Evaluate the expression. sin − 1 1 \sin^{-1}1 sin − 1 1

In this problem, we're asked to evaluate the expression at the inverse sine of 1. Remember, we can think of this as, okay, the sine of what angle is equal to 1 and we're trying to find what angle makes that true. Now looking at my unit circle I know my value has to be in between negative pi over 2 and positive pi over 2 And because I'm taking the inverse sign of a positive value, I know that my answer has to be in quadrant 1 where all of my sine values are positive. Now I also know that the sine of pi over 2 is 1 because this is one of my quadrantal angles. So this is my solution here, pi over 2, because the sine of pi over 2 is equal to 1. Now my solution is done and we are good to go. Thanks for watching and let me know if you have questions.

Evaluate the expression. sin ⁡ − 1 1 \sin^{-1}1 sin − 1 1

− π 2 -\frac{\pi}{2} − 2 π

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

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Multiple Choice

Evaluate the expression.
$\sin^{-1}1$

$0$

$\frac{\pi}{2}$

$\pi$

$-\frac{\pi}{2}$

Verified step by step guidance

Understand the problem: We need to evaluate the expression $ \sin^{-1}(1) $. This involves finding the angle whose sine is 1.

Recall the definition of $ \sin^{-1}(x) $: The inverse sine function, $ \sin^{-1}(x) $, returns the angle $ \theta $ such that $ \sin(\theta) = x $ and $ \theta $ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Identify the angle: The sine of $ \frac{\pi}{2} $ is 1, which means $ \sin^{-1}(1) = \frac{\pi}{2} $.

Verify the range: Ensure that $ \frac{\pi}{2} $ is within the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which it is.

Conclude the evaluation: The expression $ \sin^{-1}(1) $ evaluates to $ \frac{\pi}{2} $.