Inverse Circular Functions: Problems based on trigonometric functions of inverse trigonometric functions

Inverse trigonometric functions are the inverse functions of the trigonometric functions with restricted domains. They are used to determine the angle that corresponds to a given trigonometric value. The principal inverse trigonometric functions are:

$\arcsin(x)$: Inverse sine function
$\arccos(x)$: Inverse cosine function
$\arctan(x)$: Inverse tangent function
$\arccot(x)$: Inverse cotangent function
$\arcsec(x)$: Inverse secant function
$\arccsc(x)$: Inverse cosecant function

Important Points and Differences

Function	Range	Inverse Function	Domain of Inverse Function	Range of Inverse Function
$\sin(x)$	$[-1,1]$	$\arcsin(x)$	$[-1,1]$	$[-\frac{\pi}{2}, \frac{\pi}{2}]$
$\cos(x)$	$[-1,1]$	$\arccos(x)$	$[-1,1]$	$[0, \pi]$
$\tan(x)$	$(-\infty,\infty)$	$\arctan(x)$	$(-\infty,\infty)$	$(-\frac{\pi}{2}, \frac{\pi}{2})$
$\cot(x)$	$(-\infty,\infty)$	$\arccot(x)$	$(-\infty,\infty)$	$(0, \pi)$
$\sec(x)$	$(-\infty,-1] \cup [1,\infty)$	$\arcsec(x)$	$(-\infty,-1] \cup [1,\infty)$	$[0, \pi] \setminus {\frac{\pi}{2}}$
$\csc(x)$	$(-\infty,-1] \cup [1,\infty)$	$\arccsc(x)$	$(-\infty,-1] \cup [1,\infty)$	$[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus {0}$

Formulas Involving Inverse Trigonometric Functions

The following are some important formulas that involve inverse trigonometric functions:

$\sin(\arccos(x)) = \sqrt{1 - x^2}$
$\cos(\arcsin(x)) = \sqrt{1 - x^2}$
$\tan(\arcsin(x)) = \frac{x}{\sqrt{1 - x^2}}$
$\tan(\arccos(x)) = \frac{\sqrt{1 - x^2}}{x}$
$\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$
$\cos(\arctan(x)) = \frac{1}{\sqrt{1 + x^2}}$

Examples to Explain the Important Points

Example 1: Evaluate $\sin(\arccos(\frac{1}{2}))$

Using the formula $\sin(\arccos(x)) = \sqrt{1 - x^2}$, we can evaluate the expression:

$$ \sin(\arccos(\frac{1}{2})) = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$

Example 2: Find the value of $\tan(\arcsin(\frac{3}{5}))$

Using the formula $\tan(\arcsin(x)) = \frac{x}{\sqrt{1 - x^2}}$, we get:

$$ \tan(\arcsin(\frac{3}{5})) = \frac{\frac{3}{5}}{\sqrt{1 - (\frac{3}{5})^2}} = \frac{\frac{3}{5}}{\sqrt{1 - \frac{9}{25}}} = \frac{\frac{3}{5}}{\sqrt{\frac{16}{25}}} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} $$

Example 3: Determine $\cos(\arctan(1))$

Using the formula $\cos(\arctan(x)) = \frac{1}{\sqrt{1 + x^2}}$, we find:

$$ \cos(\arctan(1)) = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

Example 4: Solve for $x$ if $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$

This is a property of the inverse sine and cosine functions. For any $x$ in the domain $[-1,1]$, the following identity holds:

$$ \arcsin(x) + \arccos(x) = \frac{\pi}{2} $$

Therefore, the equation is true for all $x$ in the domain $[-1,1]$.

Example 5: Compute $\sin(\arctan(\frac{1}{\sqrt{3}}))$

Using the formula $\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$, we calculate:

$$ \sin(\arctan(\frac{1}{\sqrt{3}})) = \frac{\frac{1}{\sqrt{3}}}{\sqrt{1 + (\frac{1}{\sqrt{3}})^2}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{1 + \frac{1}{3}}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{\frac{4}{3}}} = \frac{1}{2} $$

In conclusion, understanding the relationships between trigonometric functions and their inverses is crucial for solving problems involving inverse circular functions. By using the formulas and properties outlined above, one can tackle a wide range of problems in this domain.