Answer:
[tex]sin 2 \theta = \frac{4 \sqrt5}{9}[/tex]
Step-by-step explanation:
Identities used :
[tex]1. sin^2 \theta = 1 - cos^2\theta\\\\2. tan^2 \theta + 1 = sec^2 \theta\\\\3.cos \theta = \frac{1}{sec \theta} \\\\4. sin 2 \theta = 2 sin\theta cos\theta[/tex]
[tex]Given \ tan \theta = -\frac{2 \sqrt5}{5}\\[/tex]
[tex]we \ know \ tan ^2 \theta + 1 = sec^2\theta[/tex]
[tex](-\frac{2 \sqrt 5}{5})^2 + 1 = sec^2 \theta\\\\\frac{4 \times 5}{25 } + 1 = sec^2 \theta \\\\\frac{20 + 25}{ 25} = sec^2 \theta \\\\\frac{45}{25} = sec^2 \theta \\\\sec \theta = \sqrt{ \frac{45}{25}}\\\\sec \theta = \sqrt{ \frac{9 \times 5}{5 \times 5}} \\\\sec \theta = \frac{3 \sqrt {5}}{5}\\\\[/tex]
[tex]We \ know \ cos \theta = \frac{1}{ sec \theta}[/tex]
[tex]So , \cos \theta = \frac{5}{3 \sqrt5}[/tex]
[tex]Next \ sin^2 \theta = 1 - cos^2 \theta[/tex]
[tex]= 1 - (\frac{5}{3 \sqrt5})^2\\\\=1 - \frac{25}{9 \times 5}\\\\=1 - \frac{25}{45}\\\\=\frac{45 - 25}{45}\\\\=\frac{20}{45}\\\\=\frac{4}{9}[/tex]
[tex]sin \theta = \sqrt{ \frac{4}{9}} = \frac{2}{3}[/tex]
[tex]Find \ sin 2 \theta[/tex]
[tex]sin 2 \theta = 2 (sin \theta \times cos \theta)[/tex]
[tex]= 2 \times \frac{2}{3} \times \frac{5}{3\sqrt5}\\\\= \frac{4 \times \sqrt5 \times \sqrt5}{9 \times \sqrt 5}\\\\=\frac{4 \sqrt5}{9}[/tex]
