Question

If cos θ = $\frac{5}{13}$ and 270° < θ < 360°, evaluate sin($\frac{\theta}{2}$) and cos($\frac{\theta}{2}$).

Answer 1

Given that

270° < θ < 360°

Dividing it by 2, we get

270°/2 < θ/2 < 360/2

135° < θ/2 < 180°

So θ/2 lies in quadrant 2.

Sin θ/2 will be positive and Cos θ/2 will be negative.

Kindly see the Attachment for detailed Answer.

Answer 2

Answer:

Step-by-step explanation:

Formula used:

$cosA=1-2{sin}^2(\frac{A}{2})\\\\cosA=2{cos}^2(\frac{A}{2})-1$

Given:

$cos\theta=\frac{5}{13}\\\\1-2{sin}^2(\frac{\theta}{2})=\frac{5}{13}\\\\1-\frac{5}{13}=2{sin}^2(\frac{\theta}{2})$

$\frac{8}{13}=2{sin}^2(\frac{\theta}{2})\\\\\frac{8}{26}={sin}^2(\frac{\theta}{2})$

$\frac{2\sqrt{2}}{\sqrt{13}\sqrt{2} }=sin(\frac{\theta}{2})$

$sin(\frac{\theta}{2})=\frac{2}{\sqrt{13}}$

$[since \frac{\theta}{2} lies\: in \:II quad.]$

$Now,\\cos\theta=\frac{5}{13}\\2{cos}^2(\frac{\theta}{2})-1=\frac{5}{13}\\2{cos}^2(\frac{\theta}{2})=\frac{5}{13}+1\\2{cos}^2(\frac{\theta}{2})=\frac{18}{13}$

${cos}^2(\frac{\theta}{2})=\frac{9}{13}\\cos(\frac{\theta}{2})=-\frac{3}{\sqrt{13}}$