$ ext{Find the tangent values for specific angles on the unit circle}$

To find the tangent values for specific angles on the unit circle, we can use the fact that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Let’s find the tangent value for $\frac{\pi}{4}$:

$\tan(\frac{\pi}{4}) = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})}$

We know that:

$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

Thus,

$\tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$

Therefore, $\tan(\frac{\pi}{4}) = 1$.

To find the tangent values for specific angles on the unit circle, we can use the relation $ an( heta) = frac{sin( heta)}{cos( heta)}$.

Let’s find the tangent value for $frac{3pi}{4}$:

$ an(frac{3pi}{4}) = frac{sin(frac{3pi}{4})}{cos(frac{3pi}{4})}$

We know that:

$sin(frac{3pi}{4}) = frac{sqrt{2}}{2}$

$cos(frac{3pi}{4}) = -frac{sqrt{2}}{2}$

Thus,

$ an(frac{3pi}{4}) = frac{frac{sqrt{2}}{2}}{-frac{sqrt{2}}{2}} = -1$

Therefore, $ an(frac{3pi}{4}) = -1$.

To find the tangent value for specific angles on the unit circle, we use $ an( heta) = frac{sin( heta)}{cos( heta)}$.

Let’s find the tangent value for $frac{5pi}{6}$:

$ an(frac{5pi}{6}) = frac{sin(frac{5pi}{6})}{cos(frac{5pi}{6})}$

We know that:

$sin(frac{5pi}{6}) = frac{1}{2}$

$cos(frac{5pi}{6}) = -frac{sqrt{3}}{2}$

Thus,

$ an(frac{5pi}{6}) = frac{frac{1}{2}}{-frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3}$

Therefore, $ an(frac{5pi}{6}) = -frac{sqrt{3}}{3}$.

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