Sure! Let's start by understanding the given information:
1. [tex]\(\tan \alpha = \frac{8}{15}\)[/tex]
2. [tex]\(180^\circ < \alpha < 270^\circ\)[/tex] (This implies [tex]\(\alpha\)[/tex] is in the third quadrant.)
In the third quadrant:
- Sine ([tex]\(\sin \alpha\)[/tex]) is negative.
- Cosine ([tex]\(\cos \alpha\)[/tex]) is negative.
- Tangent ([tex]\(\tan \alpha\)[/tex]) is positive (as it is [tex]\( \frac{\text{sin}}{\text{cos}} \)[/tex] and both are negative).
From the given [tex]\(\tan \alpha = \frac{8}{15}\)[/tex]:
a. Finding [tex]\(\sin \frac{\alpha}{2}\)[/tex]
[tex]\( \sin \frac{\alpha}{2} \)[/tex] can be calculated using the half-angle formula:
[tex]\[ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}} \][/tex]
Since [tex]\(180^\circ < \alpha < 270^\circ\)[/tex], [tex]\(\frac{\alpha}{2}\)[/tex] is in the second quadrant (this is because [tex]\(\frac{180^\circ}{2} < \frac{\alpha}{2} < \frac{270^\circ}{2} \Rightarrow 90^\circ < \frac{\alpha}{2} < 135^\circ \)[/tex]). In the second quadrant, sine is positive.
Using the given conditions and calculations, we have:
[tex]\[ \sin \frac{\alpha}{2} = -0.9701425001453319 \][/tex]
b. Finding [tex]\(\cos \frac{\alpha}{2}\)[/tex]
[tex]\( \cos \frac{\alpha}{2} \)[/tex] can be calculated using the half-angle formula:
[tex]\[ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}} \][/tex]
In the second quadrant, cosine is positive.
Using the given conditions and calculations, we have:
[tex]\[ \cos \frac{\alpha}{2} = 0.242535625036333 \][/tex]
c. Finding [tex]\(\tan \frac{\alpha}{2}\)[/tex]
[tex]\( \tan \frac{\alpha}{2} \)[/tex] can be calculated using the relationship:
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \][/tex]
With the previously determined values:
[tex]\[ \tan \frac{\alpha}{2} = -3.9999999999999996 \][/tex]
Therefore, the exact values are:
a. [tex]\( \sin \frac{\alpha}{2} = -0.9701425001453319 \)[/tex]
b. [tex]\( \cos \frac{\alpha}{2} = 0.242535625036333 \)[/tex]
c. [tex]\( \tan \frac{\alpha}{2} = -3.9999999999999996 \)[/tex]
