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[hops]

Ah.  So there isn't a way to find theta given cos(theta) without getting lucky with triangles.

Well it's not that hard to draw a triangle with an approximately the given ratio. Just arbitrarily choose the length of one side and use the ratio to find the length of the other side.

There should also be some algorithms but I doubt they'll be easy to use by hand. Honestly "without a calculator" questions are just plain stupid.

Edit: You could use the Taylor series for arccos if you hate yourself.

[Culise]

Edit: You could use the Taylor series for arccos if you hate yourself.

This kind of sums up my opinion on Taylor series in general.  I mean, I appreciate their uses; I just hate them with the heat of one million fires. 

But it looks like calculators do use either a Taylor series or CORDIC (COordinate Rotation DIgitial Computer) algorithm as well, unless someone preloaded it with a table of values.  Unfortunately, looking up how they do it seems to have just returned us right back to evil, tricksy Taylor series. ^_^

[hops]

They essentially are applied for computers to be able to calculate elementary fucntions, they're not great for hand calculation IMO. Also I can't imagine how you could preload a calculator with a table of values for like a billion possible inputs.

[Culise]

They essentially are applied for computers to be able to calculate elementary fucntions, they're not great for hand calculation IMO. Also I can't imagine how you could preload a calculator with a table of values for like a billion possible inputs.

Put in common inputs (say, every percent or tenth of a percent point), and interpolate algebraically for any values that aren't in the table directly, straight out of the days of Pythagoras and Ptolemy.  I can't imagine it's done much given how cheap processing is and the high relative inaccuracy compared to algorithmic methods, to be honest, but it was mentioned as a non algorithm-based method that I thus included for the sake of completeness. 

[McTraveller]

Can someone offer me some hint on how to approach this? I can solve the equation to give me the variables in term of alpha, but I don't know how to proceed from that.

My hint is: What is a test to determine if a system of linear equations has a solution?

EDIT: made the hint slightly more obvious

[frostshotgg]

This kind of sums up my opinion on Taylor series in general.  I mean, I appreciate their uses; I just hate them with the heat of one million fires. 

This more or less covers every form of math after freshman calculus. At some point you hit the line  where people stopped developing mathematics because it was beautiful and fun, and started making everything because it was useful, and things got real ugly real fast. Polar form and its cousins spherical and cylindrical, various transforms, taylor series... I assume if you go the pure math route you eventually reach the point where things stop being useful so all that's left is to be artistic or whatever, but there's a very large bloc of math that's dirty and ugly and best left to the local computing cluster.

[Helgoland]

Actually if you start with Linear Algebra and work your way from there to Topology or Algebraic Geometry or somesuch field, you can avoid all the applied stuff entirely.

[hops]

Can someone offer me some hint on how to approach this? I can solve the equation to give me the variables in term of alpha, but I don't know how to proceed from that.

My hint is: What is a test to determine if a system of linear equations has a solution?

EDIT: made the hint slightly more obvious

I don't think there's anything apart from consistency of the equations, and again, I don't know how that applies here.

[hops]

https://imgur.com/a/s5qAa

Call me an idiot, but the answer to my question really doesn't seem to me to be as obvious as some of you are acting like it is. Please, actually help me? This attitude of "It's so obvious so I'm not going to explain anything to you, you stupid pleb" is kind of the reason why people think math is so hard.

EDIT: I know I made a mistake, that's still beside the point.

[hops]

Plus trying to solve for alpha that way just gives me a 4th order quadratic...

Just please, please, please tell me what I need to do. Your hints aren't really helping me learn, it's just making me more confused.

[prefuzek]

Yikes, I don't envy you, with those calculations. I'm not particularly good at linear algebra, but here are my thoughts:

You've pretty much solved it already. You've shown that the system is inconsistent when alpha = 1, because subbing that into your 6th matrix bottom row gives 0 = -6. For all other alpha, your final matrix gives a solution, just by changing it back into a system of equations. (Note that you never divide by zero when alpha != 1.)

[Helgoland]

Cin: Consider calculating the determinant. It works even if you've got variables in your matrix and not just straight numbers.

[da_nang]

Cin: Consider calculating the determinant. It works even if you've got variables in your matrix and not just straight numbers.

To be fair, the determinant only tells you if the system of equations has a unique solution, not that it has a solution. Checking rank is a better method.

Cinder, you practically have solved it. If you wanted to make it cleaner, checking the rank is basically ensuring that if there is a row of the system matrix (rows of the first three columns from the left) that can be all zero when in row echelon form (you practically have it, you just need to swap the first and second row), then the same row of the augmented matrix (what you have in the gaussian elimination) should also be all zeros when applying the same elementary row operations. Seeing as you haven't done any potential divisions by zero after the first five steps, this is easily done after the fifth step as you only need to check the third row, i.e. check what happens if alpha = 1 at that step.

If the condition doesn't hold, then you know that alpha != 1 for it to be solvable.

[prefuzek]

(you practically have it, you just need to swap the first and second row)

Just wanted to point out to avoid confusion that it actually is in row echelon form on the second page.

[Helgoland]

Determinant's quicker though, and you can calculate by hand the few cases where it doesn't immediately give you the existence of a solution. It's an easier algorithm to follow, is what I'm saying.