Step-by-step explanation:
We are given a theta degree measure of sin in a specific interval. We are asked to find half of that degree measure.
The interval it ask us to find it in is between 2 pi and 5 pi over 2.
That is 360 degrees to 450 degrees. If you use reference angles, that is between the 0 degrees and 90 degrees. So that means our measure is going to be in the 1st quadrant.
Replace x with theta
[tex] \sin( \frac{x}{2} ) = + - \sqrt{ \frac{1 - \cos(x) }{2} } [/tex]
We dont know cos x but we can use the pythagorean trig theorem to find cos x.
[tex] \sin {}^{2} (x) + \cos {}^{2} (x) = 1[/tex]
[tex] \sin {}^{2} ( \frac{4}{5} ) + \cos {}^{2} (x) = 1[/tex]
[tex] \frac{16}{25} + \cos {}^{2} (x) = 1[/tex]
[tex] \cos {}^{2} (x) = 1 - \frac{16}{25} [/tex]
[tex] \cos {}^{2} (x) = \frac{9}{25} [/tex]
[tex] \cos(x) = \frac{3}{5} [/tex]
Cosine in 1st quadrant is positive so 3/5 is cos x.
Replace 3/5 for cos x.
[tex] + - \sqrt{ \frac{1 - \frac{3}{5} }{2} } [/tex]
[tex] + - \sqrt{ \frac{2}{10} } [/tex]
The answer is
[tex] + - \sqrt{ \frac{2}{10} } [/tex]
