Thien D. answered • 07/25/19
Duke MS Engineering Grad For Math Tutoring
1) sin(3θ-45) = - √(3) / (2)
You were able to get 95 and 115, so you should be able to get the rest.
Remember, that there are different coterminal angles that will result in the same value when taking the sine of these angles.
We can derive a formula to find these coterminal angles.
Notice that sin(240°) = sin(300°) = -√(3)/2.
(a) 3θ - 45 = 240 + 360*n, where n is an integer.
Solving for θ: θ = 95 + 120*n. (Stop at n = 2, since θ would be greater than 360 degrees if n = 3)
n | 0 | 1 | 2
----------------------
θ | 95 | 215 | 335
(b) 3θ - 45 = 300 + 360*n, where n is an integer.
Solving for θ: θ = 115 + 120*n
n | 0 | 1 | 2
-------------------------
θ | 115 | 235 | 355
Therefore, the answer is θ = 95, 115, 215, 235, 335, 355 degree.
2) Cos 3θ = -1/2
From the same 30-60-90 triangle, we know that cos(60°) = 1/2. Cosine is negative in Quadrant II (90 to 180 degrees) and in Quadrant III (180 to 270 degrees). We subtract 60 degrees from 180 degrees to move it to Quadrant II, and we add 60 degrees to 180 degrees to move it to Quadrant III. This results in: cos(120°) = cos(240°) = -1/2.
From here, we can work on the coterminal angles, similarly to before.
(a) 3θ = 120 + 360*n, where n is an integer.
Solving for θ: θ = 40 + 120*n. (Stop at n = 2, since θ would be greater than 360 degrees if n = 3)
n | 0 | 1 | 2
-------------------------
θ | 40 | 160 | 280
(b) 3θ = 240 + 360*n, where n is an integer.
Solving for θ: θ = 80 + 120*n
n | 0 | 1 | 2
-------------------------
θ | 80 | 200 | 320
Therefore, the answer is θ = 40, 80, 160, 200, 280, 320 degree
