Evaluate the expression. cos ( cos − 1 ( − 3 ) ) \cos\left(\cos^{-1}\left(-\sqrt3\right)\right) cos ( cos − 1 ( − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> ) )

this problem, we're asked to evaluate the expression, the cosine of the inverse cosine of negative square root of 3. Now, here, you may be tempted to just cancel that cosine and inverse cosine. But remember, you can't do that. You have to make sure that everything is in the correct interval. So let's go ahead and start solving this the way we would any composite trig function starting with that inside function, which is the inverse cosine of the negative square root of 3. Now something that you may notice here is that if we go and look at our interval for inverse cosine, our values that we take the inverse cosine of can only be in between negative one and one, and the angles that we get out can only be between 0 and pi. Now here we're trying to take the inverse cosine of negative root 3. Now negative root 3 will actually give you something like negative 1.73 with a bunch more decimal places, but it's safe to say that this is not within the interval from negative one to 1. So we actually can't take the inverse cosine of the negative square root of 3 at all and we don't have an answer here. Now if you don't notice that right away, that's okay because you may just come over to your unit circle and look for a value for which the cosine is equal to negative square root of 3. But you would not find a value for which that is true because there is not one. So you might not notice that interval right off the bat. But safe to say, you would probably realize that you cannot find the inverse cosine of the negative square root of 3 even if you come over to your unit circle. So here, because this value is not within our specified interval from negative one to positive one, we cannot get an answer here. And if you type this in your calculator, you would actually get an error. So here, we have no answer because you simply cannot take the inverse cosine of negative root 3. Thanks for watching, and let me know if you have questions.

0. Functions

Inverse Trigonometric Functions

Calculus

0. Functions

Inverse Trigonometric Functions

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Multiple Choice

Evaluate the expression.
$\cos\left(\cos^{-1}\left(-\sqrt3\right)\right)$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\pi$

Undefined

Verified step by step guidance

Understand the problem: We need to evaluate the expression $ \cos(\cos^{-1}(-\sqrt{3})) $. The $ \cos^{-1} $ function, also known as the inverse cosine or arccosine, returns an angle whose cosine is the given number.

Recall the domain and range of the $ \cos^{-1} $ function: The domain of $ \cos^{-1}(x) $ is $[-1, 1]$, and its range is $[0, \pi]$. This means $ \cos^{-1}(x) $ can only accept values between $-1$ and $1$.

Identify the issue: The expression $ \cos^{-1}(-\sqrt{3}) $ is problematic because $-\sqrt{3} \approx -1.732$, which is outside the domain $[-1, 1]$ of the $ \cos^{-1} $ function.

Conclude the evaluation: Since $-\sqrt{3}$ is not within the domain of the $ \cos^{-1} $ function, the expression $ \cos(\cos^{-1}(-\sqrt{3})) $ is undefined.

Summarize the result: The expression cannot be evaluated because it involves taking the inverse cosine of a value outside its valid domain, leading to an undefined result.