Question

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

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}}

eq-\frac{1}{\sqrt{2}}\), \(\sin0^{\circ} = 0
eq-\frac{1}{\sqrt{2}}\)

• 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}}

eq-\frac{1}{\sqrt{2}}\), \(\sin135^{\circ}=\frac{1}{\sqrt{2}}
eq-\frac{1}{\sqrt{2}}\)

Answer:

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