Question

If cos(0) = -15/17 and 0 is in Quadrant II, then what is sin (0/2)?Give an exact answer, using radicals as needed. Rationalize the denominator and simplify your answer completely.

Answer 1

Given that Θ is in the second quadrant

`\cos \theta=-(15)/(17)`

To Determine:

`\sin ((\theta)/(2))`

Solution:

Using Identity

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

Substitute cos Θ into the formula

`\begin{gathered} \sin ((\theta)/(2))=\sqrt[]{(1-\cos\theta)/(2)} \\ \sin ((\theta)/(2))=\sqrt[]{(1-(-(15)/(17)))/(2)} \\ \sin ((\theta)/(2))=\sqrt[]{(1+(15)/(17))/(2)} \end{gathered}`
`\sin ((\theta)/(2))=\pm\sqrt[]{((17+15)/(17))/(2)}=\pm\sqrt[]{((32)/(17))/(2)}=\pm\sqrt[]{(32)/(17)*(1)/(2)}=\pm\sqrt[]{(16)/(17)}`
`\begin{gathered} \sin ((\theta)/(2))=\frac{\pm4}{\sqrt[]{17}}=\frac{\pm4}{\sqrt[]{17}}*\frac{\sqrt[]{17}}{\sqrt[]{17}} \\ \sin ((\theta)/(2))=\frac{\pm4\sqrt[]{17}}{17} \end{gathered}`

Hence, the final answer is

`\sin ((\theta)/(2))=\frac{\pm4\sqrt[]{17}}{17}`