A problem that seems deceptively easy. (1 Viewer)

[Lazarus]

Consider this image:

Note that the question specifically asks for the answer as a fraction or a surd.

 

[McLake]

So you mean I can't have sin^-1 in the answer?

 

[Lazarus]

That's right. =P

 

[BlackJack]

Transforming a trigonometric number into a rational/irrationl? Never heard of how to do it before...

 

[Lazarus]

mm it stumped me as well.

The only info you're really given is the length of the sides, which happen to be a Pythagorean triplet, but I couldn't figure out what that meant in terms of the angles of the triangle (if anything).

 

[McLake]

I've tried:

Double angle
Sin rule
^2

None have worked so far ...

It's not a tricky polynomial thing, is it?

 

[McLake]

notice sin 60 or cos 30 = (sqrt3)/2 which squares to 3/4. Is this relevent???

cotx = 3/4
cosx/sinx = 3/4

....

 

[Dumbarse]

by using banked tracks, pythagorean identities, and noting that the area was a constant 6, i found the angle to equal 53.13010235 degrees :giddy:

 

[Lazarus]

that's actually radians, dumbarse, and it's in neither fraction nor surd form

 

[Dumbarse]

i solved it yet my calculater could not handle the monstrosity and complexity of the fraction answer
its actually degrees

 

[Lazarus]

well I'm just a dumbarse

 

[Dumbarse]

and i'm just a lazarus
u may call me god if u want

 

[Lazarus]

lol =P

 

[Dumbarse]

or u could just call me a dumarse considering i cannot solve this stoopid thing, four pages of working and ive gotten nowhere

have u got the answer how to do it laz?

it seems too simple but i cannot isolate theta by itself without it becoming irrational

 

[Lazarus]

I don't know how to do it... a guy just starting 4u asked me via ICQ.


It was being discussed in the chatroom the other night... Ex-Nor said that entering ArcSin(4/5) into Mathematica showed it to be transcendental.


Yet I've been assured that the question does have a proper answer.

 

[marsesbars]

Fraction or surd? Well I might have a way to start... Using the power series for sinx, (big image sorry)

Let sinx=4/5
Solve the polynomial (I don't know how! ) and one root will be the value we want.
Thing is, you'll also get an infinite number of other solutions, because a polynomial of degree n has n solutions, and this polynomial has infinite degree... this is troubling

 

[McLake]

Nice idea marsbars, but I don't think that will yeild an answer (infinite terms become too tricky) ....

 

[Lazarus]

Ahh!! Actually, I don't think we need to solve the polynomial... if you take the series expansion for sin x and simply let x = 4/5, the RHS of the expansion should just be equal to sin 4/5. And so maybe, with a few manipulations, we can work out the sum, or something... *wonders off to try a few things*

edit: hmm on second thought, expansions are usually used for approximating the answer, not obtaining exactly. hmm.

 

[marsesbars]

Heh, it's the damndest thing, all I need is the power series for inverse sine!

put x=4/5 and voila!

 

[Lazarus]

Hmm I suppose a series of fractions would have to be acceptable as well

nice, marsesbars...