Inverse Sine Function
As you can see the graph of y=sin x does not pass the horizontal line test on its own.
When y=sin x restricts the domain to the interval [-π/2, π/2] it passes the horizontal line test and can now have an inverse function.
The inverse sine function is represented by y=arcsin x or y= sin-1 x
Remember: the sin-1 x is the angle (or number) whose sine is x
Evaluating the Inverse Sine Function
So lets say you have a problem like this:
sin-1(-1/2)
You are trying to find the angle whose sine is -1/2
If you look at the Unit Circle you will notice that at -π/6 the sin is -1/2
sin-1(-1/2)= -π/6
Inverse Cosine Function
In the cosine function, the domain is restricted to the interval [0,π] to make it 1-1 (pass the horizontal line test).
The inverse cosine function is represented by y=arccos x or y= cos-1 x
Remember: the cos-1 x is the angle (or number) whose cos is x
Evaluating the Inverse Cosine Function
You evaluate the inverse of cosine functions the exact same way as you do with inverse of sine functions.
Ex: arccos(-1)
Angle whose cosine is -1?? π
arccos(-1)= π
Inverse Tan Function
In the tangent function, the domain is restricted to the interval (-π/2, π/2) to make it 1-1 (pass the horizontal line test).
The inverse tangent function is represented by y=arctan x or y= tan-1 x
Remember: the tan-1 x is the angle (or number) whose tan is x
Evaluating the Inverse Tangent Function
You evaluate the inverse of tangent functions the exact same way as you do with inverse of sine and cosine functions.
ALSO remember that the domain and ranges of the inverse functions are reverse from the parent function
EX:
Sin x
D: [-π/2, π/2]
R: [-1, 1]
Sin-1 x
D: [-1, 1]
R: [-π/2, π/2]
Lastly remember that when you see:
cos(arccos x)
the cos and arccos cancel out making
cos(arccos x) = x
*This is the same for sine and tangent as well
