Question

If sin(0) = 112and O is in Quadrant II, then what is sin )?

Answer 1

`\frac{3\sqrt[]{13}}{2}`

1) Considering that we need to use the half-angle formula for that angle:

`\sin ((\theta)/(2))=\pm\sqrt[]{(1-\cos(\theta))/(2)}`

2) But before that, we need to find the cosine of theta using the Fundamental Trigonometric Identity:

`\begin{gathered} \sin ^2(\theta)+\cos ^2(\theta)=1 \\ \cos (\theta)=\sqrt[]{1-\sin ^2(\theta)} \\ \cos (\theta)=\sqrt[]{1-((12)/(13))^2_{}} \\ \cos (\theta)=\sqrt[]{1-(144)/(169)} \\ \cos (\theta)=\pm\sqrt[]{(25)/(169)}=\pm(5)/(13) \end{gathered}`

Note that since the angle is in Quadrant II, then we can state that

the cosine of theta has a negative value: -5/13

2.2) Let's plug into that and find out the value of sine (theta/2):

`\begin{gathered} \sin ((\theta)/(2))=\pm\sqrt[]{(1-\cos(\theta))/(2)} \\ \sin ((\theta)/(2))=\pm\sqrt[]{(1-(-(5)/(13)))/(2)}=\frac{3\sqrt[]{13}}{2} \end{gathered}`

And that is the answer