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.

If cos0 2425 and 0 is in Quadrant I then what is cos02 Simplify your answer completely rationalize the denominator and enter it in fractional form class=

Respuesta :

The given information is:

[tex]\begin{gathered} \cos (\theta)=\frac{24}{25} \\ \theta\text{ is in quadrant I} \end{gathered}[/tex]

cos (theta/2) is given by:

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

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

[tex]\begin{gathered} \cos (\frac{\theta}{2})=\sqrt[]{\frac{1+\frac{24}{25}}{2}} \\ \cos (\frac{\theta}{2})=\sqrt[]{\frac{\frac{25+24}{25}}{2}} \\ \cos (\frac{\theta}{2})=\sqrt[]{\frac{\frac{49}{25}}{2}} \\ \cos (\frac{\theta}{2})=\sqrt[]{\frac{49}{25\times2}} \\ \cos (\frac{\theta}{2})=\sqrt[]{\frac{49}{50}} \\ \cos (\frac{\theta}{2})=\frac{\sqrt[]{49}}{\sqrt[]{50}} \\ \cos (\frac{\theta}{2})=\frac{7}{\sqrt[]{50}} \\ \cos (\frac{\theta}{2})=\frac{7}{\sqrt[]{50}}\cdot\frac{\sqrt[]{50}}{\sqrt[]{50}} \\ \cos (\frac{\theta}{2})=\frac{7\sqrt[]{50}}{50} \\ \cos (\frac{\theta}{2})=\frac{7\sqrt[]{25\times2}}{50} \\ \cos (\frac{\theta}{2})=\frac{7\cdot\sqrt[]{25}\cdot\sqrt[]{2}}{50} \\ \cos (\frac{\theta}{2})=\frac{7\cdot5\cdot\sqrt[]{2}}{50} \\ \text{Simplify 5/50} \\ \cos (\frac{\theta}{2})=\frac{7\sqrt[]{2}}{10} \end{gathered}[/tex]