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[Solved] Algebraic Solutions to Graphical Trig Problems? (mander.xyz)

submitted 8 months ago* (last edited 8 months ago) by DarkNightoftheSoul@mander.xyz to c/math@lemmy.ml

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I've been knocking out the trig problems in this section with minimal difficulty so far, but I've run straight into a brick wall on this "Algebraic" part. I'm asked to find sin(x)=0 between [0,2π). If I graphed the unit circle this would be a trivial exercise to show sin(θ)=0 when θ=0 or π.

Where I have trouble is- I'm very explicitly being told here that the solution is ALGEBRAIC, and I'm struggling to figure out a way to rearrange sin(x)=0 to come up with the known answer. Further, unit circles are not in this chapter, they wouldn't likely ask me to exercise a skill taught in another chapter. What am I missing?

It's not just 31, either. Looking ahead at eg 37, I can easily show sin(-x) = -sin(x) on a unit circle. I could maybe fuck around with inverse trig ratios but those are in section 3- this is only section 1.

Help me out here, drop a hint, share a link: how do I solve sin(x)=0 on [0,2π), but algebraically? I suspect it's something glaringly obvious and/or very very simple I've overlooked.

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[–] myslsl@lemmy.world 3 points 8 months ago* (last edited 8 months ago) (2 children)

There's nothing here that tells you that you need to use some technique from algebra explicitly as far as I can see?

Edit: I misread. See my follow up comment too.

The usual trig functions like sine and cosine are famous examples of transcendental functions so I very seriously doubt there is some clever algebra trick you're missing that the author intends you to do.

I'd assume your instructor (or the author) is expecting you to use your geometric knowledge about the unit circle on these if you haven't covered inverse trig functions yet. But I also can't really read their mind, so your best bet might be to just directly ask them?

Edit: This could also depend on what trigonometric identities you know so far too.

[–] myslsl@lemmy.world 2 points 8 months ago* (last edited 8 months ago) (1 children)

Okay, I see. I'm fucking blind and did not see the words "algebraic" literally at the top of the screenshot.

For what it is worth, they could just be referring to how they are representing the problems they are asking rather than the form of the intended solutions with that.

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago* (last edited 8 months ago)

You think... because they wrote the instructions in the form let f(x) = sin(x)... that that... that's what they mean... by algebraic?

god help me I fear you may be right...

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago (1 children)

I'm studying straight out of the book in preparation for CLEP precalculus.

The header for this section of exercises is "ALGEBRAIC"[emphasis not mine]. The previous section of exercises was GRAPHICAL. The previous section of exercises required me to either graph or interpret sinusoidal equations of the form y=Asin(Bx-C)+D (or cos). Given that, and given that previous chapters had similar sections and headings where I was expected to find the solutions algebraically, I was expecting a more... algebraic solution.

I know sin, cos, tan, csc, sec, cot, and strictly speaking I know (or am at least passably familiar with) inverse sine, cosine and tangent from khan academy, but the author wouldn't expect that.

I very seriously doubt there is some clever algebra trick you’re missing that the author intends you to do.

That's pretty much where I'm at now, probably overthinking it. I suppose this could be a typo, they meant GEOMETRIC instead of ALGEBRAIC. I've spent probably about half as much time unfucking typos in this book as actually working exercises.

[–] myslsl@lemmy.world 3 points 8 months ago* (last edited 8 months ago) (1 children)

I realized in retrospect I misread the header so I apologize for that.

I'm still betting they aren't expecting a true algebraic or analytic solution here. Things like finding max/min points, finding arbitrary particular values of trig functions, solving trigonometric equations and so on can be notoriously hard in the absence of geometric reasoning/intuitions.

Later on if you decide to study calculus you might eventually see the sine and cosine functions defined rigorously via infinite series. That may sound convoluted, but part of the purpose of doing that is because of the difficulties of the issues mentioned above. Basic sounding facts like: What is sin(0.1234)? are not so easy to answer where you are at but can be dealt with more conveniently using these kinds of tools from calculus.

