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Homework 3
Section 7.3 #20: Find the exact value of

$\cos(0)$ .
Solution: In the unit circle, the point that goes with angle

$0$ radians is

$(1,0)$ . Since the cosine reads the

$x$ -coordinate of points on the unit circle, we must conclude that

$\cos(0)=1$ .

Section 7.3 #21: Find the exact value of

$\cos \left( \dfrac{\pi}{6} \right)$ .
Solution: In the unit circle, the point that goes with angle

$\dfrac{\pi}{6}$ radians is

$\left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)$ . Since the cosine reads the

$x$ -coordinate of points on the unit cicle, we must conclude that

$\cos \left( \dfrac{\pi}{6} \right)=\dfrac{\sqrt{3}}{2}$ .

Section 7.3 #24: Find the reference angle for the angle

$-170^{\circ}$ .
Solution: The angle is in quadrant III, since it is less than

$-90^{\circ}$ but greater than

$-180^{\circ}$ . Let us first express this negative angle as a posistive angle: to do that recall that a full rotation is

$360^{\circ}$ , so adding a full rotation yields the same terminal side of the angle (and hence the same angle -- remember that an angle is merely a pair of rays, by definition). So

$-170^{\circ} \stackrel{\mathrm{as \hspace{2pt} angles}}{=} -170^{\circ}+360^{\circ}=190^{\circ}.$ To find the reference angle, we note that all angles in quadrant III have a reference angle given by subtracting half a rotation, i.e.

$\mathrm{ref \hspace{2pt} angle}=190^{\circ}-180^{\circ}=10^{\circ}.$

Section 7.3 #29: Find the reference angle for the angle

$\dfrac{2\pi}{3}$ .
Solution: This angle lies in quadrant II and so its reference angle is given by

$\mathrm{reference \hspace{2pt} angle}=\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}.$

Section 7.3 #44: Find the reference angle, the quadrant of the terminal side, and the sine and cosine of

$\dfrac{5\pi}{3}$ .
Solution: The angle

$\dfrac{5\pi}{3}$ lies in quadrant IV. To see this for sure, let us compare to the angle

$\dfrac{3\pi}{2}$ (which is the angle that separates quadrant III and quadrant IV) and to the angle

$2\pi$ . To compare them, we need a common denominator:

$6$ . So rewrite

$\dfrac{5\pi}{3}=\dfrac{10\pi}{6}$ ,

$\dfrac{3\pi}{2}=\dfrac{9\pi}{6}$ , and

$2\pi = \dfrac{12\pi}{6}$ . Sice

$9 < 10 < 12$ , it follows that

$\dfrac{9\pi}{6} < \dfrac{10\pi}{6} < \dfrac{12\pi}{6}$ , in other words,

$\dfrac{3\pi}{2} < \dfrac{5\pi}{3} < 2\pi$ . This shows that

$\dfrac{5\pi}{3}$ is in quadrant IV.

Since it is in quadrant IV, its reference angle is

$2\pi-\dfrac{5\pi}{3}=\dfrac{6\pi}{3}-\dfrac{5\pi}{3}=\dfrac{\pi}{3}$ . From this we know that the sine of

$\dfrac{5\pi}{3}$ must be negative and the cosine of

$\dfrac{5\pi}{3}$ must be positive, hence

$\sin \left( \dfrac{5\pi}{3} \right) = -\sin \left( \dfrac{\pi}{3} \right)=-\dfrac{\sqrt{3}}{2},$ and

$\cos \left( \dfrac{5\pi}{3} \right) = +\cos \left(\dfrac{\pi}{3} \right) = \dfrac{1}{2}.$

Section 7.3 #53: If

$\sin(t)=-\dfrac{1}{4}$ and

$t$ is in the third quadrant, find

$\cos(t)$ .
Solution: Since

$t$ is in the third quadrant, we must conclude that

$\cos(t)$ is negative. Since

$\sin(t)=-\dfrac{1}{4}$ , this means that the

$y$ -coordinate of the point associated with the angle

$t$ is

$-\dfrac{1}{4}$ . To find the cosine of

$t$ means to find the

$x$ -coordinate corresponding to this

$y$ -coordinate. To do that recall that the unit circle obeys the formula

$x^2+y^2=1$ , and we are told that

$y=-\dfrac{1}{4}$ . Plugging that value in yields

$x^2 + \left( -\dfrac{1}{4} \right)^2 = 1.$ In other words,

$x^2+\dfrac{1}{16}=1$ , so

$x = \pm \sqrt{ \dfrac{15}{16} }= \pm\dfrac{\sqrt{15}}{4}$ . But since

$t$ is in quadrant III, we must take the negative solution. Therefore we see that

$\cos(t) = -\dfrac{\sqrt{15}}{4}.$ Section 7.3 #71: In the following picture, find the value of the sine and the cosine at

$t$ :

Solution: Since sine is the

$y$ -coordinate and cosine is the

$x$ -coordinte, we see

$\sin(t)=-0.5966,$ and

$\cos(t)=0.803.$

Section 7.3 #87: Use a calculator to evaluate

$\cos \left( 98^{\circ} \right)$ .
Solution: Calculate (make sure the calculator is in degree mode)

$\cos \left( 98^{\circ} \right) \approx -0.13917.$

Section 7.3 #88: Use a calculator to evaluate

$\cos(310^{\circ})$ .
Solution: Calculate (make sure the calculator is in degree mode)

$\cos(310^{\circ})\approx 0.64278.$