Respuesta :
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]
The values are √1/6, √5/6 and √1 / √5.
What is Half- angle identities?
Half-angles in half angle formulas are usually denoted by θ/2, x/2, A/2, etc and the half-angle is a sub-multiple angle. The half angle formulas are used to find the exact values of the trigonometric ratios of the angles like 22.5
popular half angle identities:
- Half angle formula of sin: sin A/2 = ±√[(1 - cos A) / 2]
- Half angle formula of cos: cos A/2 = ±√[(1 + cos A) / 2]
- Half angle formula of tan: tan A/2 = ±√[1 - cos A] / [1 + cos A] (or) sin A / (1 + cos A) (or) (1 - cos A) / sin A
For example
- sin A = 2 sin(A/2) cos(A/2)
- cos A = cos2 (A/2) - sin2 (A/2) (or)
= 1 - 2 sin2 (A/2) (or)
= 2 cos2(A/2) - 1
- tan A = 2 tan (A/2) / (1 - tan2(A/2))
Given:
cos x = 2/3
we know,
sin( [tex]\theta[/tex]/2)= √(1- cos [tex]\theta[/tex] )/2 and cos( [tex]\theta[/tex]/2)= √(1 + cos [tex]\theta[/tex] )/2
So,
sin( x/2)= √(1- 2/3 )/2
= √1/6
cos(x/2)= √(1+2/3 )/2
cos(x/2)= √5/6
and, tan(x/2) = sin (x/2) / cos (x/2)
tan(x/2) = √1/6 / √5/6
tan(x/2) = √1 / √5
Learn more about this concept here:
https://brainly.com/question/2254104
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