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I'm trying to do some calculations here, but for some reason Mathematica starts using numerical approximations that are no good for my work. Specifically:

T[t_] := {(-Sin[t])/(Sqrt[1 + Cos[t/2]^2]), 
(Cos[t])/(Sqrt[1 + Cos[t/2]^2]), (Cos[t/2])/(Sqrt[1 + Cos[t/2]^2])}

B[t_] := {(-0.5 Cos[t] Sin[t/2] + Cos[t/2] Sin[t])/Sqrt[
  1.625 + 0.375 Cos[t]], (- Cos[t/2] Cos[t] - 
   0.5 Sin[t/2] Sin[t])/Sqrt[1.625 + 0.375 Cos[t]], 1/Sqrt[
  1.625 + 0.375 Cos[t]]}

Cross[B[t], T[t]]

After using Simplify, I get:

{((1.4142135623730951 + 1.4142135623730951 Cos[t/2]^2) Cos[t] + 0.3535533905932738 Sin[t]^2)/( Sqrt[1.625 + 0.375 Cos[t]] Sqrt[ 3 + Cos[t]]), ((1 + Cos[t/2]^2 - 0.25 Cos[t]) Sin[t])/( Sqrt[0.8125 + 0.1875 Cos[t]] Sqrt[3 + Cos[t]]), (0.7071067811865476 Sin[t/2])/(Sqrt[1.625 + 0.375 Cos[t]] Sqrt[3 + Cos[t]])}

I know for a fact that: $$1.4142135623730951 \approx \sqrt{2} \\ 0.7071067811865476 \approx \frac{\sqrt{2}}{2} \\ 0.3535533905932738 \approx \frac{\sqrt{2}}{4}$$

When I write Sqrt[2], (Sqrt[2]/2) and (Sqrt[2]/4) instead of these numbers, I am unable to proceed with these calculations, and I keep getting the warning: enter image description here

etc. Can someone tell me how to solve this?


I didn't know what tag or specific title to put here, so if you have a better idea, feel free to edit it.

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  • $\begingroup$ Try defining B[t] := Rationalize@ {(-0.5 Cos[t] Sin[t/2] + Cos[t/2] Sin[t])/Sqrt[ 1.625 + 0.375 Cos[t]], (- Cos[t/2] Cos[t] - 0.5 Sin[t/2] Sin[t])/Sqrt[1.625 + 0.375 Cos[t]], 1/Sqrt[ 1.625 + 0.375 Cos[t]]} instead $\endgroup$ Commented Aug 24, 2014 at 21:35
  • $\begingroup$ It did not work :( $\endgroup$
    – Ivo Terek
    Commented Aug 24, 2014 at 21:38
  • $\begingroup$ There's a backtick there that shouldn't be. My guess is that you copy-pasted Sqrt[2] over the real, but the real has a trace backtick which you didn't see, or mind. Remove that backtick. $\endgroup$
    – Gleno
    Commented Aug 24, 2014 at 21:53
  • $\begingroup$ What difference do they make? When initially typing, I just typed everything without the backticks and had no problems at all. $\endgroup$
    – Ivo Terek
    Commented Aug 24, 2014 at 21:55
  • 1
    $\begingroup$ When typing reals backticks indicate precision. To be honest, I find them mostly useless. $\endgroup$
    – Gleno
    Commented Aug 24, 2014 at 21:58

1 Answer 1

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T[t_] := {(-Sin[t])/(Sqrt[1 + Cos[t/2]^2]), 
          (Cos[t])/(Sqrt[1 + Cos[t/2]^2]), 
          (Cos[t/2])/(Sqrt[1 + Cos[t/2]^2])}

B[t_] := Rationalize@{(-0.5 Cos[t] Sin[t/2] + Cos[t/2] Sin[t]) Sqrt[1.625 + 0.375 Cos[t]], 
                      (-Cos[t/2] Cos[t] - 0.5 Sin[t/2] Sin[t])/Sqrt[1.625 + 0.375 Cos[t]], 
                       1/Sqrt[1.625 + 0.375 Cos[t]]}

Cross[B[t], T[t]] // FullSimplify // TeXForm

$\left\{-\frac{12 \cos (t)+\cos (2 t)+3}{2 \sqrt{\cos (t)+3} \sqrt{3 \cos (t)+13}},-\frac{\sin (t) (\cos (t)+6)}{\sqrt{\cos (t)+3} \sqrt{3 \cos (t)+13}},-\frac{2 \sin \left(\frac{t}{2}\right)}{\sqrt{\cos (t)+3} \sqrt{3 \cos (t)+13}}\right\}$

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  • 2
    $\begingroup$ Cross[B[t], T[t]] // Simplify will give a significantly simpler form. $\endgroup$
    – Bob Hanlon
    Commented Aug 24, 2014 at 23:38
  • $\begingroup$ @BobHanlon Thanks. Done $\endgroup$ Commented Aug 24, 2014 at 23:43
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    $\begingroup$ @belisarius I just noticed your new photo. Albeit Argentina produced & produces so many wonderful imaginations, I'm not aware of the fact that it was ever engaged in boxing :) $\endgroup$
    – eldo
    Commented Aug 25, 2014 at 0:48
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    $\begingroup$ @eldo Argentina has a long history of very good boxers and world champions. In particular Nicolino Locche was in my book one of the quickest defensive boxers. If you like boxing, take a look at this round, from the fight when he won the world championship. (Ps: Paul Fuji was crying in impotence that night when he left the ring) $\endgroup$ Commented Aug 25, 2014 at 1:43

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