In this discussion, we evaluate inverse sine.
When working with inverse trig functions, in particularly, inverse sine, the equation is given in the form,
y = sin-1(x).
Always check to see if the x-value satisfies the domain and the y-values satisfies the range.
The domain of y = sin-1(x) is [-1,1].
The Range of y = sin-1(x) is [-\frac{\pi}{2}, \frac{\pi}{2}]
One must only choose the value for y that satisfies the range.
If you want to know how to derive the domain and range, you can go to the following blog post.
Graphing Inverse Sine and Identifying the Domain and Range
Example
Given y = sin^{-1}(-\frac{1}{2})
Find the exact value for y.
🔷Observe the Domain
Recall that the domain of sin-1(x) is [-1,1].
Since x = -\frac{1}{2} is in the interval [-1, 1], the domain is satisfied.
Now take the sine of both sides to get sin(y) = -\frac{1}{2}.
Now sine of “what” gives -\frac{1}{2}?
Remember to observe the “b” coordinate when working with sine, given coordinates (a, b).
Thus, when y = \frac{7\pi}{6}, -\frac{5\pi}{6}, \frac{11\pi}{6}, -\frac{\pi}{6}
we get
sin(\frac{7\pi}{6}) = -\frac{1}{2}
sin(-\frac{5\pi}{6}) = -\frac{1}{2}
sin(\frac{11\pi}{6}) = -\frac{1}{2}
sin(-\frac{\pi}{6}) = -\frac{1}{2}
🔷Observe the Range
Remember that the final answer must to stay within the range of sin-1(x).
The range of y = sin-1(x) is [-\frac{\pi}{2}, \frac{\pi}{2}], which is the right region of the unit circle.
🔷Choose the Answer
Since \frac{11\pi}{6} is greater than the range, the exact answer is
y = -\frac{\pi}{6}.
Thus, sin(-\frac{\pi}{6}) = -\frac{1}{2}
means that -\frac{\pi}{6} = sin^{-1}(-\frac{1}{2})
Hence,
y = \frac{\pi}{6}
Done!
Let me know if you have any questions, or if this reading helps in the comments!
🔑 Practice is the key! If you would like more examples, practice problems with the answers, quizzes with the answers, and more regarding inverse trig functions, consider taking my Math Course on Inverse trig functions!
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