The questions being asked here are also just kind of typical knee jerk facts that most people want students coming out of a trig class to just know.

I think your reasoning geometrically seems very on the right track. Appealing to the unit circle or the graph of y=sinx for these feels correct in the sense of what a trig student would be expected to know coming out of or during a trig course.

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago

Thank you that's very reassuring. I've marked this solved with your help.

[–] freshcow@lemmy.world 2 points 8 months ago (1 children)

You are solving for x. To separate x from the sin function, you should use the inverse sin function (or arcsin). Do this to both sides of the equation and you will be left with x = arcsin(0)

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago* (last edited 8 months ago) (1 children)

Okay, like I get that, but I don't think that's the procedure intended by the author. For one, it will only give me one of the two correct solutions in this ~~range~~ domain. For another, I'm not supposed to know about inverse functions yet, that's two sections ahead.

[–] jdnewmil@lemmy.ca 2 points 8 months ago (1 children)

Nope. When you use sqrt to solve y=x^2 , you are expected to recognize that sqrt is only the inverse of the right branch of the parabola. Likewise, arcsin is only the inverse of the sine limited to +/-pi/2, so you have to use it intelligently as a tool, not blindly as a black box.

[–] myslsl@lemmy.world 2 points 8 months ago

The point they made was correct. Arcsine by itself only gives one of the two solutions to sinx=0. It seems like they already realize that they need to use arcsine carefully if they use it at all.

[–] wuphysics87@lemmy.ml 2 points 8 months ago (2 children)

For unit circle:

x^2 + y^2 = 1

in general

cos A = x/ r

sin A = y / r

Here:

cos A = x and sin A = y

From first equation

x^2 + sin^2 A = 1

sin A = sqrt ( 1- x^2)

sin A = 0 for n PI for all integers n

[–] myslsl@lemmy.world 3 points 8 months ago* (last edited 8 months ago) (1 children)

Rewriting the problem as solving sin(A)=0 and then claiming outright that A must be an integer multiple of pi doesn't really help as far as I can tell, since that is just the original problem with x exchanged for A?

[–] wuphysics87@lemmy.ml 1 points 8 months ago (1 children)

It does if you claim to know cos (A) = 1. Like I said, fast and loose. The question as given is illposed. You have to know something. If not, why not ask a philispohical question like what is trigonometry even?

[–] myslsl@lemmy.world 1 points 8 months ago* (last edited 8 months ago)

It does if you claim to know cos (A) = 1.

My issues with this are: Your solution did not originally claim this, it is not stated anywhere in the problem and it leads to exactly the same kind of foundational issues in the context of showing "algebraically" why cos(x)=1 at integer multiples of 2pi now.

The question as given is illposed. You have to know something. If not, why not ask a philispohical question like what is trigonometry even?

Agreed. It's at least vague/misleading. This is apparently for a precalc clep exam, so the only real sane definition a student would know to fall back on here would be geometric definitions for sine and cosine. What I think the intent of the problem is, is to build intuition on knee jerk facts about sine/cosine rather than something particularly formal?

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago (1 children)

I see how this would prove x = 1 when sin A = sqrt (1 - x^2) = 0 , but I don't see how to find the relevant angle from this. How did you go from sin A = sqrt (1 - x^2) to sin A = 0 when A = nπ where n is any integer?

[–] wuphysics87@lemmy.ml 2 points 8 months ago (1 children)

I'm playing a bit loose and fast, but think about it like this. Imagine you have a triangle in the 1st quandrant with unit hypotenuse. The triangle collapses to the x axis, giving x = 1 and theta = 0. The consequencea for sin 0 can be determined from this

[–] DarkNightoftheSoul@mander.xyz 1 points 8 months ago

Gotcha, thanks.

[–] wuphysics87@lemmy.ml 1 points 8 months ago (1 children)

Taylor expansion

[–] DarkNightoftheSoul@mander.xyz 2 points 8 months ago

I'm like, 85% sure they didn't mean me to use that, but it's reassuring to know that I could if I really, really wanted to.