Inverse Trigonometry
If we are given an angle measurement, for example 𝜋⁄6, then we can evaluate the
function of it.
𝜋
sin = ________
6
𝜋
“The sine of 6 is equal to what?”
Inversely, if you are given the value of the sine function, for example ½, then the
challenge is to name the angle measurement.
sin 𝑥 =
1
2
“The sine of what angle is equal to ½?”
We could answer:
1
𝜋
“The angle whose sine is 2 is 6 .”
The algebraic abbreviation for that sentence is
1 𝜋
arcsin =
2 6
arcsin x is called the inverse sine function. It is the angle whose sine is x. Strictly
speaking, it is the “arc” length of the unit circle (in radians) whose sine is x.
Remember, 𝑦 = sin 𝑥 is a function. It has a domain (all real numbers) and a range
(-1, 1). It also has an inverse. The inverse is 𝑦 = arcsin 𝑥.
Do you remember inverses?
If 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = arcsin 𝑥, then:
𝑓(𝑔(𝑥)) = 𝑥
sin(arcsin(𝑥)) = 𝑥
𝑔(𝑓(𝑥)) = 𝑥
arcsin(sin(𝑥)) = 𝑥
Another way to write arcsin 𝑥 is sin−1 𝑥. These two notations mean exactly the same
thing: the inverse of the sine function.
Problems: What does each of the following imply? (The first one is completed for
you.)
1.
arccos 𝑁 = 𝜃 implies 𝑁 = cos 𝜃
2.
arctan 𝑡 = 𝛽 implies _______________________________
3.
arcsec 𝑢 = 𝑎 implies _______________________________
𝜋
4.
arccot 1 = 4 implies _______________________________
Example 1: Evaluate arcsin
√2
…
2
“The angle whose sine is
√2
.”
2
√2
Solution: “ 2 is the sine of what angle?”
√2
= sin _______
2
Range
𝜋
We have a problem here. 4 is not the only angle whose sine is
√2
.
2
This means that
√2
arcsin 2 has more than one answer (actually, it has infinite answers). List 3 other
angles (these may or may not be listed on your unit circle chart) that have a sine of
√2
2
:
_______________
_________________
_________________
So what’s the problem? The inverse trig relations (like arcsin) aren’t functions. They
have inputs with more than one output. Their graphs don’t pass the vertical line
test. What do we do about this? We restrict the range, the outputs. We restrict
them to the angles with the SMALLEST ABSOLUTE VALUE.
𝜋
√2
𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑠 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑤ℎ𝑜𝑠𝑒 𝑠𝑖𝑛𝑒 𝑖𝑠
4
2
Example 2: Evaluate arcsin (−
value whose sine is −
√2
2
√2
).
2
What is the angle with the smallest absolute
?
arcsin(−𝑥) = −arcsin(𝑥)
√2
√2
arcsin (− ) = − arcsin ( )
2
2
𝜋
=−
4
Therefore, in order for 𝑦 = arcsin 𝑥 to be a function, we must limit the range to be
𝜋
𝜋
− ≤𝑦≤
2
2
Problems: Evaluate the following (leave your answer in radians).
1.
arcsin 0
2.
arcsin 1
3.
sin−1(−1)
√3
4.
sin−1 ( 2 )
5.
sin−1 (−
6.
arcsin (− 2)
√3
)
2
1