MATH SOLVE

10 months ago

Q:
# if cosx = 2/3 and x is in quadrant 4 find: sin(x/2), cos(x/2), and tan(x/2) please show steps

Accepted Solution

A:

since we have the cosine of the angle, for the half-angle identities is all we need in this case, since

[tex]\bf \textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \qquad cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\\\ tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \boxed{\pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases}[/tex]

so, let's plug them in, and we know the cos(x) is 2/3

[tex]\bf sin\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1-\frac{2}{3}}{2}}\implies \pm\sqrt{\cfrac{\frac{1}{3}}{2}}\implies \pm\sqrt{\cfrac{1}{6}}\implies \pm \cfrac{\sqrt{1}}{\sqrt{6}} \\\\\\ \textit{and rationalizing that }\pm\cfrac{1}{\sqrt{6}}\cdot \cfrac{\sqrt{6}}{\sqrt{6}}\implies \pm\cfrac{\sqrt{6}}{6}\\\\ -------------------------------[/tex]

[tex]\bf cos\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1+\frac{2}{3}}{2}}\implies \pm\sqrt{\cfrac{\frac{5}{3}}{2}}\implies \pm\sqrt{\cfrac{5}{6}}\implies \pm\cfrac{\sqrt{5}}{\sqrt{6}} \\\\\\ \textit{and rationalizing that }\pm\cfrac{\sqrt{5}}{\sqrt{6}}\cdot \cfrac{\sqrt{6}}{\sqrt{6}}\implies \pm\cfrac{\sqrt{30}}{6}\\\\ -------------------------------[/tex]

[tex]\bf tan\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1-\frac{2}{3}}{1+\frac{2}{3}}}\implies \pm\sqrt{\cfrac{\frac{1}{3}}{\frac{5}{3}}}\implies \pm \sqrt{\cfrac{1}{5}}\implies \stackrel{rationalized}{\pm \cfrac{\sqrt{5}}{5}}[/tex]

[tex]\bf \textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \qquad cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\\\ tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \boxed{\pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases}[/tex]

so, let's plug them in, and we know the cos(x) is 2/3

[tex]\bf sin\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1-\frac{2}{3}}{2}}\implies \pm\sqrt{\cfrac{\frac{1}{3}}{2}}\implies \pm\sqrt{\cfrac{1}{6}}\implies \pm \cfrac{\sqrt{1}}{\sqrt{6}} \\\\\\ \textit{and rationalizing that }\pm\cfrac{1}{\sqrt{6}}\cdot \cfrac{\sqrt{6}}{\sqrt{6}}\implies \pm\cfrac{\sqrt{6}}{6}\\\\ -------------------------------[/tex]

[tex]\bf cos\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1+\frac{2}{3}}{2}}\implies \pm\sqrt{\cfrac{\frac{5}{3}}{2}}\implies \pm\sqrt{\cfrac{5}{6}}\implies \pm\cfrac{\sqrt{5}}{\sqrt{6}} \\\\\\ \textit{and rationalizing that }\pm\cfrac{\sqrt{5}}{\sqrt{6}}\cdot \cfrac{\sqrt{6}}{\sqrt{6}}\implies \pm\cfrac{\sqrt{30}}{6}\\\\ -------------------------------[/tex]

[tex]\bf tan\left( \frac{x}{2} \right)=\pm\sqrt{\cfrac{1-\frac{2}{3}}{1+\frac{2}{3}}}\implies \pm\sqrt{\cfrac{\frac{1}{3}}{\frac{5}{3}}}\implies \pm \sqrt{\cfrac{1}{5}}\implies \stackrel{rationalized}{\pm \cfrac{\sqrt{5}}{5}}[/tex]