Math 12 Pre-calculus
Working with Inverse trig functions
1.
Remember, you take the sine of an angle to get a ratio
sin(angle)  ratio
You take the arcsine of a ratio to get an angle
angle  sin 1 (ratio)
Below are the graphs of y = sec(x) , y = csc(x), and y = cot(x). On the empty grids below, sketch the full range of the inverse
of each of these functions. Then propose a cut and define a range of principal values.
2.
2. Assume only principal values apply, and find the simplest exact value of each of the following
a)
cos1   12 
e)
cot sin 1
h)
sec cot 1 
k)
15
sin  cos1  17
  tan 1  125 

 


3
2
11
2

 
b)
sin 1
f)
sin tan 1  3
i)
cos  arctan 15 
2
2


l)
b)
 
tan 1 

sin  tan 1  x  
d)
cot cos 1
  
x 2 9
x
d)

csc  arctan   23  
j)
cos  2sin 1  54  
cos  cos1  12   sin 1   12  
m)

 
sin cos 1 
g)
3. Re-write as an algebraic expression of x.
a)
3
3
sin  2cos 1  53  
 
b)
tan  arccos  x  
c)
sec arcsin
e)
cos  2 tan 1 ( x) 
f)
sin 2arccos( x) 
x
x2  4
4. What the heck, one for the road: Spiral of triangles… The hypotenuse of the first right triangle
(see figure) has a length of . In the second, the hypotenuse is , in the third it’s . The pattern
continues. In this sequence, which triangle will intersect the first triangle at a point other than the
vertex? (I think this one is a toughie…)
Possible answers:
Can’t really draw the graphs here. It’s interesting to see arcsec(x) and arccsc(x) have ranges in two
chunks. It’s also interesting to note that arccot(x) can be defined two different ways – and you
could make an argument for both.

140
221
2
3
2 x 1  x2


6
½
x2  9
3
7
25
3
2
x

x2  1
3
3

4
1  x2
x
Ha! I haven’t done that one yet. I said that it was a toughie….
226
226
3

2
x2  4
2
165
11
20
29

13
2
1  x2
x2  1
3
2