EVALUATING COMPOSITES - TRIG FUNCTIONS AND INVERSE TRIG
FUNCTIONS
paul vaz
The table below lists some composites of trig functions and inverse trig functions:
sin 1
sin( sin 1 x)
sin 1 (sin x)
cos( sin 1 x)
sin 1 (cos x)
tan( sin 1 x)
sin 1 (tan x)
cos 1
sin( cos 1 x)
cos 1 (sin x)
cos( cos 1 x)
cos 1 (cos x)
tan( cos 1 x)
cos 1 (tan x)
tan 1
sin( tan 1 x)
tan 1 (sin x)
cos( tan 1 x)
tan 1 (cos x)
tan( tan 1 x)
tan 1 (tan x)
csc
csc( sin 1 x)
csc( cos 1 x)
csc( tan 1 x)
sec
sec( sin 1 x)
sec( cos 1 x)
sec( tan 1 x)
cot
cot( sin 1 x)
cot( cos 1 x)
cot( tan 1 x)
sin
cos
tan
We will refer to any composite as: outside function(inside function)
Example:
In tan( sin 1 x), 'tan' is the outside function, and ' sin 1 ' is the inside function.
Inside function is an inverse:
Generally, for all composites in which the inside function is an inverse, we evaluate the
composite for a given 'x' using a substitution for the inside function. The substitution and
the definition of the inverse, leads to the construction of a right triangle that helps
evaluate the composite easily.
Example: Evaluate tan( sin 1 0.8)
Solution: Let p = sin 1 0.8 (Substitution)
Therefore,
sin p = 0.8
(Definition)
Construction of right triangle:
1
0.8
angle p
0.6
( Using the Pythagorean, the third side of the triangle is 1  0.82  0.6 )
Therefore,
tan( sin 1 0.8) = tan(p) = 0.8/0.6 = 1.33.
Useful Results:
1. sin( sin 1 x) = x if  1  x  1
1
2. cos( cos x) = x if  1  x  1
1
3. tan( tan x) = x if x is any real number.
Example: Evaluate sin( sin 1 0.8)
Since  1  0.8  1 ,
sin( sin 1 0.8) = 0.8.
Inside function is not an inverse:
In this case, work the inside function first, and then apply the outside function to the
result.
Example: Evaluate sin 1 ( tan

)
3
 3  1.732
Solution:
tan
Therefore,
sin 1 ( tan
3


3
) = sin 1 (1.732) (Not defined)
Useful Results:
4. sin 1 (sin x) = x if 
5.
cos 1 (cos x) = x if
6.
tan 1 (tan x) = x if 

x
2
2
0  x 

2
x
Example: Evaluate tan 1 ( tan
Since




3
2

2
)


3


2
,
tan 1 ( tan

3
) =

3