Question

If cos(0) = 24/25, and 0 is in Quadrant I, then what is cos(0/2)? Simplify your answer completely, rationalize the denominator, and enter it in fractional form.

Answer 1

The given information is:

`\begin{gathered} \cos (\theta)=(24)/(25) \\ \theta\text{ is in quadrant I} \end{gathered}`

cos (theta/2) is given by:

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

In Quadrant I, cos (theta) is positive, then the answer is positive. By replacing the known values:

`\begin{gathered} \cos ((\theta)/(2))=\sqrt[]{(1+(24)/(25))/(2)} \\ \cos ((\theta)/(2))=\sqrt[]{((25+24)/(25))/(2)} \\ \cos ((\theta)/(2))=\sqrt[]{((49)/(25))/(2)} \\ \cos ((\theta)/(2))=\sqrt[]{(49)/(25*2)} \\ \cos ((\theta)/(2))=\sqrt[]{(49)/(50)} \\ \cos ((\theta)/(2))=\frac{\sqrt[]{49}}{\sqrt[]{50}} \\ \cos ((\theta)/(2))=\frac{7}{\sqrt[]{50}} \\ \cos ((\theta)/(2))=\frac{7}{\sqrt[]{50}}\cdot\frac{\sqrt[]{50}}{\sqrt[]{50}} \\ \cos ((\theta)/(2))=\frac{7\sqrt[]{50}}{50} \\ \cos ((\theta)/(2))=\frac{7\sqrt[]{25*2}}{50} \\ \cos ((\theta)/(2))=\frac{7\cdot\sqrt[]{25}\cdot\sqrt[]{2}}{50} \\ \cos ((\theta)/(2))=\frac{7\cdot5\cdot\sqrt[]{2}}{50} \\ \text{Simplify 5/50} \\ \cos ((\theta)/(2))=\frac{7\sqrt[]{2}}{10} \end{gathered}`