Trig Functions with the Unit Circle Exercise 4

Answer: \(\sin(-45^\circ)=-\dfrac{\sqrt{2}}{2}\quad\) \(\sec\left(-\dfrac{2\pi}{3}\right)=-2\)

Solution Method: To find the trig value of an angle less than 0, we find its coterminal angle on the unit circle, that means between 0 and \(360^\circ\) (or 0 and \(2\pi\)). Since coterminal angles all have the same terminal side, the trig values of coterminal angles are the same. 

Coterminal Angle: Angles in standard position (with the initial side on the positive x-axis) that have a common terminal side. Coterminal angles differ by multiples of \(360^\circ\) or 2\(\pi\) in radians.
(For more info see the coterminal angles section.)

Since we're starting with a negative angle in degrees, we're going to add a full rotation, 360 degrees, until the angle is between 0 and \(360^\circ.\) \(-45^\circ+350^\circ=315^\circ.\)
This means that \(-45^\circ\) and \(315^\circ\) are coterminal angles and \(\sin(-45^circ = \sin(315^\circ).\) The coordinate at \(315^\circ\) is \(\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right).\) Sine is equal to the \(y\)-coordinate, so \(\sin(-45^\circ)=-\frac{\sqrt{2}}{2}.\)

For \(\sec\left(-\frac{2\pi}{3}\right),\) we want to find the coterminal angle to \(-\frac{2\pi}{3}\). So we add \(2\pi\) to \(\frac{2\pi}{3}\): 
    \begin{equation*}
        -\frac{2\pi}{3} +2\pi= -\frac{2\pi}{3}+\frac{6\pi}{3} = \frac{4\pi}{3}
    \end{equation*}
So \(-\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) are coterminal and \(\sec\left(-\frac{2\pi}{3}\right)=\sec\left(\frac{4\pi}{3}\right).\) The coordinate at \(\frac{4\pi}{3}\) is \(\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).\) Secant is the reciprocal of cosine and cosine is the x-coordinate, so we have
    \begin{equation*} 
        \sec\left(-\frac{2\pi}{3}\right) = \sec\left(\frac{4\pi}{3}\right) = \frac{1}{\cos{\frac{4\pi}{3}}} = \frac{1}{-\frac{1}{2}}=-2
    \end{equation*}