For a function to have an inverse function, it must be one to one(must pass the horizontal line test). The graphs of sin and cosine are clearly not functions- that is, they do not pass the horizontal line test. However the domains have been limited for the functions to be one to one.

Sin graphs if the domain is x= -1 to 1 or y= -п/2 to п/2 then sin(sin^-1x)= x and sin-1(siny)=y

Cosine Graphs:
if the domain is x=-1 to 1 or y=0 to п then cos(cos^-1x)=x and cos-1(cos y)=y
external image cosine_graph.gif

Tangent Graphs
if x= a real number and x is between - п/2 and п/2 then tan(tan^-1x)=x and tan-1(tany)=y


Evaluating the Inverse Sine Functions!
Examples: find exact value
a.) arc sin (-1/2)
First, draw a reference triangle and label theta. Because sine= opposite/ hypotenuse, you label -1 as the opposite and 2 as the hypotenuse on your graph. According to the 30, 60, 90 triangle it is clear that the angle theta is -30 degrees. -30 degrees = -pi/6 ...... or because sin(-pi/6)=-1/2 you can see thaet arcsin (-1/2)= -pi/6


Evaluating Compositions of Functions!

For example,
1.) sin(arctan 4/3)
First, draw a reference triangle. Next, label theta and the opposite side 4 and the adjacent side 3. By pythagorean theorem you know that the hypotenuse is 5. Now find the sin of that reference triangle. Take the opposite side over the hypotenuse. The answer should be 4/5.

2.) tan(arcsin(-1/external image 00e61be02b9ecec6a3fd923288029dbd.png2)
First, draw a reference triangle in Quadrant IV, because it is negative. Then, label theta and the opposite side -1, and the hypotenuse external image 00e61be02b9ecec6a3fd923288029dbd.png2. Now label the adjacent side1, because a 45, 45, 90 triangle has sides: 1,1, external image 00e61be02b9ecec6a3fd923288029dbd.png2. Now take the tangent of this reference triangle to evaluate the composition of the function for the answer. The opposite side, -1 over the adjacent side, 1 = -1.


tan(arccos x/2)
First, draw a reference triangle in the first quadrant. Label theta and the adjacent side x, and the hypotenuse 2. Label the opposite side any variable. For example, a. Now, use the pythagorean theorem to solve for a.
a^2 + x^2 = 2^2
a= external image 00e61be02b9ecec6a3fd923288029dbd.png4-x^2.
Now, find the tangent of this reference triangle. Tangent= (external image 00e61be02b9ecec6a3fd923288029dbd.png4-x^2)/x










Using A Calculator
To evaluate arc functions with a graphing calculator, press 2nd and then the function that you want. For arc secent, cosecent, and cotangent, it's one divived by the desired arc function.

Inverse Properties
The value that lies in the domain of the arcfunction will often come out as the answer. For example, tan[arctan(x)], the answer is x.

Compositions of Functions
Similar to the inverse property, these problems are set up in a form such as tan[arccos(x)]. To evaluate tan[arccos(2/3)]:
1. Draw a reference triangle illustrating the problem. Set the domain of the tangent function(arccos(2/3)) equal to theta, because it represents an angle.
2. Because theta is equal to arccos(2/3), cos(theta) is equal to 2/3.
3. From there you can determine which quadrant it will be in. Because cos(theta) is positive, theta is in the first quadrant.
4. From there, using Cos=Adj/Hyp, you can determine that the adjacent line has a value of 2 and the hypotenuse has a value of 3.
5. Next, find the third side using pathagoreum theorum(a^2+b^2=c^2). Root (2^2+x^2=3^2) is equal to root(5).
6. Now an equation can be determined. This equation is derived from tan=Opp/adj, which is [root(5)/2]