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Q. 5.2.5

Trigonometric functions using reference angles

Find the sine and cosine for each angle using reference angles.

a. θ = 120°      b. θ = 7π/6      c. θ = 690°       d. θ = -7π/4

Verified Solution

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