Download Section 6.6: The Inverse Trigonometric Functions (i.e. “Arctrigs

Section 6.6: The Inverse Trigonometric Functions (i.e. “Arctrigs”)
Defining sin−1 , cos−1 , and tan−1 (aka arcsin or asin, etc.)
See the other handout on my website for pictures.
Recall: When y = f (θ) is one-to-one (passes the Horizontal Line Test), y = f (θ) ⇐⇒ θ = f −1 (y).
ISSUE: The trigs are not one-to-one functions! They are periodic!
FIX: Restrict θ to come from an interval where the trig is one-to-one. (See drawings in class.)
When computing trig−1 (x), we always use Quad I for positive inputs.
For negative inputs, sin−1 and tan−1 use Quadrant IV, whereas cos−1 uses Quadrant II.
To keep θ as close to 0 as possible, Quadrant IV uses negative angles from −π/2 to 0: so θ = −θR .
Inverse function
θ = sin−1 (x) means:
θ = cos−1 (x) means:
θ = tan−1 (x) means:
Original
x = sin(θ)
x = cos(θ)
x = tan(θ)
Domain
with x in [−1, 1]
with x in [−1, 1]
with x in (−∞, ∞)
Range
and θ in [−π/2, π/2]
and θ in [0, π]
and θ in (−π/2, π/2)
Make sure your calculator is using radians when you use these functions.
RULE OF THUMB: When x is positive, trig−1 (x) is always Quad I; it’s a reference angle! When x is
negative, work out θR = arctrig(|x|) first, and then either
θ = π − θR for Quad II
θ = −θR for Quad IV
or
Ex 1: Compute the exact values of these expressions (they should be special or quadrantal):
(a) sin−1 (−1)
√
(b) sin−1 ( 3/2)
(c) arccos(−1/2)
(d) atan(−1)
Inverse Trigs And Trigs Canceling
We have to be careful with something like sin−1 (sin(θ)) or sin(sin−1 (x)). Why? sin−1 isn’t the “true inverses”
of sin; it only inverts one portion! You’ll need to pay attention to the domain and range of the inverses.
Computing inverse first: Make sure the input x is in domain, otherwise your answer is undefined.
Sample: sin(sin−1 (0)) = 0 but sin(sin−1 (2)) is undefined .
The difference: 0 is in domain of sin−1 (i.e. range of sin) and 2 is not.
Ex 2: Find the exact value when the expression is defined.
(a) sin(sin−1 (1/4))
(b) cos(cos−1 (−3))
(c) tan(tan−1 (24))
Computing inverse second: With trig−1 (trig(θ)), ask if θ is actually in the range of the inverse! If not:
1. Find θ’s reference angle θR . (Use θR = 0 for east-west poles or θR = π/2 for north-south poles.)
2. What’s the sign of trig(θ)? This will tell you whether to use Quad I (answer θR ) or Quad II (answer
π − θR ) or Quad IV (answer −θR ).
Basically, you’re switching one solution θ of trig(θ) = x for another which IS in range of trig−1 .
Sample: arcsin(sin(3π/2)) 6= 3π/2 since 3π/2 isn’t in the range [−π/2, π/2]. We use θR = π/2. Since sine
is negative, the answer is −θR , or −π/2 .
Ex 3: Find the exact value when the expression is defined.
(a) tan−1 (tan(−π/6))
(b) cos−1 (cos(6π/7))
(c) arcsin(sin(13π/12))
(d) arccos(cos(6π/5))
(e) atan(tan(9π/8))
Using Inverse Trigs With a DIFFERENT Trig
With things like sin(cos−1 (1/2)), think of the arctrig as being an angle: θ = trig−1 (a). Figure out:
1. θ’s quadrant (use the sign of a and type of inverse trig)
2. a reference triangle (write trig(θR ) = |a| with SOHCAHTOA)
Once you have the reference triangle and quadrant, you can make any trig(θ) you want!
Sample: sin(cos−1 (−1/2))... look at this as sin(θ) where θ = cos−1 (−1/2). Since cos is negative, cos−1 uses
√
Quad II for θ. We get cos(θR ) = 1/2, so we make a triangle with adj = 1 and hyp = 2, so you get opp = 3.
√
In Quad II, sin is positive, so sin(θ) = opp/hyp = 3/2 .
Ex 4: Find the exact value of these expressions.
√
(a) cos(sin−1 (− 3/2))
(b) tan(2 tan−1 (4/3))
HINTS: For (b), first say θ = tan−1 (4/3), so this makes tan(2θ). We don’t have a tangent double-angle
formula, but we can find sin(2θ) and cos(2θ) and then divide them. DON’T pull out the 2!
The same kinds of skills can also apply with variable expressions. A good thing to keep in mind is that when
θ is acute, all trigs are positive, and then trig(θ) = x ⇐⇒ x = trig−1 (θ).
Ex 5:
(a) Write cos(arctan(x/3)) in terms of x (without using any trigonometry in your answer).
NOTE: Don’t worry about signs here... Whether or not arctan uses Quad I or IV, cos is positive.
(b) If sin(θ) = 4x and θ is acute, write θ + sin(2θ) in terms of x.
HINT: First, to get θ on its own from sin(θ), that’s why we use inverses: θ = sin−1 (4x). Next, to get
sin(2θ), use double-angle because sin(θ) and cos(θ) are easier to find.
Equation Solving using Inverse Trigs
Before today, our basic trigs trig(θ) = a had to use special or quadrantal angles. Now, we can handle
anything! In the reference step, go from trig(θR ) = a to θR = trig−1 (a), and all the rest of the steps are the
same! (If a doesn’t lead to a special or quadrantal angle, leave trig−1 (a) unsimplified.)
Ex 6: Find all solutions for x in [0, π) (NOT 2π!) to these equations:
(b) cos2 (x) = 1/5
(a) sin(x) = 3 cos(x)
HINT for (a): First divide cos(x); this produces the equation tan(x) = 3.
6.6B: Using Arctrigs in Word Problems (part 1)
In Section 5.7, we solved word problems by drawing right triangles and using SOHCAHTOA. However, we
were finding sides, not angles, in those situations. That’s because we couldn’t solve for θ unless its trig value
was special.
Now that we have arctrigs, we can also find unknown angles from two sides! The inverse trig of ANY positive
ratio is automatically acute. Thus, many word problems with unknown angles feature unsimplified inverse trigs.
Ex 7: In the isosceles triangle in the bottom figure,
find the base angle θ in radians.
NOTE: Split this into two right triangles.
Ex 8: In the figure on the right, angles ∠CAO and
∠BCO are right angles. Determine ∠AOB in radians.
HINT: The angle you want can be split into two
pieces (∠AOC and ∠COB) which each come from a
different right triangle. These two right triangles have
a common side; you should find its length.
To be continued next class...