Section 5.7 cont: Harder Elevation/Depression Problems
These problems only use SOHCAHTOA, but they introduce some extra complications:
• One uses speed! Recall (speed) = (distance) / (time).
• One problem draws two right triangles: a small one inside of a big one.
Ex 1: An airplane flying at an altitude of 30, 000 feet passes directly over a fixed object on the ground. One
minute later, the angle of depression of the object is 38◦ . Approximate the speed of the airplane to the nearest
mile per hour.
Note: Since the answer needs to use miles per hour, convert your units.
Ex 2: A boat has a 40◦ elevation angle to a lighthouse. It then sails 100 feet away and has a 20◦ elevation angle, as shown in the figure. Find the height h
of the lighthouse.
Hints: There’s a big right triangle that contains
a small one. Use SOHCAHTOA to get each base in
terms of h. Since (big base) = 100 + (small base), you
can substitute to get a linear equation for h.
Section 6.2: More Complicated Trig Equations
When we studied Section 5.4, we studied basic trig equations trig(θ) = a in an interval. Now, we will solve
harder equations with unknown angles. There are three new sources of difficulty.
1. No goal interval: you get infinitely many solutions
2. Complicated angle: θ is a function of x, like in tan(3x + π/4) = 0
3. Putting trigs in other equation types: e.g. factoring problems like (sin x − 1)(cos x) = 0
Change 1: Full Solutions
When no interval is given, we want the full solution, which has infinitely many answers across the whole
line. Find the solutions in one unit circle (0 to 2π) and add or subtract ANY number of 2π’s to get all the
coterminal answers.
Notation: If n stands for any integer, we write θ = (initial solution) + 2πn. Think: “add any number of 2π’s
to my answer.” (Don’t forget that n can be negative, corresponding to subtracting 2π’s.)
Sample: Say sin(x) = 1/2. In Quads I and II, you get x = π/6, 5π/6. Full solution:
x = π/6 + 2πn, 5π/6 + 2πn for any integer n
Half-circle periods: You may also find scenarios where symmetry shows two equally-spaced answers in the
unit circle... just write one of the answers and add or subtract π’s! (For instance, tan and cot always produce
this situation... why?)
Sample: Say tan(x) = 1. In Quads I and III, you get x = π/4, 5π/4, which are π apart. Rather than taking
both π/4 and 5π/4 and adding 2π’s, it’s more efficient to take π/4 and add π’s: x = π/4 + πn .
Ex 3: Find all solutions of these equations. (a) sin(x) = −1 (b) cos(x) = 0
Hint for (b): The answers are a half-circle apart, so you add +πn to one value instead of +2πn to both.
Change 2: Substituting θ for Complex Angles
If
1.
2.
3.
4.
the unknown angle is a function, like trig(f (x)) = a, then let’s use a variable to replace the function!
Say we substitute θ = f (x), so we get trig(θ) = a.
Find θ’s FULL solution (including the n’s).
Plug back in f (x) for θ; now solve for x’s full solution (with n’s). The period may change!
Finally, if you have an interval, see which n values give you valid answers in the interval. (How many
periods do you add or subtract?)
Sample: Say sin(2x + 1) = −1, so we’ll use θ = 2x + 1. Solve sin(θ) = −1 (Ex 3(a)) to get θ = 3π/2 + 2πn.
Now, go back to θ = 2x + 1: 2x + 1 = 3π/2 + 2πn. Subtract 1 and divide by 2 to get
x = (3π/4 − 1/2) + πn (you divide the 2πn by 2 as well!)
If you want all x in [−π, 2π), then the values of n = −1, 0, 1 give answers in that interval:
x = −π/4 − 1/2, 3π/4 − 1/2, 7π/4 − 1/2 . (i.e. You can subtract one π, do nothing, or add one π.)
Ex 4: Find all solutions (i.e. full solution) of the following equations.
(a) cos(x − π/11) = 1
(b) sin(2x − π) =
√
3/2
(c) cos(ln x) = 0
Ex 5: In parts (a) and (b) of Example 4, find the solutions in [0, 2π).
