question 12 (1 point)\nwhich pair of angles between 0° and 360° satisfies sin θ = -1/√2?\n a) 45° and 0°\n…

question 12 (1 point)\nwhich pair of angles between 0° and 360° satisfies sin θ = -1/√2?\n a) 45° and 0°\n b) 225° and 315°\n c) -45° and 225°\n d) 45° and 135°

question 12 (1 point)\nwhich pair of angles between 0° and 360° satisfies sin θ = -1/√2?\n a) 45° and 0°\n b) 225° and 315°\n c) -45° and 225°\n d) 45° and 135°

Answer

Explanation:

Step1: Recall the sine values of special angles

We know that (\sin45^{\circ}=\frac{1}{\sqrt{2}}), (\sin135^{\circ}=\sin(180 - 45)^{\circ}=\frac{1}{\sqrt{2}}), (\sin225^{\circ}=\sin(180 + 45)^{\circ}=-\frac{1}{\sqrt{2}}), (\sin315^{\circ}=\sin(360 - 45)^{\circ}=-\frac{1}{\sqrt{2}}).

Step2: Check each option

  • Option a: (\sin45^{\circ}=\frac{1}{\sqrt{2}}\neq-\frac{1}{\sqrt{2}}), (\sin0^{\circ} = 0\neq-\frac{1}{\sqrt{2}})
  • Option b: (\sin225^{\circ}=-\frac{1}{\sqrt{2}}), (\sin315^{\circ}=-\frac{1}{\sqrt{2}})
  • Option c: (- 45^{\circ}) is not in the range (0^{\circ}\leq\theta\leq360^{\circ})
  • Option d: (\sin45^{\circ}=\frac{1}{\sqrt{2}}\neq-\frac{1}{\sqrt{2}}), (\sin135^{\circ}=\frac{1}{\sqrt{2}}\neq-\frac{1}{\sqrt{2}})

Answer:

B. (225^{\circ}) and (315^{\circ})