Evaluate the expression.sin⁡−1(cos⁡2π3)\sin^{-1}\left(\cos\frac{2\pi}{3}\right)sin−1(cos32π)

− π 6 -\frac{\pi}{6} − 6 π

Evaluate the expression. sin ⁡ − 1 ( cos ⁡ 2 π 3 ) \sin^{-1}\left(\cos\frac{2\pi}{3}\right) sin − 1 ( cos 3 2 π )

In this problem, we're asked to evaluate the expression that the inverse sine of the cosine of 2 pi over 3. Now because this is a composite function, we wanna look at that inside function first, which is the cosine of 2 pi over 3. Now coming over here to my unit circle, I don't have to worry about my intervals right now because this is just a regular trig function, not an inverse one. So I can just come to my unit circle and look at 2 pi over 3. For 2 pi over 3, I know that my cos is equal to negative one half. So that gives you my answer to that inside trig function, so negative one half. Now I'm left to find the inverse sine of negative one half. Now because this is an inverse trig function, now we do need to consider the interval, which for the inverse sine is between negative pi over 2 and positive pi over 2. So I want to find an angle for which the sine is equal to negative one half within this interval. Remember that when working with inverse trig functions, we can also think of this as, okay, the sine of what angle is equal to negative one half? So we're looking for that angle within this interval. Now within this interval, I know that over here at 11 pi over 6, the sine is going to be equal to negative one half. But remember, our interval is from negative pi over 2 to positive pi over 2. And 11 pi over 6 is not within that interval as it is currently written. So we need to think of this as instead of being 11pi over 6, consider going clockwise around your unit circle. This angle is also negative pi over 6 instead of 11 pi over 6. So that gives me my actual answer within my specified interval. So the inverse sine of the cosine of 2 pi over 3 is negative pi over 6. And that's my final answer. Thanks for watching and let me know if you have questions.

Evaluate the expression. sin ⁡ − 1 ( cos ⁡ 2 π 3 ) \sin^{-1}\left(\cos\frac{2\pi}{3}\right) sin − 1 ( cos 3 2 π )

− π 6 -\frac{\pi}{6} − 6 π

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions

Precalculus

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions

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Multiple Choice

Evaluate the expression.
$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

$\frac{\pi}{6}$

$\frac{5\pi}{6}$

$\frac{\pi}{3}$

$-\frac{\pi}{6}$

Verified step by step guidance

First, understand that the problem involves evaluating the inverse sine function, $ \sin^{-1} $, of the cosine of an angle. The expression is $ \sin^{-1}(\cos(\frac{2\pi}{3})) $.

Recall that the cosine of $ \frac{2\pi}{3} $ is equivalent to the cosine of $ \pi - \frac{\pi}{3} $, which is $ -\cos(\frac{\pi}{3}) $. Since $ \cos(\frac{\pi}{3}) = \frac{1}{2} $, it follows that $ \cos(\frac{2\pi}{3}) = -\frac{1}{2} $.

Now, we need to find the angle $ \theta $ such that $ \sin(\theta) = -\frac{1}{2} $ and $ \theta $ is in the range of $ \sin^{-1} $, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The angle $ \theta $ that satisfies $ \sin(\theta) = -\frac{1}{2} $ within this range is $ -\frac{\pi}{6} $.

Thus, the evaluated expression $ \sin^{-1}(\cos(\frac{2\pi}{3})) $ is $ -\frac{\pi}{6} $.

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Struggling with Precalculus?

Evaluate the expression.sin⁡−1(cos⁡2π3)\sin^{-1}\left(\cos\frac{2\pi}{3}\right)sin−1(cos32π​)

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Evaluate the expression.
$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$