Answer:
[tex]\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }[/tex]
[tex]\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}[/tex]
[tex]\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaystyle \sin(x)=-\frac{21}{29}[/tex]
Where x is in QIII.
First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:
[tex]a=\sqrt{29^2-21^2}=20[/tex]
So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.
Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.
And if x is in QIII, this means that:
[tex]180<x<270[/tex]
So:
[tex]\displaystyle 90 < \frac{x}{2} < 135[/tex]
Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.
1)
Recall that:
[tex]\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}[/tex]
Since x/2 is in QII, this will be positive.
Using the above information, cos(x) is -20/29. Therefore:
[tex]\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}[/tex]
Simplify:
[tex]\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}[/tex]
2)
Likewise:
[tex]\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }[/tex]
Since x/2 is in QII, this will be negative.
Using the above information, cos(x) is -20/29. Therefore:
[tex]\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }[/tex]
Simplify:
[tex]\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}[/tex]
3)
Finally:
[tex]\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}[/tex]
Therefore:
[tex]\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}[/tex]
