# Sin2 ProofMay 1, 2015

The Sin2 Proof

Once again… What times are there (and what are they) such that the angle between the hour, minute, and second hands are the same. That is, the angle between h and m, h and s, s and m are 120o or 240o       (  2/3 * Pi or 4/3 * Pi). Or put another way, the hour, minute, and second hands cut the face of the clock into 3 equal pieces/areas. The location of the hands must represent a legitimate time. That is, they can not be simply placed to arbitrary positions such as 12, 4, and 8. Also, the clock doesn’t “tick” but rather it runs smoothly (more on this a little later). See the bottom portion of the Clock Figure just above.

We’ll keep in mind that Sin2(120o) = Sin2(240o).     For a proof see Proof 120/240  below. Then click your browser’s back button.

To solve the problem, we start the clock at 12:00:00 and let it run. That is, all hands start at “zero degrees”. At any time, the position of all of the hands, in either degrees or radians, is described by the following:

h = m/12
m = m
s = 60m

That is, the second hand travels at a rate of 60 times the minute hand, and the minute hand travels at a rate of 12 times the hour hand. The clock runs “smoothly” and doesn’t “tick” so h, m, and s can be any (positive) value… not just integers!

Rule 1

Sin2(AngleBetweenTwoHandsAsViewed)

Is equal to

Sin2(TotalAngleTraversedByOneHand – TotalAngleTraversedByTheOtherHand)
See Proof 2 – Viewed vs Traversed below.   Then click your browser’s back button.

Or, put another way, the Sin2 of the angle between two hands “as you view it” is equal to the Sin2 of the total angle traversed by one hand minus the total angle traversed by the other. For example, the Sin2 of the angle between the second and minute hands (as viewed) is the same as Sin2 (s-m) where s is the total angle traversed by the second hand at some time and m is the total angle traversed by minute hand at that same time. The angles s and m can be greater than 360o . In fact, s will always be greater than 360o after just 60 seconds and m will start to exceed 360o after 60 minutes. h will pass 360o after 12 hours. Again, See the proof.

Since s travels faster than m or h we have…

For the angle between the second and minute hands…

Sin2(s-m) = Sin2(120 or 240)

and since Sin2(240) = Sin2(120) we have

Sin2(s-m) = Sin2(120).

We have the same situation for the angle between the second and hour hands…

Sin2(s-h) = Sin2(120 or 240)

and since Sin2(240) = Sin2(120) we have

Sin2(s-h) = Sin2(120).

Or… Sin2(s-m) = Sin2(s-h) = Sin2(120)

Note that angles s-m and s-h are both positive since s is always greater than both m and h… but that doesn’t really matter. Also, as you’ll find later, we don’t even need to calculate anything relative to m-h (angle between the minute and hour hand).

Rule 2

The only angles which satisfy Sin2(x) = Sin2(120) are:

x = N*Pi + (+-120)

This is on page 162 of my trig book but it should be obvious from the Figure. The general rule is…

for Sin2(x) = Sin2(y) then x = N*Pi + (-1)Ny

From now on let’s work in degrees. Using all degrees instead of radians we have…

x = N*180 +-120

Working first on angle s-m we have Sin2(s-m) = Sin2(120)

Applying Rule 2 and letting s-m = x

and N1= some integer, we have:

s-m          =    N1*180 +-120
60m – m      =    N1*180 +-120
59m          =    N1*180 +-120

m            =    N1*180       +-120
~~~~~~  +   ~~~~~~~
59           59

Now we work on angle s-h again using Rule 2…

s-h                =    N2*180 +-120
60m – m/12         =    N2*180 +-120

(720m/12) – (m/12) =    N2*180 +-120

(719/12)m              =    N2*180 +-120

m                      =    12*(+-120)    N2*180*12
__________ +  _________
719          719

Now let’s combine the two formulas that equate m.

M   =     +-120     +   N1*180
~~~~~~~       ~~~~~~~~
59             59

m   =    12*(+-120) +   N2*180*12
~~~~~~~~~~     ~~~~~~~~~
719            719

Therefore:

+-120   +   N1*180     12*(+-120)  +   N2*180*12
~~~~~       ~~~~~~  =  ~~~~~~~~~~      ~~~~~~~~~
59           59          719            719

719*(+-120)   +   719*180*N1       =   59*12(+-120)   +   59*12*180*N2

719*180*N1     –    59*12*180*N2   =   59*12(+-120)   –   719*(+-120)

180(719N1 – 708N2)                 =   59*12(+-120)   –   719*(+-120)

180*(719N1 – 708N2)                =   120(708*(+-1)  – 719*(+-1))

(719N1 – 708N2)                    = 120(+-708 +-719)
~~~
180

(719N1 – 708N2) = 2/3 * (+-708 +-719)

There are four combinations of values for +-708 and +-719. They are 1427, -1427, 11, and -11. None of these are evenly divisible by 3 so..

INTEGER = NON-INTEGER

We reached a contradiction… an integer can’t possibly equal a non-integer,

therefore it’s not possible for s-m = s-h = (120 or 240)

Therefore, there do not exist any times such that

the angle between the 3 hands of a clock are all 120 (or 240) degrees.

Supplemental Proofs Used Above Follow

Proof 120/240

——————-

Sin(A)    = Sin(Pi – A)               Rule P1_1
Sin(A)    = Sin(180 – A)
Sin(-A)   = -Sin(A)                   Rule P1_2
Sin(240)  = Sin(180 – 240)             P1-1
Sin(240)  = Sin(-60)
Sin(240)  = -Sin(60)                   P1_2
Sin(240)  = -Sin(180 – 60)             P1_1
Sin(240)  = -Sin(120)
Sin2(240)  = (-1)2Sin2(120)
Sin2(240)  = Sin2(120)
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Proof 2 – Viewed vs Traversed

———————————————
Sin(a+b) = Sin(a)Cos(b) + Cos(a)Sin(b)
Sin(a-b) = Sin(a)Cos(b) – Cos(a)Sin(b)
Sin(s-m) = Sin(s)Cos(m) – Cos(s)Sin(m)
S0 = angle of s relative zero (or remainder of s/360)
m0 = angle of m relative zero (or remainder of m/360)
s = s0 + 360Ns and m = m0 + 360Nm
where Ns and Nm are some integers.
Sin(s-m) = Sin(s0 + 360Ns – m0 – 360Nm)
Sin(s-m) = Sin( s0 – m0 – 360(Nm – Ns) )
For convenience, let Nm-Ns = some integer N
Sin(s-m) = Sin(s0-m0)Cos(360N) – Cos(s0-m0)Sin(360N)
Cos(360N) = +1
Sin(360N) = 0
Sin(s-m) = Sin(s0-m0) * (+1) – Cos(s0-m0) * 0
Sin(s-m) = (+1)Sin(s0-m0)
Sin2(s-m) = Sin2(s0-m0)
Also, since Sin(-a) = -Sin(a)
Sin2(s-m) = Sin2(m-s) = Sin2(s0-m0) = Sin2(m0-s0)

********* END of Sin2 Proof **********

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