$Unit Circle and Trigonometric Functions$

Consider the unit circle centered at the origin. Let point $P$ have coordinates $(x, y)$ on the unit circle such that the angle $\theta$ formed by the positive x-axis and the radius to $P$ is in the fourth quadrant. If the secant and tangent of $\theta$ are given as $\sec \theta = 5$ and $\tan \theta = -\frac{4}{3}$, find the coordinates of point $P$.

Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\cos \theta = \frac{1}{5}$. Since $x^2 + y^2 = 1$ for any point on the unit circle:

$x^2 + y^2 = 1$

Given that $x = \cos \theta = \frac{1}{5}$, we find $y$ using $\tan \theta = \frac{y}{x}$:

$y = \tan \theta \cdot x = -\frac{4}{3} \cdot \frac{1}{5} = -\frac{4}{15}$

Therefore, the coordinates of point $P$ are:

$P = \left( \frac{1}{5}, -\frac{4}{15} \right)$

Given a point $A$ on the unit circle in the fourth quadrant with coordinates $(a, b)$ such that $csc heta = -4$ and $cot heta = -frac{1}{2}$. Find the exact coordinates of point $A$.

Since $csc heta = frac{1}{sin heta}$, we have $sin heta = -frac{1}{4}$. Using $cot heta = frac{cos heta}{sin heta}$, we find $cos heta$:

$cot heta = frac{cos heta}{sin heta} = -frac{1}{2}$

Thus,

$cos heta = cot heta cdot sin heta = -frac{1}{2} cdot -frac{1}{4} = frac{1}{8}$

The coordinates of point $A$ are:

$A = left( frac{1}{8}, -frac{1}{4}
ight)$

Suppose point $B$ lies on the unit circle in the fourth quadrant and can be expressed as $(b, c)$. If $cos phi = frac{2}{7}$ and $sin phi = -frac{sqrt{45}}{7}$, determine the coordinates of point $B$.

Using the Pythagorean identity for the unit circle:

$b^2 + c^2 = 1$

Given $cos phi = frac{2}{7}$ and $sin phi = -frac{sqrt{45}}{7}$, the coordinates of point $B$ are:

$B = left( frac{2}{7}, -frac{sqrt{45}}{7}
ight)$

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