Pre-Calculus
Lesson 28:
Inverse Circular Functions
1. The following are some facts about generic (non-circular) inverse functions:
a) If a function 𝑓 is one-to-one, then 𝑓 has an inverse function 𝑓 −1
b) In a one-to-one function, each 𝑥-value corresponds to only one 𝑦-value and each 𝑦value corresponds to only one 𝑥-value
c) The domain of 𝑓 is the range of 𝑓 −1 , and the range of 𝑓 is the domain of 𝑓 −1
d) The graphs of 𝑓 and 𝑓 −1 are reflections of each other about the line 𝑦 = 𝑥
e) To find 𝑓 −1 (𝑥) from 𝑓(𝑥), follow these steps:
1) Replace 𝑓(𝑥) with 𝑦 and interchange 𝑦 and 𝑥
2) Solve for 𝑦
3) Replace 𝑦 with 𝑓 −1 (𝑥)
2. Inverse Sine Function
𝜋
𝑦 = sin−1 𝑥 or 𝑦 = arcsin 𝑥 means that 𝑥 = sin 𝑦, for − 2 ≤ 𝑦 ≤
𝜋
2
3. Finding Inverse Sine Values
Ex.
Find 𝑦 in each equation.
1
(a) 𝑦 = arcsin 2
(b) 𝑦 = sin−1 (−1)
(a) We ask ourselves what angle in the interval [−
𝜋
2
(c) 𝑦 = sin−1(−2)
𝜋
1
, ] yields a sine value of .
2
2
1
2
If we rewrite the equation thusly, sin 𝑦 = , and then look at our unit circle, we see that the
𝜋
angle 6 meets both criteria, so that is our answer.
Alternatively, you could type this into your calculator using the “2𝑛𝑑 ” button and then pressing
the “sin” button to get sine inverse.
(b) Take the same steps as in (a). Rewrite the equation as sin 𝑦 = −1 and look at your unit
𝜋
circle (or just type it in your calculator). We find that 𝑦 = − 2 is our answer.
(c) Rewrite the equation as sin 𝑦 = −2. Remember the Range of the sine curve is [−1, 1],
therefore, since −2 is outside of that interval, sin−1(−2) does not exist.
4. Inverse Cosine Function
𝑦 = cos −1 𝑥 or 𝑦 = arccos 𝑥 means that 𝑥 = cos 𝑦, for 0 ≤ 𝑦 ≤ 𝜋
5. Finding Inverse Cosine Values
Ex.
Find 𝑦 in each equation.
(b) 𝑦 = cos−1 (−
(a) 𝑦 = arccos 1
√2
)
2
(a) Rewrite the equation as cos 𝑦 = 1. Again, using the unit circle or the calculator, we find that
𝑦 = 0.
(b) Rewrite the equation as cos 𝑦 = −
√2
.
2
Using the same steps, we find that 𝑦 =
3𝜋
4
or 135°.
6. Inverse Tangent Function
𝜋
𝑦 = tan−1 𝑥 or 𝑦 = arctan 𝑥 means that 𝑥 = tan 𝑦, for − 2 ≤ 𝑦 ≤
𝜋
2
.
7. Inverse Function Values
Range
Interval
Inverse Function
Domain
𝑦 = sin−1 𝑥
[−1, 1]
𝑦 = cos −1 𝑥
𝑦 = tan−1 𝑥
[−1, 1]
(−∞, ∞)
𝑦 = csc −1 𝑥
(−∞, −1] ∪ [1, ∞)
𝑦 = sec −1 𝑥
(−∞, −1] ∪ [1, ∞)
𝑦 = cot −1 𝑥
(−∞, ∞)
𝜋 𝜋
[− , ]
2 2
[0, 𝜋]
𝜋 𝜋
(− , )
2 2
𝜋 𝜋
[− , ] , 𝑦 ≠ 0
2 2
𝜋
[0, 𝜋], 𝑦 ≠
2
(0, 𝜋)
Range
Quadrants of Unit
Circle
I and IV
8. Finding Inverse Function Values (Degree-Measured Angles)
Ex.
Find the degree measure of 𝜃 in the following:
(a) 𝜃 = arctan 1
(b) 𝜃 = sec −1 2
𝜋 𝜋
(a) Convert the radian interval to degrees: (− 2 , 2 ) = (−90°, 90°)
Rewrite the equation as tan 𝜃 = 1, so by the unit circle or a calculator, 𝜃 = 45°
1
(b) Rewrite the equation as sec 𝜃 = 2, so cos 𝜃 = 2, which means 𝜃 = 60°
I and II
I and IV
I and IV
I and II
I and II
9. Finding Function Values Using Definitions of the Trigonometric Functions
Ex.
Evaluate each expression without using a calculator.
3
2
(a) sin (tan−1 )
(b) tan (cos−1 (−
5
))
13
3
(a) Treat tan−1 2 as 𝜃
3
𝑦
So tan 𝜃 = 2 = 𝑥 , now we use the Pythagorean theorem to find the length of the hypotenuse.
𝑥2 + 𝑦2 = 𝑟2
22 + 3 2 = 𝑟 2
4 + 9 = 𝑟2
13 = 𝑟 2
√13 = 𝑟
sin 𝜃 =
(b) Treat cos−1 (−
So cos 𝜃 = −
5
13
=
5
) as 𝜃
13
𝑥
, and cosine
𝑟
3
√13
=
3√13
13
inverse is negative in quadrant II, we can now use the
Pythagorean theorem to find the value of 𝑦.
𝑥2 + 𝑦2 = 𝑟2
(−5)2 + 𝑦 2 = 132
25 + 𝑦 2 = 169
𝑦 2 = 144
𝑦 = 12
(it is positive because it is in quadrant II)
tan 𝜃 = −
12
5
10. Finding Function Values Using Identities
Ex.
Evaluate each expression without using a calculator.
1
3
2
5
(a) cos (arctan √3 + arcsin )
(b) tan (2 arcsin )
1
(a) If we let arctan √3 be 𝐴 and we let arcsin 3 be 𝐵. So we can use the identity cos(𝐴 + 𝐵).
1
1
= cos(arctan √3) cos (arcsin ) − sin(arctan √3) sin (arcsin )
3
3
𝜋 2√2
𝜋 1
= cos (
) − sin ( )
3 3
3 3
1 2√2
√3 1
= (
( )
)−
2 3
2 3
=
2
2√2 − √3
6
(b) Let 𝐴 = arcsin 5 , then you have tan 2𝐴 so we can use the double angle identity.
2 tan 𝐴
=
1 − tan2 𝐴
2
2(
)
√21
=
2 2
1−(
)
√21
4
4
= √21 = √21
4
17
1−
21
21
4√21
= 21
17
21
4√21 21
=
∙
21 17
4√21
=
17