WildTrig4: Five main laws of rational trigonometry

[footstep002]

This may be a silly question, but as I heard from the previous video triple quad formula is (internal large interval + internal small interval - whole interval) squared equal 4 times (internal large interval times internal small interval), and the triple quad formula from this video seems different. Instead, the triple quad formula in this video says (internal large interval + whole interval - internal small interval), which is p_3 - q_2 - p_1. Am I missing somthing here?

[njwildberger]

@footstep002 I think your question is answered amongst the earlier questions.

[horstmueller1000]

What is not "simple" or "natural" to me - at least, it does not come naturally to me - is when the hight is outside the triangle. The same difficulty with me, when I look at a spread and make it open more than the perpendicular...

[njwildberger]

@horstmueller1000 It does take some getting used to after all our experience with angles. Keep in mind that the spread is defined between two lines, not between two rays. It might be instructive to look at the spread protractor posted by M. Ossmann online.

[horstmueller1000]

the question naturally arises , why the triple quad formular can be used the way as described (+a + b - c)^2 = 4ab. This is a great excercise, given the hints in wildtrig 2 (?) where the simple case is proven - but the same ideas apply, when all other options are written down. It is a lovely exercise.

[varungargg8]

P3<Q2 how come sister formula is valid in cross law

[yuraaa1990]

Q1+Q2-Q3 have to be equal to zero , nope ?!

[njwildberger]

@yuraaa1990 Only if the triangle has perpendicular sides, ie Pythagoras' thm applies.

[Eichboson]

When proving the Triple Spread Formula, you write s1/Q1 as 1/D. But if s1=0 (so we'd have a degenerate triangle), there is no such D. Obviously, the formula still holds, but I think one ought to mention that possibility.

[GlorifiedTruth]

I finally got the correct spreads for the angles, but I had to use irrational numbers to figure out the perpendicular, and that seemed to violate the spirit of things. I hope I am still allowed in the fraternity...?

[njwildberger]

Hi GlorifiedTruth, Not quite sure what you mean by "figure out the perpendicular".

The Cross law allows you to find the three spreads of a triangle once you know the three quadrances. In fact you can just find one of the spreads, and then the other two via the Spread law. Why don't you try that?

[GlorifiedTruth]

@njwildberger - D'oh. Of course. It works out easier that way... and after all I imagine that's what the Cross law is for.

What I was doing before was drawing a line perpendicular up from the A1A3 side up to A2 so that I'd have a hypotenuse that I could square to get my S1 and S3. Even as I was doing it I knew I was not being true to rational trig. Maybe after I watch the applications video I'll have better sense about applying the various laws. Thanks for the reply!

[Anonymystik]

Spread law is the same as sin law squared. This spread law is therefore the weaker conclusion of sin law, because squaring the equation adds some new roots.

[DrDoubleplusungood]

3:30 "because the triangles are symmetrical, we can also extend it to equaling s2/q2"

I'm having trouble understanding this part. I'm not sure in what manner you mean symmetrical since the sides/quadrances of the triangle have different lengths. If s2 is defined as the ratio of the opposite quadrance to the hypotenuse quadrance, how can you construct these quadrances from the diagram?

[njwildberger]

The argument is symmetrical in that it does not depend on which pair of spreads and opposite quadrances we look at. I derived s1/Q1=s3/Q3 but I could just as well have derived s1/Q1=s2/Q2. Putting these two together we get

s1/Q1=s2/Q2=s3/Q3

which is the Spread law.

[DrDoubleplusungood]

Thanks, I understand what you mean now.

[sixbillionmorons]

Hi, Dr. Wildberger. You're probably getting sick of me, but I'm stuck on algebra here again. (I'm a grad student; this is embarrassing.)

At 5:05 you use the Triple-Quad Formula in the form (P3+Q2-P1)^2=4(P3)(Q2), but the largest quadrance of three in the diagram is Q2. When I watch you derive the TQF in another video, it seems clear to me that the odd sign in the squared term has to go on the large quadrance. If I'm making a bad assumption here, could you point me in the right direction please?

[sixbillionmorons]

Well, guess what: I worked through it with side-lengths and found out once again that you were right and I was wrong. (But in my defense, I would never have intuitively guessed you could rearrange it like that and still have it be true.) Sorry for bothering you with my feats of ineptitude. Keep up the good work.

[njwildberger]

The general Triple Quad Formula has the form (x+y-z)^2=4xy and the important thing is that the two variables involved in the right hand side occur with the same sign in the left hand side. Perhaps it's easier to remember in the equivalent form (x+y+z)^2=2(x^2+y^2+z^2).

[QuantumMaths]

@sixbillionmorons

In other words, P3, Q2, and P1 are *symmetric* and those totally interchangeable amongst one another as long as the general form is followed. Although Wildberger gave the negative sign to the largest quadrance in his initial derivation, he didn't have to.

[QuantumMaths]

Holy crap I just realized you posted that comment three years ago...

[Bloodsaberxy]

you kind of lost me on the cross law, a bit more examples wouldve been useful

[joziasdelrios]

Hello wildberger. I like you work a lot...

I'd like to ask: is there a development of the Rational Trigonometry laws for R^3 ? is it straight-forward?

Thank you!

[njwildberger]

Rational trig can also be developed in R^3 but it is theoretically more work. It turns out that also one needs some additional concepts, basically the idea of a `solid spread' to replace `solid angles'.