chapter 3 radian measure and the unit circle\nfind the exact value of s in the given interval that has the…

chapter 3 radian measure and the unit circle\nfind the exact value of s in the given interval that has the given circular - function value. see example 4(b).\n67. \\frac{\\pi}{2},\\pi; \\sin s = \\frac{1}{2}\n68. \\frac{\\pi}{2},\\pi; \\cos s=-\\frac{1}{2}\n69. \\pi,\\frac{3\\pi}{2}; \\tan s = \\sqrt{3}\n70. \\pi,\\frac{3\\pi}{2}; \\sin s=-\\frac{1}{2}\n71. \\frac{3\\pi}{2},2\\pi; \\tan s=-1\n72. \\frac{3\\pi}{2},2\\pi; \\cos s = \\frac{\\sqrt{3}}{2}\nfind the exact values of s in the given interval that satisfy the given condition.\n73. 0,2\\pi); \\sin s=-\\frac{\\sqrt{3}}{2}\n74. 0,2\\pi); \\cos s=-\\frac{1}{2}\n75. 0,2\\pi); \\cos^{2}s = \\frac{1}{2}\n76. 0,2\\pi); \\tan^{2}s = 3\n77. -2\\pi,\\pi); 3\\tan^{2}s = 1\n78. -\\pi,\\pi); \\sin^{2}s = \\frac{1}{2}\nsuppose an arc of length s lies on the unit circle x^{2}+y^{2}=1, starting at the point (1,0) and terminating at the point (x,y). (see figure 12, repeated in the margin.) use a calculator to find the approximate coordinates for (x,y) to four decimal places.\n79. s = 2.5\n80. s = 3.4\n81. s=-7.4\n82. s=-3.9\nconcept check for each value of s, use a calculator to find \\sin s and \\cos s, and then use the results to decide in which quadrant an angle of s radians lies.\n83. s = 51\n84. s = 49\n85. s = 65\n86. s = 79\nconcept check each graphing calculator screen shows a point on the unit circle. find the length, to four decimal places, of the shortest arc of the circle from (1,0) to the point.\n87. x^{2}+y^{2}=1\n88. x^{2}+y^{2}=1\nthe unit circle x^{2}+y^{2}=1\nfigure 12 (repeated)

chapter 3 radian measure and the unit circle\nfind the exact value of s in the given interval that has the given circular - function value. see example 4(b).\n67. \\frac{\\pi}{2},\\pi; \\sin s = \\frac{1}{2}\n68. \\frac{\\pi}{2},\\pi; \\cos s=-\\frac{1}{2}\n69. \\pi,\\frac{3\\pi}{2}; \\tan s = \\sqrt{3}\n70. \\pi,\\frac{3\\pi}{2}; \\sin s=-\\frac{1}{2}\n71. \\frac{3\\pi}{2},2\\pi; \\tan s=-1\n72. \\frac{3\\pi}{2},2\\pi; \\cos s = \\frac{\\sqrt{3}}{2}\nfind the exact values of s in the given interval that satisfy the given condition.\n73. 0,2\\pi); \\sin s=-\\frac{\\sqrt{3}}{2}\n74. 0,2\\pi); \\cos s=-\\frac{1}{2}\n75. 0,2\\pi); \\cos^{2}s = \\frac{1}{2}\n76. 0,2\\pi); \\tan^{2}s = 3\n77. -2\\pi,\\pi); 3\\tan^{2}s = 1\n78. -\\pi,\\pi); \\sin^{2}s = \\frac{1}{2}\nsuppose an arc of length s lies on the unit circle x^{2}+y^{2}=1, starting at the point (1,0) and terminating at the point (x,y). (see figure 12, repeated in the margin.) use a calculator to find the approximate coordinates for (x,y) to four decimal places.\n79. s = 2.5\n80. s = 3.4\n81. s=-7.4\n82. s=-3.9\nconcept check for each value of s, use a calculator to find \\sin s and \\cos s, and then use the results to decide in which quadrant an angle of s radians lies.\n83. s = 51\n84. s = 49\n85. s = 65\n86. s = 79\nconcept check each graphing calculator screen shows a point on the unit circle. find the length, to four decimal places, of the shortest arc of the circle from (1,0) to the point.\n87. x^{2}+y^{2}=1\n88. x^{2}+y^{2}=1\nthe unit circle x^{2}+y^{2}=1\nfigure 12 (repeated)

Answer

Explanation:

Step1: Recall unit - circle trigonometry

We know the values of trigonometric functions for standard angles on the unit - circle. For example, $\sin\theta=\frac{y}{r}$, $\cos\theta=\frac{x}{r}$, and $\tan\theta=\frac{y}{x}$ (where $r = 1$ for the unit - circle $x^{2}+y^{2}=1$).

Step2: Solve problem 67

We know that $\sin s=\frac{1}{2}$. The angles for which $\sin\theta=\frac{1}{2}$ are $\theta=\frac{\pi}{6}+2k\pi$ and $\theta=\frac{5\pi}{6}+2k\pi,k\in\mathbb{Z}$. Since $s\in[\frac{\pi}{2},\pi]$, then $s = \frac{5\pi}{6}$.

