sin(x)= -3/5, x is in third quadrant
In right angle triangle ,
Opposite side of angle x = -3
Hypotenuse = 5
Now we find out adjacent side of angle x so that we can find out cosx
We use pythagorean theorem
[tex]hypotenuse^2 = opposite ^2 + adjacent^2[/tex]
[tex]5^2 = (-3)^2 + x^2[/tex]
25 = 9 +x^2
x^2 = 16, so x= 4
adjacent side = 4
cos (x) = [tex]\frac{adjacent}{hypotenuse} = \frac{4}{5}[/tex]
Cos is negative in third quadrant so cos(x) = [tex]-\frac{4}{5}[/tex]
Now we use double angle formula
[tex]cos(\frac{x}{2})= \sqrt{\frac{1+cosx}{2}} =\sqrt{\frac{1+\frac{4}{5}}{2}}= \frac{\sqrt{90}}{10}[/tex]
[tex]sin(\frac{x}{2})= \sqrt{\frac{1-cosx}{2}} =\sqrt{\frac{1-\frac{4}{5}}{2}}= \frac{\sqrt{10}}{10}[/tex]
[tex]tan(\frac{x}{2})=\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}[/tex]
[tex]tan(\frac{x}{2})= \frac{\frac{\sqrt{10}}{10}}{\frac{\sqrt{90}}{10} }[/tex]
=[tex]\frac{\sqrt{10}}{\sqrt{90}} =\frac{1}{3}[/tex]
Tan is positive in third quadrant so final answer is [tex]\frac{1}{3}[/tex]
