Answer: [tex]\bold{-\dfrac{7\sqrt2}{10}}[/tex]
Step-by-step explanation:
It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0). Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ [tex]cos\ \dfrac{\theta}{2} <0[/tex]
It is given that sin θ = [tex]-\dfrac{7}{25}[/tex] ⇒ y = -7 & hyp = 25
Use Pythagorean Theorem to find "x":
x² + y² = hyp²
x² + (-7)² = 25²
x² + 49 = 625
x² = 576
x = 24
Use the "x" and "hyp" values to find cos θ:
[tex]cos\ \theta=\dfrac{x}{hyp}=\dfrac{24}{25}[/tex]
Lastly, input cos θ into the half angle formula:
[tex]cos\bigg(\dfrac{\theta}{2}\bigg)=\pm \sqrt{\dfrac{1+cos\ \theta}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{1+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{25}{25}+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{49}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{49}{50}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\\\\\\.\qquad \quad =\pm \dfrac{7\sqrt2}{10}[/tex]
Reminder: We previously determined that the half-angle will be negative.
