Answer:
(a) [tex]sin 2\theta = -\frac{5\sqrt{11} }{18}[/tex]
(b)[tex]cos 2\theta= -\frac{7}{18}[/tex]
(c)[tex]tan 2\theta=[/tex][tex]\frac{5\sqrt{7} }{11}[/tex]
Step-by-step explanation:
If [tex]sin \theta =\frac{5}{6} , 90\leq \theta\leq 180[/tex]
Using Pythagoras,
Opposite=5, Hypotenuse =6, Adjacent=?
[tex]6^2=5^2+Adj^2\\Adj^2=36-25=11\\Adjacent=\sqrt{11}[/tex]
In the Second Quadrant,
[tex]sin \theta =\frac{5}{6} , cos \theta =-\frac{\sqrt{11} }{6}, Tan \theta =-\frac{5 }{\sqrt{11}}[/tex]
(a) [tex]sin 2\theta=2sin\theta cos\theta=2 X \frac{5}{6} X -\frac{\sqrt{11} }{6} = -\frac{5\sqrt{11} }{18}[/tex]
(b)[tex]cos 2\theta= cos^2\theta-sin^2\theta=(-\frac{\sqrt{11} }{6})^2-(\frac{5}{6})^2= -\frac{7}{18}[/tex]
(c)[tex]tan 2\theta=\frac{2tan\theta}{1-tan^2\theta}[/tex]
[tex]tan 2\theta=\frac{2(-\frac{5 }{\sqrt{11}})}{1-(-\frac{5 }{\sqrt{11}})^2} =\dfrac{-\frac{10 }{\sqrt{11}}}{1-\frac{25}{11}} =\dfrac{-\frac{10 }{\sqrt{11}}}{-\frac{14}{11} }=\frac{5\sqrt{7} }{11}[/tex]
