a. Figure 5.29(a) shows the terminal side of 120° in quadrant II and its reference angle 60°. Since sin θ > 0 and cos θ < 0 for any angle θ in quadrant II, sin(120°) > 0 and cos(120°) < 0. So to get cos(120°) from cos(60°), we must prefix a negative sign. We have

\sin \left(120^{\circ}\right)=\sin \left(60^{\circ}\right)= \frac{\sqrt{3}} {2}           and           \cos \left(120^{\circ}\right)=-\cos \left(60^{\circ} \right)= -\frac{1}{2}

b. Figure 5.29(b) shows the terminal side of 7π/6 in quadrant III and its reference angle π/6. Since sin θ < 0 and cos θ < 0 for any angle θ in quadrant III, we must prefix a negative sign to both sin(π/6) and cos(π/6). So

\sin \left(\frac{7 \pi}{6}\right)=-\sin \left(\frac{\pi}{6}\right)=-\frac{1}{2}            and           \cos \left(\frac{7 \pi}{6}\right)=-\cos \left(\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}

c. The angle 690° terminates in quadrant IV (where sin θ < 0 and cos θ > 0) and its reference angle is 30°, as shown in Fig. 5.29(c). So

\sin \left(690^{\circ}\right)=-\sin \left(30^{\circ}\right)=-\frac{1}{2}          and           \cos \left(690^{\circ}\right)=\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}

d. The angle -7π/4 terminates in quadrant I (where sinθ > 0 and cosθ > 0) and its reference angle is π/4, as shown in Fig. 5.29(d). So

\sin \left(-\frac{7 \pi}{4}\right)=\sin \left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}              and           \cos \left(-\frac{7 \pi}{4}\right)=\cos \left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}