed.
I approached this quite differently. Since the title of the video only asks for existence, I did the following. Let f(x)=sin^2x+sin^2(2x)-sin^2(3x). This function is clearly continuous. If we can find points where it's >=0 and
This is actually a very neat proof, but you are missing some important details in your proof. Your proof works only if the two points we find for the IVT are on the open range (0,pi/3), other wise you can a redundant case (all sides are of length zero), or cases with negative valued sides. Still, this is the most creative solution i saw for this problem, so props to you.
Michael, i think you need to a) fix the thumbnail b) don’t stop before eliminating degenerate triangles c) comment on the application of the sine rule .
I realised the issue with degenrate cases from the get go. but as for wether this was a good place to stop, without discussing the degenrate cases, I think it WAS a good place to stop, as it generated discussion and thinking about it in the comments. Which is a good thing for people to do, isn;'t it? And I have a feeling he's done solutions with such degenrate cases before so he'd be treading old ground.
The discussion below is as interesting and educational as the problem itself. (I wish his problems were better motivated, but that might be too hard given the need for something new each day.)
This was probably the worst video I ever watched from you. The thumbnail is wrong, the derivation of double angle identity felt so redundant (the triple thing would suffice) and you don't comment on the acceptable values of sin(x) as a side length at all.
I feel like some checking of the final results would be useful, since there really aren't 3 solution, let alone 7 (as the video suggests):
- All three solutions in the case of x=0,+/- pi give us a degenerate triangle with side lengths (0,0,0), which some might not consider as a proof for the existence of a triangle with the given form.
- Both of the solution in the case x=+/- pi give a triangle with one side of length zero, and another side with a negative length, the possibilities being either (1,0,-1) or (-1,0,1). again, both of these sets of side length uphold the Pythagorean theorem, but they do not in any way prove the existence of a triangle with the wanted form (if we allow for side lengths o zero, this case be corrected by taking the absolute value of each side).
- Only in the case of x=+/- pi/6, +/- 5*pi/6 do we get an actual solution. The given values of x correspond to side lengths of (1/2.sqrt(3)/2,1), (-1/2,-sqrt(3)/2,-1),(1/2,-sqrt(3)/2,1),(-1/2,sqrt(3)/2,-1). All solutions except one have negative side length (which again, might make sense if placed on some grid, but has absolutely no manning in the strictly geometric problem).
I would say that a correct solution to the question in this video would be: Yes, a solution does exist, which is a triangle with side lengths (1/2.sqrt(3)/2,1), given only by x=pi/6 (assuming x on the interval (-pi, pi)).
I get that the point of the video is mostly to show the beautiful connection between complex numbers and the trigonometric multi-angle formulas for sin and cos. But still, it is important to remember what the initial question was.
In other words, that was not a good place to stop. Good analysis!
@@Jullan-Mthx
Thank you for finishing the video by resolving the part that makes it interesting. Failing to answer the actual question would give a low score on an exam.
I agree. I didn't do the analysis you did, I just thought: x=0, that's not a triangle
As it was about right triangles on a plane which must have internal angles adding up to 180 degrees or pi radians then
5 pi /6 + pi /2 = 8 pi /6 > pi can be dismissed leaving the answer x= pi/6 as you suggested.
umm is the thumbnail wrong? ^^v
Yes, the hypotenuse should be sin(3x), not sin(2x)
My guess, that's a bait
It's wrong but also on purpose I guess : D
Click bait thumb nail = not watching the video nor anything else from this person.
@@tedr.5978 or two other farmore likely scenarios...
1) it's a legitimate mistake, which happens.
2) it's deliberate in order to get people thinking for themselves. (akin to the deliberate mistakles in ye olde meccano instruction booklets)
If you give up watching these videos on the basis of this thumbnail, maybe maths isn't for you?
Triangle sides must obviously be positive, so sin(x), sin(2x) and sin(3x) cannot be negative or zero.
If x = pi/2 then sin(2x) = 0, which doesn't make a triangle.
The only possible case is when x=pi/6. The triangle sides are sin(x) = 1/2, sin(2x) = sqrt(3)/2, sin(3x) = 1.