Step3: Solve problem 68

We know that $\cos s=-\frac{1}{2}$. The angles for which $\cos\theta=-\frac{1}{2}$ are $\theta=\frac{2\pi}{3}+2k\pi$ and $\theta=\frac{4\pi}{3}+2k\pi,k\in\mathbb{Z}$. Since $s\in[\frac{\pi}{2},\pi]$, then $s=\frac{2\pi}{3}$.

Step4: Solve problem 69

We know that $\tan s=\sqrt{3}$. The angles for which $\tan\theta=\sqrt{3}$ are $\theta=\frac{\pi}{3}+k\pi,k\in\mathbb{Z}$. Since $s\in[\pi,\frac{3\pi}{2}]$, then $s=\frac{4\pi}{3}$.

Step5: Solve problem 70

We know that $\sin s=-\frac{1}{2}$. The angles for which $\sin\theta=-\frac{1}{2}$ are $\theta=\frac{7\pi}{6}+2k\pi$ and $\theta=\frac{11\pi}{6}+2k\pi,k\in\mathbb{Z}$. Since $s\in[\pi,\frac{3\pi}{2}]$, then $s=\frac{7\pi}{6}$.

Step6: Solve problem 71

We know that $\tan s=-1$. The angles for which $\tan\theta=-1$ are $\theta = \frac{3\pi}{4}+k\pi,k\in\mathbb{Z}$. Since $s\in[\frac{3\pi}{2},2\pi]$, then $s=\frac{7\pi}{4}$.

Step7: Solve problem 72

We know that $\cos s=\frac{\sqrt{3}}{2}$. The angles for which $\cos\theta=\frac{\sqrt{3}}{2}$ are $\theta=\frac{\pi}{6}+2k\pi$ and $\theta=\frac{11\pi}{6}+2k\pi,k\in\mathbb{Z}$. Since $s\in[\frac{3\pi}{2},2\pi]$, then $s=\frac{11\pi}{6}$.

Step8: Solve problem 73

We know that $\sin s=-\frac{\sqrt{3}}{2}$. The angles for which $\sin\theta=-\frac{\sqrt{3}}{2}$ are $\theta=\frac{4\pi}{3}+2k\pi$ and $\theta=\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}$. Since $s\in[0,2\pi)$, then $s=\frac{4\pi}{3}$ or $s=\frac{5\pi}{3}$.

Step9: Solve problem 74

We know that $\cos s=-\frac{1}{2}$. The angles for which $\cos\theta=-\frac{1}{2}$ are $\theta=\frac{2\pi}{3}+2k\pi$ and $\theta=\frac{4\pi}{3}+2k\pi,k\in\mathbb{Z}$. Since $s\in[0,2\pi)$, then $s=\frac{2\pi}{3}$ or $s=\frac{4\pi}{3}$.

Step10: Solve problem 75

If $\cos^{2}s=\frac{1}{2}$, then $\cos s=\pm\frac{\sqrt{2}}{2}$. When $\cos s=\frac{\sqrt{2}}{2}$, $s=\frac{\pi}{4}+2k\pi$ or $s = \frac{7\pi}{4}+2k\pi$. When $\cos s=-\frac{\sqrt{2}}{2}$, $s=\frac{3\pi}{4}+2k\pi$ or $s=\frac{5\pi}{4}+2k\pi$. Since $s\in[0,2\pi)$, then $s=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$.

Step11: Solve problem 76

If $\tan^{2}s = 3$, then $\tan s=\pm\sqrt{3}$. When $\tan s=\sqrt{3}$, $s=\frac{\pi}{3}+k\pi$. When $\tan s=-\sqrt{3}$, $s=\frac{2\pi}{3}+k\pi$. Since $s\in[0,2\pi)$, then $s=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$.

Step12: Solve problem 77

If $3\tan^{2}s = 1$, then $\tan^{2}s=\frac{1}{3}$, so $\tan s=\pm\frac{\sqrt{3}}{3}$. When $\tan s=\frac{\sqrt{3}}{3}$, $s=\frac{\pi}{6}+k\pi$. When $\tan s=-\frac{\sqrt{3}}{3}$, $s=\frac{5\pi}{6}+k\pi$. Since $s\in[-2\pi,\pi)$, then $s=-\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6}$.

Step13: Solve problem 78

If $\sin^{2}s=\frac{1}{2}$, then $\sin s=\pm\frac{\sqrt{2}}{2}$. When $\sin s=\frac{\sqrt{2}}{2}$, $s=\frac{\pi}{4}+2k\pi$ or $s=\frac{3\pi}{4}+2k\pi$. When $\sin s=-\frac{\sqrt{2}}{2}$, $s=\frac{5\pi}{4}+2k\pi$ or $s=\frac{7\pi}{4}+2k\pi$. Since $s\in[-\pi,\pi)$, then $s=-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}$.

Answer:

  1. $s=\frac{5\pi}{6}$
  2. $s=\frac{2\pi}{3}$
  3. $s=\frac{4\pi}{3}$
  4. $s=\frac{7\pi}{6}$
  5. $s=\frac{7\pi}{4}$
  6. $s=\frac{11\pi}{6}$
  7. $s=\frac{4\pi}{3},\frac{5\pi}{3}$
  8. $s=\frac{2\pi}{3},\frac{4\pi}{3}$
  9. $s=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$
  10. $s=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
  11. $s=-\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6}$
  12. $s=-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}$