High School Mathematics-Physics SMILE Meeting
Mathematics: Trigonometry

27 February 2001 Walter McDonald (CPS Substitute Teacher; VA X-Ray Technician)
showed how to make an indirect measurement of the height H of a building, using trigonometry. He moved a distance D from the building, and measured the angle in the right triangle between the top and bottom, as shown:

```
|
| *
H       |   *
|     *
|       *
|         *
|        q  *
|_____________*___
D
```
It follows from trigonometry that
H = D tan q.
He illustrated the method by measuring the height of the ceiling, by measuring out D = 16 feet, and then measuring the angle q: [lying flat on the floor to get an accurate reading!] to be q = 30o. Thus,
H = D tan q = 16 feet * tan 30o = 9.2 feet.
Very good, Walter.

showed us this graph of the trigonometric functions [sine, cosine, tangent, cotangent, secant and cosecant] which was obtained from the Microsoft Encarta Encyclopedia.

• The sine and cosine are periodic with period 360o or 2p radians.
• The tangent is periodic with periodic with period 180o or p radians.
• The tangent has an asymptote at ± p/2 radians.

21 March 2006: Fred Schaal (Lane Tech HS, mathematics)               Graphing Inverse Trig Functions
Fred
noted that books don't tell how to graph inverse trig functions. Fred figured out a pretty good way to do it with a TI-83 calculator. He projected the Ti-83 screen on the wall for us to follow along with his method. He had written a little program to plot these functions. First Fred plotted the inverse sin and the plot looked reasonable. But the same method with the inverse cosine did not give a curve that made sense. Some adjustments were made in the graph scale, as suggested by the friendly crowd, and the inverse cosine curve looked better. Fred then tried the inverse tangent and it also looked reasonable. Interesting, Fred!

18 April 2006: Porter Johnson (IIT Physics)                 Regular Pentagons and Pentagrams
Porter used the fact that the angles q  = 36o and q =72o both satisfy the relation sin 5q = 0 to determine the value of cos q for each angle.

He used the basic double angle formulas

sin 2q = 2 sin q cos q
cos 2q = 2 cos2q - 1
to show that
sin 4q = 2 sin 2q cos 2q = sin q ( 8 cos3q - 4 cos q )
cos 4q = 2 cos22q - 1 = 8 cos4q - 8 cos2q + 1
He then used the addition formula
sin (q + f) =  sin q cos f + cos q sin f
with f = 4q to obtain
sin 5q = sin q cos 4q + cos q sin 4q
= sin q ( 16 cos4q - 12 cos2q + 1 )
In our case, sin 5q = 0, whereas sin ¹ 0. Thus,
16 cos4q - 12 cos2q + 1 = 0
Setting z = cos2q, we may convert this equation into a quadratic equation for z:
16 z2 -12 z + 1 = 0
The solutions of this quadratic equation are
cos2q = z = (3 ± Ö5 ) / 8
In fact, one can take the (positive) square roots of this expression to obtain
cos q = Ö:z = (Ö5 ± 1 ) / 4
The specific expressions are
cos 36o = (Ö5 + 1 ) / 4 = 0.809016994...
cos 72o = (Ö5 - 1 ) / 4 = 0.309016994...
Note, in particular, that
cos 36o = cos 72o + 1 / 2
These angles play a crucial role in construction of a regular pentagon and a regular decagon.  For additional information see  http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html and http://www.cut-the-knot.org/pythagoras/pentagon.shtml.

You can also use these angles to make a pentagramhttp://en.wikipedia.org/wiki/Pentagram. By the way, is it true that pentagrams are always located somewhere on the bodies of werewolves? (See Werewolf: Detection and Preventionhttp://www.zerotime.com/night/detect.htm.)