Well spotted
If we allowed for degenerate triangles, then you'd need to go a step further in the pi/2 case, and the 0 case actually works.
How about x=13pi/6 ? Triangle sides would be the same as for x=pi/6
@@Apollorion Yes, of course, since sin is 2pi periodic you trivially have an infinite family of solutions, but upon restricting to a reasonable domain, say -pi to pi, you don't have any redundant ones.
This was not a good place to stop.
When talking about triangle side lengths, we should only consider the positive values of sin(x). So only sin(x) = 1/2 and sin(x) = 1.
From there we get x = pi/6, 5pi/6, or pi/2, a.k.a. 30, 90, 150 degrees.
BUT!
When x = 90 degrees, 2x = 180 degrees => sin (2x) = 0, not a side length.
When x = 150 degrees, 2x = 300 degress => sin (2x) = -SQRT(3)/2 < 0, not a side length.
Finally, when x = 30 degrees, 2x = 60 degrees, 3x = 90 degrees,
so we get sin(x) = 1/2, sin(2x) = sqrt(3)/2, sin(3x) = 1,
and 1/2^2 + (sqrt(3)/2) = 1/4 + 3/4 = 1 = 1^2.
So there is only 1 valid x = pi/6 a.k.a. 30 degrees, and not the 3+4+2 = 9 solutions shown on the board.
Oh it was - the amount of engagement on this solution has been an education in itself.
At 7:18, that should be 3 sin(x)-4 sin^3(x).
Glad I’m not the only one to have noticed!
Came to say this. Haven't finished the video yet but I hope this doesn't effect the end result 🤣
Oh lol he fixed it after 10 seconds 😂
Using the sine rule with angles A, B, C
sin(x)/sin(A) = sin(2x)/sin(B) = sin(3x)/sin(C)
So x=A=30 deg, B=60 deg, C=90 deg is a solution
Came down to comments to write exactly this.
This is epic I used the sine sum and difference formula by moving sin^2x to the right. This is ingenious and you already know which angle is the right one
best clickbait thumbnail so far..worked like magic.
I clicked only because it showed hypotenuse to be sin(2x)
Are we counting degenerate cases? Because if sin(x) = 0, then all we'd get is a point...
We also can't have lengths of 1,0,-1 for x=pi/2
x = pi/6 seems to be the only solution that would really make sense as the others would give you either side lengths of 0 or negative side lengths which wouldn't make an actual triangle.
If we only accept positive numbers as the sides, excluding 0 and negative values, then sinx = 1/2 and x = pi/6 + 2npi is the only solution, the length of the three sides will be 1/2, sqrt(3)/2 and 1 respectively
I don't think x = 0 or +/- pi counts as a right triangle - side lengths would be (0,0,0). With x = +/- pi/2 you get side lengths (1,0,1) and (-1,0,-1) respectively - again, hard to call those "real" right triangles. Restricting to positive real values for the sides, pi/6 (giving sides 1/2, \sqrt{3}/2, 1) is the only one.
I kept doing the algebra in my head trying to understand where the extra -2sin³x came from. 😅
I would imagine the sides as vectors and for -π/6 the triangle is mirrored but you got point
The thumbnail I saw was an isosceles triangle with a right angle in the bas which hurts my brain
It was not denoted as the right angle though...
Let the short side = 0, then both angles = 90degrees. Easy peasy!
Easy problem. For ANY triangle with angles A,B,C sin(A)=Sin(B+C) , sin(B)=sin(A+C), & sin(C) = sin(A+B),
because A+B+C=180 degrees. For Michael's problem 1x+2x+3x= 6x = 180, thus x = 30 degrees => 30-60-90 triangle
Euclidean terms: the measure of an exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles.
Thumbnail is possible 5x=180=> x=36 => Isosceles triangle 36-72-72
What?
@@3dindian what what? maybe you need sin(z)=sin(Pi-z). for angle z, pi-z is the exterior angle
of these, the only valid (non-degenerate) triangle is the one with x=π/6, and we get the triangle with sides 1/2, √3/2, 1. you might notice that the hypotenuse is 1, so the small angle there is actually x. 2x=π/3=π/2-π/6, so sin(2x) = sin(π/2-x)=cos(x). 3x=π/2.
