Question

If sin theta = 3 sqr of 2 over 5 determine theta over 2

Answer 1

Given:

`\sin \theta=\frac{3\sqrt[]{2}}{5}`
`\begin{gathered} \sin 2\theta=2\sin \theta\cos \theta \\ \sin 2((\theta)/(2))=2\sin ((\theta)/(2))\cos ((\theta)/(2)) \\ \sin \theta=2\sin ((\theta)/(2))\cos ((\theta)/(2)) \\ \frac{3\sqrt[]{2}}{5*2}=\sin ((\theta)/(2))\cos ((\theta)/(2)) \end{gathered}`
`\begin{gathered} \cos ^2\theta=1-\sin ^2\theta \\ \cos ^2\theta=1-(\frac{3\sqrt[]{2}}{5})^2 \\ \cos ^2\theta=1-(18)/(25) \\ \cos ^2\theta=(7)/(25) \end{gathered}`
`\cos \theta=\frac{\sqrt[]{7}}{5}`