Caution: In this “θ substitution” process, you must find the FULL θ solution before returning to x. If
you switch back variables too early, you may miss solutions, because x and θ may have different periods.
Change 3: Other Equations with Trigs Substituted
Remember equations like e2x − ex − 2 = 0? These were quadratics in ex , i.e. we substituted u = ex to get a
more usual quadratic u2 − u − 2 = 0. (After we finish getting u, we then replace u back with ex to solve for x.)
The strategies are similar here! First solve for the trig on its own. (If it helps, substitute u = trig(x).)
When you’re done find trig(x)’s values, treat each one as its OWN trig equation. Solve all the parts!
Some common tactics:
• Factoring: Get 0 on one side and pull out common factors. Set each factor to 0.
Sample: cos3 (x)+cos2 (x) = 0 factors as cos2 (x)(cos(x)+1) = 0. Thus, either cos2 (x) = 0 or cos(x)+1 = 0.
Our basic trig equations are cos(x) = 0 and cos(x) = −1.
• Quadratics with trig2 (x) and trig(x):1 Substitute u = trig(x), and find the roots for u (by Quadratic
Formula or factoring). Return back to trig(x) and solve for x.
Sample: cos2 (x) − 3 cos(x) + 2 = 0 is quadratic in cos(x), so say u = cos(x). Thus, u2 − 3u + 2 = 0 =
(u − 1)(u − 2), which gives you u = 1 or u = 2. Return to x: cos(x) = 1 or cos(x) = 2.
√
2
Ex 6: Find all solutions to − 3 tan(α)
√+ tan (α) = 0.
Hint: You get tan(α) = 0 or tan(α) = 3; the first one has quadrantal answers. Both answers use period π.
Ex 7: Find all solutions to cot2 (x)√− 1/3 = 0 in [0, 2π).
Hint: You end up with tan(x) = ± 3. Both signs lead to the same reference angle, so you have solutions in
every quadrant!
Ex 8: Solve tan2 (x) sin(x) = sin(x) for x in [0, 2π).
Hint: Factor; you find tan(x) = ±1 or sin(x) = 0.
Ex 9: Solve 2 sin2 (x) − 5 sin(x) + 2 = 0 in [0, 2π).
Hint: This is quadratic in sin(x). You get sin(x) = 1/2, 2, but sin(x) = 2 has no solutions.
Extra Practice (outside of class)
These problems show some skills that the earlier problems didn’t show. Answers can be found on a bonus
handout on my website.
Ex 10: Find all solutions to csc5 (x) = 4 csc(x) in [0, 2π).
Getting started: Write as csc5 (x) − 4 csc(x) = 0, so csc(x)(csc4 (x) − 4) = 0. Therefore, csc(x) = 0 or
4
csc (x) = 4. The first equation has no solutions (write it as sin(x) = 1/0). For the
√ other, it’s easiest to compute
√
the fourth root of 4 by taking square roots twice: 41/4 = (41/2 )1/2 = 21/2 = 2. This means csc(x) = ± 2.
(Why ±?)
Ex 11: Find all solutions to cos2 (x) − sin2 (x) = 1.
Getting started: You want this equation to have only ONE type of trig function; we’ll use the Pythagorean
identity sin2 + cos2 = 1! One option is to substitute sin2 = 1 − cos2 , so we obtain
cos2 (x) − (1 − cos2 (x)) = 1
so
2 cos2 (x) − 1 = 1
Ex 12: Find all solutions to the equation
cos2 (x) = sin2 (x) + cos(x)
in the interval [−2π, 4π).
Getting started: Like the previous exercise, you want to make this only one type of trig. If you replace sin2
by 1 − cos2 , you get a quadratic equation in cos.
What if you tried replacing cos2 by 1 − sin2 instead? It wouldn’t work, since you have a single cos(x) left
over, so it’s not quadratic in sin.
1 Compare
this to problems with e2x and ex in them from Section 4.6!