The other two have sidelengths 0,0,0, and 1,0,1
Curious where you're going with this. Per the thumbnail, we have a triangle with sides sin(x), sin(2x) and sin(2x). This is a 36-72-72-degree isosceles triangle in planar geometry. It appears to be right in the picture, but of course that would mean sin(x) = 0 per Pythagoras and we'd have a degenerate triangle with area 0.
Edit: And the actual problem is not what's in the thumbnail. Now we clearly have a 30-60-90 triangle and sin(x) = 1/2, sin(2x) = sqrt(3)/2 and sin(3x) = 1. Still curious....
Edit 2: sin(3x) is opposite a right angle per the construction. This means sin(3x) = 1. The only angle we can have in a triangle that has sin 1 is pi/2 (as all usable angles must be strictly between 0 and pi), so 3x = pi/2. This means x = pi/6. If we accept the basics of triangles, this is the only possible solution. I mean, it was fun, but it's kinda like a theme park safari-venture ride. You wind up back where you started except with a sunburn at best.
That was not a good place to stop
"Check your answers!" - math teachers, everywhere
Only one of these solutions, x = pi/6 is acceptable for a true triangle. The constraint imposed here is to comply with the pythagorean theorem but you would also like to add the constraint of all sides been larger than zero since they are lenghts of a polygon.
I prefered to use the difference of squares, sine sum and difference formulas applied to `sin(3x)` and `sin(x)` to simplify the equation quickly
The x=π/6=30° solution, with sides sin30°=1/2, sin60°=√3/2 and sin90°=1, also happens to be a 30°-60°-90° right triangle.
It's not magic, it's literally one of the definitions of sine
try the same task with sin(3x), sin(4x) and sin(5x), since (3, 4, 5) is the most known Pythagorean triple.
Dude the thumbnail is extremely wrong
sin x can't be negative because is a length
and if you remove the answers that give a 0 for a side the only answer is pi/6
Clearly sin(x)
Surely there should be a last stage in which you check all three sides of the triangle are >0?
In which case x=pi/6 is the only answer.
Thumbnail is clickbait, cause it’s wrong
The only nondegenerate solution is x = 30°, which leads to a triangle with α = 30°.
Is it a good place to stop? Maybe useful to mention that only sin(x) = 1/2 gives solutions in which all the sides of the triangle are non-zero!
If we are asking is such a right triangle ‘possible’, the one possible solution is to have the largest angle, 3x=90deg… Hence x would be pi/6. Then the sides would be 1/2, root(3)/2, and 1, which works, so it is definitely possible. Is it the only solution?
TBH I also solved it first using the double and triple angle formulae, as you did, and apart from x=pi/6+2K*pi, all of the others either are degenerate, or give negative sides… eg x=5*pi/6 gives sides: 1/2, -root(3)/2, and 1. Nice problem though. Thank you
when Sin(x)
A bit lazy to not check the solutions against some obvious conditions... like non-degenerate triangles, or more importantly, *non-negative side lengths*
It felt like a prank, especially ending it with what felt like a wink and a nod in the "definitely make this a right triangle" line.
@GroundThing hmm, maybe - even so, I think that approach isn't ideal, since now it's assuming the whole viewership will pick up on the joke, and pick up on the fact that not all the cases work
Engagement farming with intentional mistakes in thumbnail 🤮
BPRP did the same question 1 year ago. Love the different ways of solving it.
I did a simple graph of sin(3x) compared to sqrt(sin(x)^2+sin(2x)^2), the pythagorean one looks like infinite teeth, but taking the pythagorean sin function to greater lengths ended me up with a whole other rabbit hole, so thanks! also pi/6 is one of the infinite solutions.
Following some comments in different threads, I'm going to offer an alternative solution. Using the sine rule, a/sinA = b/sinB = c/sinC. on the triangle gives us a = sin(x), b = sin(2x) and c = sin(3x). Note than the acute angles A and B are clearly less than 90°, which means that arcsin(sinA) has a single value, namely A, in that range, and the same is true for B.
Now we have sin(x)/sinA = sin(2x)/sinB = sin(3x)/sinC, and the only way for that to be true is if B = 2A and C = 3A, by taking arcsin of the top and bottom of each fraction.
