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

• Let's first consider the angle $60^\circ.$ From the unit circle, we see that the angle $60^\circ$ corresponds to the point \[ \left(\dfrac{1}{2},\dfrac{\sqrt 3}{2} \right). \] Since the sine of the angle corresponds to the $y$-coordinate, we have \[ \sin 60^\circ = \dfrac{\sqrt 3}{2}.\]

• Now, let's consider the angle $30^\circ.$ From the unit circle, we see that the angle $30^\circ$ corresponds to the point \[ \left(\dfrac{\sqrt 3}{2},\dfrac{1}{2} \right). \] Since the cosine of the angle corresponds to the $x$-coordinate, we have \[ \cos 30^\circ = \dfrac{\sqrt 3}{2}. \]

Finally, we combine the results:

\begin{align*} \sin 60^\circ - \cos 30^\circ &= \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} \\[5pt] &=0 \end{align*}

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

• Let's first consider the angle $180^\circ.$ From the unit circle, we see that the angle $180^\circ$ corresponds to the point \[ \left(-1,0 \right). \] Since the cosine of the angle corresponds to the $x$-coordinate, we have \[ \cos 180^\circ = -1. \]

• Now, let's consider the angle $300^\circ.$ From the unit circle, we see that the angle $300^\circ$ corresponds to the point \[ \left( \dfrac{1}{2}, -\dfrac{\sqrt{3}}{2} \right). \] Since the cosine of the angle corresponds to the $x$-coordinate, we have \[ \cos 300^\circ = \dfrac{1}{2}. \]

Using the fact that $\sec{\theta} = \dfrac{1}{\cos\theta},$ we have \begin{align*} \sec 300^\circ &= \dfrac{1}{\cos 300^\circ} \\[5pt] &= \dfrac{1}{\left(\dfrac{1}{2}\right)} \\[5pt] &= 2. \end{align*}

Finally, we combine the results:

\begin{align*} \cos^2\left(180^\circ\right) + 2\sec^2\left(300^\circ\right) &= (-1)^2+2\left( 2\right)^2 \\[5pt] &= 1 +2\left(4\right) \\[5pt] &= 1 +8 \\[5pt] &= 9 \end{align*}

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

• Let's first consider the angle $\dfrac{\pi}{3}.$ From the unit circle, we see that the angle $\dfrac{\pi}{3}$ corresponds to the point \[ \left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right). \] Since the sine of the angle corresponds to the $y$-coordinate, we have \[ \sin\left( \dfrac {\pi} {3}\right) = \dfrac{\sqrt 3}{2}. \]

• Now, let's consider the angle $\dfrac{5\pi}{3}.$ From the unit circle, we see that the angle $\dfrac{5\pi}{3}$ corresponds to the point \[ \left(\dfrac{1}{2},-\dfrac{\sqrt{3}}{2}\right). \] Since the cosine of the angle corresponds to the $x$-coordinate, we have \[ \cos\left( \dfrac {5\pi} {3}\right) = \dfrac{1}{2}. \]

Finally, we combine the results:

\begin{align*} \sin\left(\dfrac{\pi}{3}\right) - \cos\left(\dfrac{5\pi}{3}\right) &= \dfrac{\sqrt 3}{2}-\dfrac{1}{2} \\[5pt] &=\dfrac{-1+\sqrt{3}}{2} \end{align*}

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

• Let's first consider the angle $\dfrac{\pi}{2}.$ From the unit circle, we see that the angle $\dfrac{\pi}{2}$ corresponds to the point \[ (0,1). \] Since the sine of the angle corresponds to the $y$-coordinate, we have \[ \sin\left( \dfrac {\pi} {2}\right) = 1. \]

• Now, let's consider the angle $\dfrac \pi 6.$ From the unit circle, we see that the angle $\dfrac \pi 6$ corresponds to the point \[ \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right). \] Since the sine of the angle corresponds to the $y$-coordinate, we have \[ \sin\left( \dfrac {\pi} {6}\right) = \dfrac{1}{2}. \]

Finally, we combine the results:

\begin{align*} \sin^4 \left(\dfrac \pi 2\right) - \sin \left(\dfrac \pi 6\right) &= (1)^4 - \dfrac 1 2 \\[5pt] &= 1 - \dfrac 1 2 \\[5pt] & = \dfrac 1 2 \end{align*}