Trigonometry: Inverse Functions, Applications of Triangles, Law of Sine and Law of Cosine

Running Head; TRIGONOMETRY
TRIGONOMETRY
Name
Institute of Affiliation
Date

TRIGONOMETRY 2
I. Inverse Functions
1.
a. State the functions, sine, cosine and tangent. Then state the domain and range of these
functions.
Function Domain Range
Sin −∞ ≤ x ≤ ∞ ∞
Cosine −∞ ≤ x ≤ ∞ ∞
Tangent −∞ ≤ x ≤ ∞ ∞
a) Why is it so important to restrict these function’s domains to be one-to-one?
The function can find values of large numbers up to infinity, it is important to describe
how far should a person go. For instance, since some of angles are θ+2 nπ, one has to
define the maximum value of n.
c) List the inverse of three functions sine, cosine and tangent using the appropriate notation.
Function inverse Domain Range
Sin Sin-1 −1 ≤ x ≤1 2
Cosine Cosine-1 −1 ≤ x ≤1 2
Tangent Tan-1 −∞ ≤ x ≤ ∞ ∞
Then name the restricted domain and range for each of these functions.
Beside each restriction sketch a drawing of the new function.

TRIGONOMETRY 3
d) Find the inverse of (𝑥) = (1/2)Cos(x). State the domain and range of both f(x) and f^-1(x)(aka
inverse).
f ( x)=1
2 cos( x ).
2 f ( x )= cos x
= cos-1 x
Domain is −1 ≤ x ≤1 , this because the cosine of the angle can only be between -1 and 1.
and the range is given as shown
¿ 1−−1
= 2
2. Complete the following problems. (5 pts.)
a) Co s−1 1
2 = 60 (from the above triangle)
b) Si n−1−1
2
it is easier to find Si n−1 1
2 and then check in the quadrant where the sin has negative values
Si n−1−1
2 = 180+ Si n−1 1
2

TRIGONOMETRY 4
= 180 + 30
= 210
b) Tan−1− √ 3= ( the value has to be in the 2nd or 4th quadrant)
= 180 - Ta n−1 √3 ( for the value in the 2nd quadrant)
= 180 – 60
= 120
And /or
= 360 - Ta n−1 √ 3 ( for the value in the 4th quadrant)
= 360 – 60
= 300
c) cos (ta n−1 (8
3 ))
Starting with the value in the bracket first
ta n−1
( 8
3 ) = nπ +θ ( this is the general solution where n=0,1,2,3,4,5...)
= 69.44, 249.44, 429.44..........
Rewriting the question again
cos (69.44) = .......( taking the first value only)
= 0.351
d) sin(co s−1 ( 3
4 ))
Starting with the value in the bracket first