So A + 2A + 3A = 180°, making A = 30°, hence B = 60° and C = 90°. And that's the right-angled triangle we want.
[Edit:] Oops! Taking arcsin of the top and bottom of a fraction does not preserve its value, so there's a huge flaw in my argument. Oddly enough the result turns out to be true in this case, but I'm yet to figure out why.
I've replied to you on our original discussion, but i have a new wonder. If we exclude the last sentence of your comment, didn't you prove somehow that every triangle with side lengths (sin(x),sin(2x),sin(3x)) has to be a 30-60-90 triangle, which isn't correct?
I actually think you found a very cool proof, but you got the order wrong at the end., and the correct order should be something like: If we require that C=90deg then we must have A=30deg and B=60deg. Since this happens to uphold the requirement A+B+C=180deg, then this triangle can exist.
@@andyneeman4378 I thought I proved that every triangle with side lengths sin(x), sin(2x), sin(3x) _that is a right triangle_ has to be 30-60-90. but you may be right that I assumed the condition, rather than using it!
Thanks for the interesting comments.
@@RexxSchneider You should also know there is an inherent flaw in your argument, which i explained in the original thread. This has been a very fun, interesting, and respectful exchange, so thank you.
is the thumbnail typo actually ingenious clickbait or smth. i was so confused lmao
ISTM the only valid answer is sin x = 1/2. So the triangle's side lengths are 1/2, √3/2, and 1. The other values of sin x result in nonpositive side lengths.
Isn't a 30, 60, 90 triangle a trivially obvious solution?
1:50 Horseshoe!
I dont agree with the, "That's a good place to stop" statement. A good place to stop would be after feeding the answers back in and exploring the significance of the answers.
is it math wise ok to just ditch the imaginary parts ?
So this right triangle can only be the 30-60-90 right triangle
I am not sure what you call it, but the "teaser" which invites people to view this video has the hypotenuse as equal to the base as equal to 2x. Obviously it would be nice to see that fixed.
I approached this quite differently. Since the title of the video only asks for existence, I did the following. Let f(x)=sin^2x+sin^2(2x)-sin^2(3x). This function is clearly continuous. If we can find points where it's >=0 and
This is actually a very neat proof, but you are missing some important details in your proof. Your proof works only if the two points we find for the IVT are on the open range (0,pi/3), other wise you can a redundant case (all sides are of length zero), or cases with negative valued sides. Still, this is the most creative solution i saw for this problem, so props to you.
pi/6, for the given interval. Otherwise, pi/6 + 2(pi)n, where n is a natural number, is also valid. The others don’t form a valid triangle.
Right triangle: 0 < x < 1/2 π
45 45 90 would satisfay this
The thumbnail was wrong
You must factor out sin^2 x
Michael, i think you need to
a) fix the thumbnail
b) don’t stop before eliminating degenerate triangles
c) comment on the application of the sine rule .
De Moivre's formula.
wrrreeeeee wrong thumbnail piccc!!!!!111 love this triangle btw
WHy not just use the sin addition formula?
terrible place to stop
4:40 again a bug in the matrix 😀
Oh the thumbnail is not representative of the video. Bye.
Thumbnail was false.
I realised the issue with degenrate cases from the get go.
but as for wether this was a good place to stop, without discussing the degenrate cases, I think it WAS a good place to stop, as it generated discussion and thinking about it in the comments. Which is a good thing for people to do, isn;'t it?
And I have a feeling he's done solutions with such degenrate cases before so he'd be treading old ground.
Umm the thumbnail seems incorrect, and ideally sin x=0 shouldn't be a solution, that would be a degenerate case!
Worse still, try and see what the sin x = 1 case gives...
And - signs for triangle lebghts...
The discussion below is as interesting and educational as the problem itself.
(I wish his problems were better motivated, but that might be too hard given the need for something new each day.)
This was probably the worst video I ever watched from you. The thumbnail is wrong, the derivation of double angle identity felt so redundant (the triple thing would suffice) and you don't comment on the acceptable values of sin(x) as a side length at all.
Nice trigonometry gymnastics.. 😊
You got it buddy, that was it. True to the educational system, true to mathematics.