If we want to solve this problem we first need to list a few properties of trigonometric functions:
[tex]\begin{gathered} \text{cot }\theta=\frac{\cos\theta}{\sin\theta} \\ \sin^2\theta+\cos^2\theta=1 \end{gathered}[/tex]We are told that cot(θ)=1/2. Using the first equation and this data we obtain the following:
[tex]\frac{1}{2}=\frac{\cos\theta}{\sin\theta}[/tex]We multiply both sides and we get an expression for the cosine of θ:
[tex]\begin{gathered} \frac{1}{2}\sin\theta=\frac{\cos\theta}{\sin\theta}\cdot\sin\theta \\ \cos\theta=\frac{1}{2}\sin\theta \end{gathered}[/tex]Now we are going to take the second property I wrote in the begining and replace the cosine of θ with this new expression that we found:
[tex]\begin{gathered} \sin^2\theta+\cos^2\theta=\sin^2\theta+(\frac{1}{2}\sin\theta)^2=1 \\ \sin^2\theta+\frac{1}{4}\sin^2\theta=1 \\ \frac{5}{4}\sin^2\theta=1 \end{gathered}[/tex]We must solve this equation for the sine of θ. We can multiply both sides by 4/5:
[tex]\begin{gathered} \frac{4}{5}\cdot\frac{5}{4}\sin^2\theta=1\cdot\frac{4}{5} \\ \sin^2\theta=\frac{4}{5} \end{gathered}[/tex]And we apply a square root to both sides:
[tex]\begin{gathered} \sqrt{\sin^2\theta}=\sqrt{\frac{4}{5}} \\ |\sin\theta|=\frac{2}{\sqrt{5}} \end{gathered}[/tex]We are told that θ is located in quadrant I which means that its sine is positive. Therefore we get:
[tex]\sin\theta=\frac{2}{\sqrt{5}}[/tex]AnswerThen the answer is 2/√5
