50g, Trig Identities Gone Wild  Hal Bitton in Boise  03052008
Hi again folks.
I'm taking a nice little analytical trig class, and while doing my homework this evening, encountered the the question: Express Cosecant (csc) theta in terms of Cotangent (cot) theta. I was able to arrive (on paper) at the solution of:
csc x = 1/sqrt(11/(1+1/(cot x)^2)).
To confirm my solution on my 50g, with the x variable purged, I keyed in the right side of the above equation exactly as written (with the exception of substituting (tan x)^2 for 1/(cot x)^2), with the intention of evaluating it for a few different values of x. As I keyed it in using the stack (with RPN keystrokes) when I took the square root (my next to last keystroke) the machine evaluated the radical, which is no surprize. The result of that evaluation, however, floored me. It was an absolute monstrosity, that when viewed in the equation writer, took up nearly 2 screens in "text" view. Curious, I entered the final keystroke...1/X...and then "condensed" it using the eval key three or four times (my machine is set to step by step). It condensed to about half of it's original length (but still huge), and it now contained terms up through the 7th power!! Still curious, I created an x variable containing an integer degree value. When I evaluated this monster and forced a real number output, the result was correct (ie, it equaled 1/sinx, or cscx).
I then purged the x variable, so I could play around with this thing symbolically a bit more, and found that by raising the denominator (of the above equation) to the .5 power, rather than using the sqrt key, no transformation took place.
I'm not even going to try to post this monster, as it would be virtually unreadable in this format anyway, but if one of you with a 50g would care to enter in the above equation, and then prognosticate on this transformation, I would be most curious.
Best regards, Hal
PS...I tried it on my 48gx, and there was no transformation at all.
HB
Re: 50g, Trig Identities Gone Wild  Stefan Vorkoetter  03052008
For what it's worth, I tried this in Maple. Here's what I got:
> 1/sqrt(11/(1+1/(cot (x))^2));
1

/ 1 \1/2
1  
 1 
 1 + 
 2
\ cot(x) /
> simplify(%);
csgn(sin(x))

sin(x)
From Maple's documentation:
 The csgn function is used to determine in which halfplane ("left" or
"right") the complexvalued expression or number x lies. It is defined by
/ 1 if Re(x) > 0 or Re(x) = 0 and Im(x) > 0
csgn(x) = <
\ 1 if Re(x) < 0 or Re(x) = 0 and Im(x) < 0
Stefan
Re: 50g, Trig Identities Gone Wild  Jason Artz  03052008
How about csc(x)=sqrt(1+cot(x)^2) ?
Re: 50g, Trig Identities Gone Wild  Mike Morrow  03052008
I set my HP50g for EXACT mode, then entered your expression in the form:
'1/sqrt(1INV(1+SQ(TAN(X))))'
After pushing EVAL, the massive expression to which you refer appeared.
After pushing EVAL once again, that mass simplified to:
'sqrt(TAN(X)^2+1)/TAN(X)'
Pushing EVAL once again produced the same result.
It is interesting.
Mike
Re: 50g, Trig Identities Gone Wild  Mike Morrow  03052008
Which is what the HP 50g produces, expressed differently:
sqrt(TAN(X)^2+1)/TAN(X) = sqrt(1+COT(X)^2)
Mike
