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Notes 4.7 pre-calculus
Inverse Trig Functions
If domain is restricted of y = sin x to
  
  2 , 2 
the inverse sine
1
function y  sin x is the inverse of the restricted sine function.
Inverse sine function (ArcSine function):
The unique angle y in the interval  2 , 2  such that


1
sin y = x is the inverse sine (or arcsine) of x, denoted sin
arcsin x.
1
y

sin
x is [-1,1] and the range is   ,   .
The domain of
2 2


2
f x = sin -1x 
5
-5
-2
Inverse cosine function (ArcCosine function):
The unique angle y in the interval 0,  such that


cos y = x is the inverse cosine (or arccosine) of x denoted
cos 1 x
or arccos x.
x
or
1
The domain of y  cos x is [-1,1] and the range 0, 


gx = cos- 1x
2
-5
5
-2
Inverse Tangent Function(Arctangent Function):
The unique angle y in the interval  2 , 2  such that tan y = x is the


1
inverse tangent (or arctangent) of x, denoted tan x
or arctan x.
1
The domain y  tan x is (-,) and the range is  2 , 2  .

2
hx =
tan- 1x
-5
5
-2

Always true:
sin(sin 1 x)  x
cos(cos 1 x)  x
tan(tan 1 x)  x
(x here relates to a coordinate)
True on restricted domain:
sin 1 (sin x)  x
cos1 (cos x)  x
tan 1 (tan x)  x
(x here relates to an angle)
Examples:
Find the exact value of each expression without a calculator.
a)
1
sin 1  
 2
Remember, on the unit circle the y coordinate is sin . So, we
are looking for the angle where we have a y coordinate of 1 .
2
Also, the range for the inverse sin function is restricted to
[  , 2 ]. So, the angle we are looking for is 6 .
2
b) sin 1  sin  9  





Remember on the restricted domain
sin 1 (sin x)  x
So,

  
cos1 (cos x)  x
tan 1 (tan x)  x

sin 1  sin    
 9  9

Evaluating trig functions with a calculator:
Find the approximate value, express answer in degrees.
a) cos1(0.76) 139.46
b) arcsin(0.89)  62.87
Evaluating Inverse Trig Functions without a calculator:
Find the exact value of the expression without a calculator.
a) cos1  cos  1.1 


Draw an angle of -1.1 in standard position (notice that this angle is
not in the interval [0,  ]) and mark its x-coordinate on the x-axis.
The angle in the interval [0,  ] whose cosine is this number is 1.1.
So, cos1  cos  1.1  = 1.1

b)


 


1
sin cos   





3  
First we need to find the value of cos    .
3
 
cos    = 1
3 2
 

1
Now we find sin  1  .
 2
 

 

1
sin  1   
 2
6
 

So,

  


1
sin cos    =





3   6