Calculate the exact value of $ an(-frac{pi}{6})$ using the unit circle and verify by applying trigonometric identities.

Using the unit circle, first note that $-\frac{\pi}{6}$ is equivalent to $-30^\circ$. On the unit circle, this angle corresponds to the coordinates $\left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$.

Therefore, the value of $\tan(-\frac{\pi}{6})$ is given by the ratio of the y-coordinate to the x-coordinate:

$ \tan(-\frac{\pi}{6}) = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $

Verification using trigonometric identities can be done by noting that $\tan(-x) = -\tan(x)$. Hence,

$ \tan(-\frac{\pi}{6}) = -\tan(\frac{\pi}{6}) = -\frac{\sqrt{3}}{3} $

We start by recognizing that $-frac{pi}{6}$ corresponds to $-30^circ$ on the unit circle, where the coordinates are $left( frac{sqrt{3}}{2}, -frac{1}{2}
ight)$.

The tangent function is defined as the ratio of the y-coordinate to the x-coordinate:

$ an(-frac{pi}{6}) = frac{-frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}}$

For clarity, simplify $-frac{1}{sqrt{3}}$:

$ -frac{1}{sqrt{3}} cdot frac{sqrt{3}}{sqrt{3}} = -frac{sqrt{3}}{3} $

To verify, using the identity $ an(-x) = – an(x)$:

$ an(-frac{pi}{6}) = – an(frac{pi}{6}) = -frac{sqrt{3}}{3} $

We know $-frac{pi}{6}$ or $-30^circ$ on the unit circle has coordinates $left( frac{sqrt{3}}{2}, -frac{1}{2}
ight)$. Thus,

$ an(-frac{pi}{6}) = frac{-frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} $

This simplifies to

$ -frac{sqrt{3}}{3} $

Verifying with $ an(-x) = – an(x)$:

$ an(-frac{pi}{6}) = – an(frac{pi}{6}) = -frac{sqrt{3}}{3} $

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