Download it here. - UGA Math Department

Section 5.2: More coverage of some awkward questions
This document is meant to address a couple problems that have strange setups. The techniques learned in
the previous classes are useful here, but they don’t do everything. You should look at these problems as being
the exception rather than the typical case, but they do have some handy lessons.
WQ 7 #6: Getting one trig in terms of another
In the last classes, we saw techniques for expressing trig functions in terms of sin or cos. However, what if
it’s the other way around? For instance, if you want tan in terms of csc, what would you do?
The technique commonly used is called the Right Triangle Method. It takes advantage of SOHCAHTOA by
drawing a right triangle in a clever way.
1. First, write the SOHCAHTOA ratio for the trig that you want to USE. Now, write that same trig as itself
over 1.
2. Use this ratio to label two sides of a right triangle. One side will be 1, and the other will be your function
from Step 1.
3. Find the length of the third side with Pythagoras.
4. Now that you have all three sides, you can find the ratio you were looking for!
NOTE: Technically, the answer should have a ± in front of it. To know whether to use the plus or the minus,
know which quadrant you’re in.
Sample: Let’s get tan(x) in terms of sec(x).
1. Since sec is the reciprocal of cos, that means we
want
sec(x)
hyp
=
adj
1
2. Draw a right triangle whose hypotenuse is labeled sec(x) and whose adjacent side is 1.
3. Thus, the last leg is opp =
4. Finally,
p
hyp2 − adj 2 =
p
sec2 (x) − 1.
opp
tan(x) =
=
adj
p
sec2 (x) − 1
1
(Note: If we’re in quadrants where tan is positive, like Quad I or Quad III, I’d use this answer. For the
others, I’d put a minus sign out front.)
You try this: Write cos(x) in terms of tan(x).
See next page for another problem.
HW 5.2B #8: A very strange triangle in a circle
This strange problem gives you the following scenario: in the picture on the right, point P is on the
unit circle x2 + y 2 = 1. Determine the area of the
triangle ∆ABP as a function of θ.
How To Think About The Goal:
To find area, I need the lengths of the two legs of the triangle. So I need to find AP and AB.
Major Steps in the Thought Process:
1. We were told P is on the unit circle!
When we learned about trig functions in terms of x, y, r, we had an interesting remark about points on
the unit circle. For those points, we know x = cos(θ) and y = sin(θ).
In particular, though, those coordinates now tell us some lengths. The x-coordinate tells us that AP =
cos(θ). Great, that’s one leg down of our triangle!
2. Once you have one side of a right triangle AND an angle, you can use SOHCAHTOA to get whatever
other sides you need. Hence, we should try and get an angle in the triangle ∆ABP . Obviously, that θ
near the origin is going to have to be used, but how??
3. There are two geometry ways to find an angle in ∆ABP :
• Alternate Interior Angles: The angle marked with θ and the angle AP B (at the upper-right of
the figure) are called alternate interior angles in geometry. Two such angles always have the same
measure. Thus, AP B has measure θ.
• Using complements over and over: Any time two angles are stuck together to make a right angle, or
whenever two angles are acute in a right triangle, they add to 90◦ . We can keep using this over and
over to find more angles in the figure.
Since AOB makes a right angle with θ, AOB must equal 90◦ − θ (it’s the complement of θ). Next,
AOB and OAB are complements (they occur in the same right triangle), so OAB is 90◦ −(90◦ −θ) = θ.
(If you complement twice, you end up with the original angle!) Next, OAB and P AB make a right
angle, so P AB is 90◦ − θ again. Doing this complement ONE more time, you find AP B has measure
θ. (Whew!)
Clearly, the first way is quick, but it requires knowing some geometry. The second way is more elementary,
but it’s longer. It’s good to be able to understand both views, because each of these techniques can be
handy.
4. Alright, we now have the angle θ at AP B. We know one side of the triangle we want; its HYPOTENUSE
is cos(θ). (Note that it’s opposite the right angle!) Let’s use that hypotenuse, with SOHCAHTOA, to get
the legs.
For the adjacent leg BP , cos(θ) = adj/hyp, so cos(θ) = BP/ cos(θ). (NOTE cos(θ) is occurring in TWO
places here!) Solving for BP , BP = cos2 (θ).
Similarly, for the opposite leg AB, we use sin(θ) = opp/hyp = AB/ cos(θ). Hence, AB = sin(θ) cos(θ).
5. You finally have both legs!! You can now use the triangle area formula: Area = (AB)(BP )/2. I leave it
to you to write that.
This problem is much longer and harder than I would expect any test question to be... but there’s a lot of
lessons that come from it and a lot of surprises. Fight with it!