Evaluate the expression.tan−1(tan2π3)\tan^{-1}\left(\tan\frac{2\pi}{3}\right)tan−1(tan32π)

− π 3 -\frac{\pi}{3} − 3 π

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

we're asked to evaluate the expression of the inverse tangent of the tangent of 2 pi over 3. Now you may be tempted here to cancel that inverse tangent and that tangent, but don't. We don't wanna do that. We wanna start solving this the way we would any composite trig function starting with that inside function, the tangent of 2 pi over 3. Now coming over here to my unit circle, my angle 2 pi over 3, I know that the tangent of this angle is going to be the negative square root of 3. So that gives me my answer for that inside function, negative root 3. Now I'm just left to find the inverse tangent of that value. Now remember, when working with inverse trig functions, we can also think of this as, okay, the tangent of what angle is equal to negative root 3? But specifically, what angle is equal to negative square root of 3 within our specified interval. Now for the inverse tangent, my angle has to be between negative pi over 2 and positive pi over 2. So on my unit circle, I'm only looking for a value for which my tangent is equal to negative root 3 within this interval. Now I know that for my tangent value to be negative, I need to be in this quadrant 4. So for which of these angles is the tangent equal to negative root 3? Well, for 5 pi over 3, I know that my tangent is equal to negative square root of 3. But 5 pi over 3 is actually not within my interval from negative pi over 2 to positive pi over 2. But remember, when working with the tangent, we are thinking of this bottom quadrant, quadrant 4, as being negative values. So we measure all of these angles clockwise from our zero point, and they just all represent the negative angles of our positive quadrant 1. So this 5 pi over 3 is actually the same thing as a negative pi over 3, just measured clockwise from 0 here. So the tangent of negative pi over 3 is negative root 3. So that gives me my final answer, negative pi over 3, a value that is within my specified interval. Now, if we would have just canceled that inverse tangent and tangent, we would have got 2 pi over 3, which would be completely wrong. So remember to be careful, check those intervals, and let me know if you have questions.

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

− π 3 -\frac{\pi}{3} − 3 π

0. Functions

Inverse Trigonometric Functions

Calculus

0. Functions

Inverse Trigonometric Functions

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Multiple Choice

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

$-\frac{\pi}{3}$

$\frac{\pi}{3}$

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

$\frac{5\pi}{3}$

Verified step by step guidance

Understand the problem: We need to evaluate the expression $ \tan^{-1}(\tan(\frac{2\pi}{3})) $. The $ \tan^{-1} $ function, also known as the arctangent function, returns the angle whose tangent is the given number.

Recall the range of the $ \tan^{-1} $ function: The principal value of $ \tan^{-1} $ is in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. This means the result of $ \tan^{-1} $ must be an angle within this range.

Consider the angle $ \frac{2\pi}{3} $: This angle is in the second quadrant, where the tangent function is negative. The tangent of $ \frac{2\pi}{3} $ is the same as the tangent of $ \pi - \frac{2\pi}{3} = \frac{\pi}{3} $, but with a negative sign.

Find the equivalent angle in the principal range: Since $ \tan(\frac{2\pi}{3}) = -\tan(\frac{\pi}{3}) $, we need to find an angle $ \theta $ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $ \tan(\theta) = -\tan(\frac{\pi}{3}) $. This angle is $ -\frac{\pi}{3} $.

Conclude the evaluation: Therefore, $ \tan^{-1}(\tan(\frac{2\pi}{3})) = -\frac{\pi}{3} $, which is the angle in the principal range that has the same tangent value as $ \frac{2\pi}{3} $.

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