Answer:
Solution given
Cos[tex]\displaystyle \theta_{1}=\frac{13}{15}[/tex]
consider Pythagorean theorem
[tex]\bold{Sin²\theta+Cos²\theta=1}[/tex]
Subtracting [tex]Cos²\theta[/tex]both side
[tex]\displaystyle Sin²\theta=1-Cos²\theta[/tex]
doing square root on both side we get
[tex]Sin\theta=\sqrt{1-Cos²\theta}[/tex]
Similarly
[tex]Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}[/tex]
Substituting value of [tex]Cos\theta_{1}[/tex]
we get
[tex]Sin\theta_{1}=\sqrt{1-(\frac{-13}{15})²}[/tex]
Solving numerical
[tex]Sin\theta_{1}=\sqrt{1-(\frac{169}{225})}[/tex]
[tex]Sin\theta_{1}=\sqrt{\frac{56}{225}}[/tex]
[tex]Sin\theta_{1}=\frac{\sqrt{56}}{\sqrt{225}}[/tex]
[tex]Sin\theta_{1}=\frac{\sqrt{2*2*14}}{\sqrt{15*15}}[/tex]
[tex]Sin\theta_{1}=\frac{2\sqrt{14}}{15}[/tex]
Since
In III quadrant sin angle is negative
[tex]\bold{Sin\theta_{1}=-\frac{2\sqrt{14}}{15}}[/tex]
