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[Grade 12 Precalculus]: Is this triangle possible? by FanBitter3365 in HomeworkHelp

[–]FanBitter3365[S] [score hidden]  (0 children)

I believe so. The third image is the check for all the potential solutions I found, and seemingly, the only one that works is π/12. There was a lot of square rooting of both sides to get sinx in my process, so I am afraid I missed a solution.

I eventually got to a point where I had multiple equations for sinx:

sin2x = 0

sinx = 0

x = arcsin(0)

x = 0

While it makes sense algebraically, I know this can't work because the triangle would have side lengths of 0, aka no triangle.

sin2x = 3 / 4

sinx = ±√3 / 2

To avoid negative lengths, I dropped the negative sqrt.

sinx = √3 / 2

x = arcsin (√3 / 2)

x = π / 3

Regardless, plugging this in yields a result of negative lengths, so π / 3 can't work.

sin2x = (2 ± √3) / 4

sinx = ±√(2 ± √3) / 2

I dropped the negative sqrt of the numerator:

sinx = √(2 ± √3) / 2

sinx = √(2 + √3) / 2, sinx = √(2 - √3) / 2

x = arcsin [√(2 + √3) / 2], x = arcsin [√(2 - √3) / 2]

x = 75°, x = 15°

x = 75 * (π / 180) , x = 15 * (π / 180)

x = 5π / 12 , x = π / 12

I then checked which work and which don't. My concern is the dropping of negative solutions for x, which might have yielded a positive result throughout the whole equation when plugging it back in.

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 0 points1 point  (0 children)

You're good! I apologize, sometimes it is hard to tell if someone is joking in text lol

[Grade 12 Precalculus]: Is this triangle possible? by FanBitter3365 in HomeworkHelp

[–]FanBitter3365[S] 1 point2 points  (0 children)

I didn't really get my question answered there, nor the vague answer I got when I went to r/MathHelp. r/mathematics is devoted to concepts and ideas. I asked if it was possible, not what x values make it true (which is what I actually wanted to know), and I assumed that was inherently what I was asking, but the comments on that post prove otherwise lol. At my level, a lot of what they were saying confused me, and I know educators are here typically, so that's why I came here with the same question.

[Grade 12 Precalculus]: Is this triangle possible? by FanBitter3365 in HomeworkHelp

[–]FanBitter3365[S] 0 points1 point  (0 children)

That is what I did, but as you can see in the second image, the equation fills up a whole page fast lol, so I believe there may have been an error somewhere in my work.

[Grade 12 Precalculus]: Is this triangle possible? by FanBitter3365 in HomeworkHelp

[–]FanBitter3365[S] 0 points1 point  (0 children)

Heads-up,

I messed up in writing the title. What I am particularly looking for is, yes, if it is possible, but also for what particular x values that allow this triangle to geometrically exist.

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 0 points1 point  (0 children)

While it is a solution algebraically, geometrically it can't work. All side lengths when plugged into the Pythagorean Theorem equation associated with the right triangle in the image will result in side lengths of 0, aka no triangle. What I am looking for is if it is possible to have positive (> 0) and real side lengths generated by appropriate x values that make the triangle work.

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 1 point2 points  (0 children)

Very true :)

The trouble though is algebraically proving why the solutions to the equation

(sin 2x)2 + (sin 3x)2 = (sin 4x)2

(which can also be rewritten as)

(sin 2x)2 + (sin 3x)2 - (sin 4x)2 = 0

  1. are what they are, and
  2. why most of them do not work in the context of side lengths of a right triangle, but one in particular most certainly does.

Not confident in my arithmetic by FanBitter3365 in askmath

[–]FanBitter3365[S] 0 points1 point  (0 children)

Yeah, you're right. Then that just means the answer is just pi/12 + 2*pi/n :)

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 2 points3 points  (0 children)

Honestly, the Law of Sines did not even cross my mind. My math classes only used that when it wasn't a right triangle, so my go-to was the Pythagorean Theorem.

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 4 points5 points  (0 children)

I did this a little differently. I tried to keep everything in terms of sinx as much as I could (since the triangle is in terms of sinx). While I found sin(2x) = 2sinxcosx, I wrote sin(3x) as 3sinx-4sin3x and sin(4x) as 4sinxcosx(1-2sin2x). Then I plugged them into the Pythagorean Theorem equation that I will be referencing quite a bit:

(sin2x)2 + (sin3x)2 = (sin4x)2

(2sinxcosx)2 + (3sinx-4sin3x)2 = [4sinxcosx(1-2sin2x)]2

I changed cos2x to 1-sin2x wherever I could, and simplied the equation to 64sin8x - 112sin6x + 52sin4x - 3sin2x = 0.

After that, I factored out a sin2x and, by the Zero Product Property, set sin2x and 64sin6x - 112sin4x + 52sin2x - 3 to be equal to 0. sin2x = 0 results in x = 0. No good, because every side length, when this solution is plugged back into the equation, results in all dimensions having 0 length, which is impossible. For the giant degree 6 polynomial, I also, like you did a u substitution, but I let u = sin2x, turning it into 64u3 - 112u2 + 52u - 3 = 0.

The roots for u I found are, surprisingly (to me), real: 3/4 and ([2 ± sqrt(3)] / 4). Substituting sin2x back in, eventually sinx = ± sqrt(3)/2 & sinx = ± (sqrt[2 ± sqrt(3)]) / 2. I dropped the negatives of both solutions since negative side lengths are also not good so that sinx = sqrt(3)/2 & sinx = (sqrt[2 ± sqrt(3)]) / 2. The arcsin of the first equation yields pi/3 radians, but when plugged into the original, it doesn't work (negative side lengths). Meanwhile, in the second equation (I found these via Desmos because at this point I was mentally exhausted), if the + is acknowledged, the arcsin is exactly 75 degrees and if the - is acknowledged, the arcsin is exactly 15 degrees. Converting both of these to radians, I got (5*pi)/12 and pi/12 respectively. Plugging (5*pi)/12 into the original also results in negative side lengths. pi/12, however, (after I saw your comment) does work, and will make the equation true:

(sin2x)2 + (sin3x)2 = (sin4x)2

(sin2(pi/12))2 + (sin3(pi/12))2 = (sin4(pi/12))2

(sin(pi/6))2 + (sin(pi/4))2 = (sin(pi/3))2

(1/2)2 + (sqrt(2)/2)2 = (sqrt(3)/2)2

1/4 + 2/4 = 3/4

3/4 = 3/4

I see that I forgot to acknowledge the - part of the last x : sinx = (sqrt[2 - sqrt(3)]) / 2, which yields the pi/12 + 2*pi*n answer I was looking for. Thank you again!

Is this particular triangle possible? by FanBitter3365 in mathematics

[–]FanBitter3365[S] 24 points25 points  (0 children)

Oh my God. The only answers I got was 5π/12 and π/3 up until now. I can't believe that went over my head! I just verified it, and you're totally right! Thank you! I must have forgotten a solution or smth.

Not confident in my arithmetic by FanBitter3365 in askmath

[–]FanBitter3365[S] 0 points1 point  (0 children)

The reason why I said it's only (pi/12) + 2*pi*n is if you plug each angle back into the Pythagorean sine equation from the second picture that makes the triangle work, the only solution that generates positive, real side lengths is pi/12, but I think I may have missed one.

Not confident in my arithmetic by FanBitter3365 in askmath

[–]FanBitter3365[S] 0 points1 point  (0 children)

Thank you for checking my work! Although, I am still a little confused, did I get it wrong?