Arctan square root 3 /3

Arctan square root 3 /3

The Correct Answer and Explanation is:

The expression you provided is the inverse tangent of 33\frac{\sqrt{3}}{3}33​​, written as:arctan⁡(33)\arctan\left(\frac{\sqrt{3}}{3}\right)arctan(33​​)

To find the exact value, we need to recall some fundamental trigonometric relationships. Specifically, we are looking for an angle θ\thetaθ whose tangent value equals 33\frac{\sqrt{3}}{3}33​​. That is:tan⁡(θ)=33\tan(\theta) = \frac{\sqrt{3}}{3}tan(θ)=33​​

From trigonometry, we know that:tan⁡(π6)=33\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}tan(6π​)=33​​

Thus, the angle θ=π6\theta = \frac{\pi}{6}θ=6π​ is the solution to the equation. Therefore:arctan⁡(33)=π6\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}arctan(33​​)=6π​

Explanation:

The inverse tangent function, arctan⁡(x)\arctan(x)arctan(x), gives the angle whose tangent is xxx. The tangent function, tan⁡(θ)\tan(\theta)tan(θ), for a standard 30-60-90 right triangle, has known values for the angles:

  • At θ=π6\theta = \frac{\pi}{6}θ=6π​ or 30°, the ratio of the opposite side to the adjacent side is 13\frac{1}{\sqrt{3}}3​1​, or equivalently, 33\frac{\sqrt{3}}{3}33​​.
  • In this case, we are dealing with the ratio 33\frac{\sqrt{3}}{3}33​​, which directly corresponds to the angle π6\frac{\pi}{6}6π​.

So, the solution to the inverse tangent of 33\frac{\sqrt{3}}{3}33​​ is:arctan⁡(33)=π6 radians\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \text{ radians}arctan(33​​)=6π​ radians

This is the simplest and most exact answer in radians. If you were to convert it to degrees, π6\frac{\pi}{6}6π​ radians equals 30°.

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