1.8 E: Trigonometry

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204

Question 140

A research team is modeling the vertical displacement, $H$ , of a specialized underwater sensor. The displacement is given by the function

H(\theta) = \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta

where $θ\theta$ is the tilt angle of the sensor.

Show that the expression for $H(θ)H(\theta)$ can be written as

H(\theta) = \sin \theta + \csc \theta

where $sin⁡θ≠0\sin \theta \neq 0$ and $cos⁡θ≠0\cos \theta \neq 0$ .

[4]

bi.

A technician attempts to find the tilt angles where the displacement is exactly $4.25$ units by solving the equation

\sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta = 4.25

for $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ . They produce the following solution:

Step 1: $sin⁡θ+csc⁡θ=4.25\sin \theta + \csc \theta = 4.25$

Step 2: $sin⁡θ+1sin⁡θ=174\sin \theta + \frac{1}{\sin \theta} = \frac{17}{4}$

Step 3: $4sin⁡2θ−17sin⁡θ+4=04\sin^{2} \theta - 17\sin \theta + 4 = 0$

Step 4: $sin⁡θ=4\sin \theta = 4$ or $sin⁡θ=0.25\sin \theta = 0.25$

Step 5: $θ=14.5∘,165.5∘\theta = 14.5^{\circ}, 165.5^{\circ}$

Explain why the technician should reject the value $sin⁡θ=4\sin \theta = 4$ in Step 4.

[1]

bii.

Determine if there are any other reasons, based on the original expression's domain, why solutions might need to be rejected, and state the final correct solutions for the technician's equation in the range $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ .

[3]

A technician attempts to find the tilt angles where the displacement is exactly $4.25$ units by solving the equation

\sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta = 4.25

for $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ . They produce the following solution:

Step 1: $sin⁡θ+csc⁡θ=4.25\sin \theta + \csc \theta = 4.25$

Step 2: $sin⁡θ+1sin⁡θ=174\sin \theta + \frac{1}{\sin \theta} = \frac{17}{4}$

Step 3: $4sin⁡2θ−17sin⁡θ+4=04\sin^{2} \theta - 17\sin \theta + 4 = 0$

Step 4: $sin⁡θ=4\sin \theta = 4$ or $sin⁡θ=0.25\sin \theta = 0.25$

Step 5: $θ=14.5∘,165.5∘\theta = 14.5^{\circ}, 165.5^{\circ}$

Explain why the technician should reject the value $sin⁡θ=4\sin \theta = 4$ in Step 4.

1.8 E: Trigonometry

Question 140

1.8 E: Trigonometry Questions

1.8 E: Trigonometry

Question 140

1.8 E: Trigonometry Questions