Math

Use the information given about the angle θ, 0 ≤ θ ≤ 2π, to find the exact value of the indicated trigonometric function. sin \theta=-(\sqrt(5))/(5),(3\pi)/(2)< \theta < 2\pi Find sin (\theta)/(2).

$0 ≤ θ ≤ 2π\\ sin \theta=-\frac{\sqrt5}{5},\qquad \frac{3\pi}{2}< \theta < 2\pi \qquad Find \;\;\frac{sin \theta}{2}\\$

Ihis is what you have asked....

Is this your intended question?  - I guess not  :(

I guess underneath here  is what you really meant...

$0 ≤ θ ≤ 2π\\ sin \theta=-\frac{\sqrt5}{5},\qquad \frac{3\pi}{2}< \theta < 2\pi \qquad Find \;\;sin\frac{ \theta}{2}\\ sin(2*\frac{\theta}{2})=2sin\frac{\theta}{2}cos\frac{\theta}{2}\\ sin^2(\theta)=4sin^2\frac{\theta}{2}cos^2\frac{\theta}{2}\\ sin^2(\theta)=4sin^2\frac{\theta}{2}(1-sin^2\frac{\theta}{2})\\ \frac{5}{25}=4sin^2\frac{\theta}{2}-4sin^4\frac{\theta}{2}\\ \frac{1}{5}=4sin^2\frac{\theta}{2}-4sin^4\frac{\theta}{2}\\ 1=20sin^2\frac{\theta}{2}-20sin^4\frac{\theta}{2}\\ 20sin^4\frac{\theta}{2}-20sin^2\frac{\theta}{2}+1=0\\ let\;\;y=sin^2\frac{\theta}{2}\\ 20y^2-20y+1=0\\ y=\frac{20\pm \sqrt{400-80}}{40}\\ y=\frac{20\pm \sqrt{320}}{40}\\ y=\frac{5\pm 2\sqrt{5}}{10}\\$

$\text{y must be positive so}\\ y=\frac{5+\sqrt5}{10}\\ sin^2{\frac{\theta}{2}}=\frac{5+\sqrt5}{10}\\ sin{\frac{\theta}{2}}=\sqrt{\frac{5+\sqrt5}{10}}\\~\\ \text{I should have mentioned earlier but since }\theta \text{ is in the 4th quadrant,}\\ \frac{\theta}{2}\;\text{is int the second quadrant so }sin\frac{\theta}{2}>0$

Assuming that I did not make any silly mistakes (big assumption here)

$sin{\frac{\theta}{2}}=\sqrt{\frac{5+\sqrt5}{10}}\\~\\$

Just the two silly mistakes Melody.