Find the exact values of sine, cosine, and tangent for the angle that corresponds to the point where the terminal side of angle $ heta $ intersects the unit circle at $(cos heta, sin heta)$. Given that $ heta$ is in the fourth quadrant and the poin

Given that $\theta$ is in the fourth quadrant and the point on the unit circle is $(\frac{1}{2}, -\frac{\sqrt{3}}{2})$, we can find the exact values of $\sin\theta$, $\cos\theta$, and $\tan\theta$.

First, we recognize that $(\cos\theta, \sin\theta)$ directly gives us the cosine and sine values:

$ \cos\theta = \frac{1}{2} $

$ \sin\theta = -\frac{\sqrt{3}}{2} $

To find $\tan\theta$, we use the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$:

$ \tan\theta = \frac{ -\frac{\sqrt{3}}{2} }{ \frac{1}{2} } $

$ \tan\theta = -\sqrt{3} $

Therefore, the values are:

$ \cos\theta = \frac{1}{2} $

$ \sin\theta = -\frac{\sqrt{3}}{2} $

$ \tan\theta = -\sqrt{3} $

Given $(frac{1}{2}, -frac{sqrt{3}}{2})$ on the unit circle in the fourth quadrant, we determine:

For $cos heta$:

$ cos heta = frac{1}{2} $

For $sin heta$:

$ sin heta = -frac{sqrt{3}}{2} $

For $ an heta$:

$ an heta = frac{ sin heta }{ cos heta } = frac{ -frac{sqrt{3}}{2} }{ frac{1}{2} } = -sqrt{3} $

Thus,

$ cos heta = frac{1}{2} $

$ sin heta = -frac{sqrt{3}}{2} $

$ an heta = -sqrt{3} $

Given the point $(frac{1}{2}, -frac{sqrt{3}}{2})$, calculate:

$ cos heta = frac{1}{2} $

$ sin heta = -frac{sqrt{3}}{2} $

$ an heta = -sqrt{3} $

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