How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions?

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How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions?

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Answer 1 · 2015-03-19T17:08:43

Throughout the following answer, I will assume that you are asking about trigonometry restricted to real numbers.

Using Domain of $a r c sin x$ $arc sin x$

Find $a r c sin (3)$ $arc sin (3)$ .

$3$ $3$ is not in the domain of $a r c sin$ $arc sin$ , ( $3$ $3$ is not in the range of $sin$ $sin$ ) so $a r c sin (3)$ $arc sin (3)$ does not exist.

Using Range of $a r c sin x$ $arc sin x$

Find $a r c sin (\frac{1}{2})$ $arc sin (1/2)$ .

Although there are infinitely many $t$ $t$ with $sin t = \frac{1}{2}$ $sin t = 1/2$ , the range of $a r c sin$ $arc sin$ restricts the value to those $t$ $t$ with $\frac{- π}{2} \leq t \leq \frac{π}{2}$ $(-pi)/2 <= t <= pi/2$ , So the value we want is $\frac{π}{6}$ $pi/6$ .

Using the quadrant
This is the same as using the range, but it involves thinking about the problem more geometrically.

Find $a r c sin (\frac{1}{2})$ $arc sin (1/2)$ .

$a r c sin (\frac{1}{2})$ $arc sin (1/2)$ is a number (an angle) in quadrant I or IV. It is the $t$ $t$ with smallest absolute value (the shortest path from the initial side).

Again, $a r c sin (\frac{1}{2}) = \frac{π}{6}$ $arc sin (1/2) = pi/6$ .

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How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions?

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