First, we find an angle that's coterminal with $-300^\circ$ that lies in the range $[0^\circ,360^\circ).$

\[ -300^\circ + 360^\circ = 60^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\cos\left(-300^\circ\right) = \cos 60^\circ.$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $60^\circ$ corresponds to the point $\left(\dfrac12,\dfrac{\sqrt 3}{2}\right).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos 60^\circ = \dfrac{1}{2}.$

Therefore, $\cos \left(-300^\circ\right) =\dfrac 1 2$

First, we find an angle that's coterminal with $-150^\circ$ that lies in the range $[0,360^\circ).$

\[ -150^\circ + 360^\circ = 210^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\sin\left(-150^\circ\right) = \sin 210^\circ.$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $210^\circ$ corresponds to the point $\left(-\dfrac{\sqrt 3}{2},-\dfrac12\right).$

Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin 210^\circ = -\dfrac{1}{2}.$

Therefore, $\sin \left(-150^\circ\right) =-\dfrac {1} 2$

First, we find an angle that's coterminal with $-225^\circ$ that lies in the range $[0^\circ,360^\circ).$

\[ -225^\circ + 360^\circ = 135^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\sec\left(-225^\circ\right) = \sec (135^\circ).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $135^\circ$ corresponds to the point $\left(-\dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2}\right).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos (135^\circ) = -\dfrac{\sqrt 2}{2}.$

Therefore, using the fact that $\sec\theta = \dfrac{1}{\cos\theta},$ we have \[ \sec(135^\circ)= \dfrac{1}{\cos(135^\circ)} = \dfrac{1}{\left(-\dfrac{\sqrt 2} {2}\right)} = -\dfrac {2} {\sqrt 2} = -\sqrt 2. \]

Therefore, $\sec \left(-225^\circ\right) = -\sqrt 2.$

First, we find an angle that's coterminal with $-30^\circ$ that lies in the range $[0,360^\circ).$

\[ -30^\circ + 360^\circ = 330^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\csc\left(-30^\circ\right) = \csc (330^\circ).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $330^\circ$ corresponds to the point $\left(\dfrac{\sqrt 3}{2},\dfrac12\right).$

Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin (330^\circ) = -\dfrac{1}{2}.$

Therefore, using the fact that $\csc\theta = \dfrac{1}{\sin\theta},$ we have \[ \csc 330^\circ= \dfrac{1}{\sin(330^\circ)} = \dfrac{1}{\left(-\dfrac{1} {2}\right)} = -2. \]

Therefore, $\csc \left(-30^\circ\right) = -2.$

First, we find an angle that's coterminal with $-\dfrac {5\pi}{3}$ that lies in the range $[0,2\pi).$

\[ -\dfrac{5\pi} 3 + 2\pi = \dfrac{\pi}{3}\quad{\color{green}{\checkmark}} \]

Therefore, $\tan\left(-\dfrac{5\pi} 3\right) = \tan \left(\dfrac{\pi} {3}\right).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $\dfrac\pi 3$ corresponds to the point $\left(\dfrac12, \dfrac{\sqrt 3}{2}\right).$

• Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos \left( \dfrac{\pi}{3}\right) = \dfrac{1}{2}.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin \left( \dfrac{\pi}{3}\right) = \dfrac{\sqrt 3}{2}.$

Therefore, using the fact that $\tan\theta = \dfrac{\sin\theta}{\cos\theta},$ we have \[ \tan \left( \dfrac{\pi}{3}\right)= \dfrac{\sin\left( \dfrac{\pi}{3}\right)}{\cos \left( \dfrac{\pi}{3}\right)} = \dfrac{\left(\dfrac{\sqrt 3}{2}\right)}{\left(\dfrac{1} {2}\right)} = \sqrt 3. \]

First, we find an angle that's coterminal with $-\dfrac \pi 3$ that lies in the range $[0,2\pi).$

\[ -\dfrac\pi 3 + 2\pi = \dfrac{5\pi}{3}\quad{\color{green}{\checkmark}} \]

Therefore, $\cot\left(-\dfrac\pi 3\right) = \cot \left(\dfrac{5\pi} {3}\right).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $\dfrac{5\pi}{3}$ corresponds to the point $\left(\dfrac12, -\dfrac{\sqrt 3}{2}\right).$

• Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos \left( \dfrac{5\pi}{3}\right) = \dfrac{1}{2}.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin \left( \dfrac{5\pi}{3}\right) = -\dfrac{\sqrt 3}{2}.$

Therefore, using the fact that $\cot\theta = \dfrac{\cos\theta}{\sin\theta},$ we have \[ \cot \left( \dfrac{5\pi}{3}\right)= \dfrac{\cos\left( \dfrac{5\pi}{3}\right)}{\sin \left( \dfrac{5\pi}{3}\right)} = \dfrac{\left(\dfrac{1}{2}\right)}{\left(-\dfrac{\sqrt 3} {2}\right)} = -\dfrac{\sqrt 3}{3}. \]

First, we find an angle that's coterminal with $\dfrac{7\pi}{3}$ that lies in the range $[0,2\pi).$

\begin{align} \dfrac{7\pi}{3} - 2\pi & = \dfrac{\pi}{3} \quad{\color{green}{\checkmark}}\\[5pt] \end{align}

Therefore, $\sin \left(\dfrac{7\pi}{3}\right) = \sin \left(\dfrac{\pi}{3}\right).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $\dfrac{\pi}{3}$ corresponds to the point $\left(\dfrac12,\dfrac{\sqrt 3}{2}\right).$

Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt 3}{2}.$

Therefore, $\sin\left(\dfrac{7\pi}{3}\right) = \dfrac{\sqrt 3}{2}.$

Finally, \[ \sin^4\left(\dfrac{7\pi}{3}\right) = \left(\dfrac{\sqrt 3}{2}\right)^4=\dfrac{9}{16}. \]

First, we find an angle that's coterminal with $480^\circ$ that lies in the range $[0,360^\circ).$

\begin{align} 480^\circ - 360^\circ & = 120^\circ\quad{\color{green}{\checkmark}} \end{align}

Therefore, $\cos (480^\circ) = \cos (120^\circ).$

Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.

From the unit circle, we see that the angle $120^\circ$ corresponds to the point $\left(-\dfrac12,\dfrac{\sqrt 3}{2}\right).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos (120^\circ) = -\dfrac{1}{2}.$

Therefore $\cos(480^\circ) = -\dfrac12.$