Question

Use the information given about the angle $0<\theta <2\mathrm{\pi }$, to find the exact value of $a.\sin 2\theta b.\cos 2\theta c.\sin \frac{\theta }{2}d.\cos \frac{\theta }{2}e.\tan \left(2\theta \right)f.\tan \frac{\theta }{2}$

$sec\theta =3,\sin \theta >0$

Short answer

Part (a). $\sin 2\theta =\frac{4\sqrt{2}}{9}$

Part (b). $\cos 2\theta =-\frac{7}{9}$

part (c). $\sin \frac{\theta }{2}=\frac{1}{\sqrt{3}}$

Part (d). $\cos \frac{\theta }{2}=\sqrt{\frac{2}{3}}$

part (e). $\tan 2\theta =-\frac{4\sqrt{2}}{7}$

part (f).$\tan \frac{\theta }{2}=\frac{1}{\sqrt{2}}$

Step-by-step solution

Part (a) Step 1. Value of sin2θ

We know that,

$$\begin{gathered} sec\theta =\frac{1}{\cos \theta } \\ \cos \theta =\frac{1}{3} \end{gathered}$$

Now,

$$\begin{gathered} \sin \theta =\sqrt{1-{\cos }^2\theta } \\ \sin \theta =\sqrt{1-{\left(\frac{1}{3}\right)}^2} \\ \sin \theta =\sqrt{1-\frac{1}{9}} \\ \sin \theta =\frac{2\sqrt{2}}{3} \end{gathered}$$

Also,

$$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ =2\times \frac{2\sqrt{2}}{3}\times \frac{1}{3} \\ =\frac{4\sqrt{2}}{9} \end{gathered}$$

part (b) Step 1. Value of cos2θ

We know that,

$$\begin{gathered} \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ ={\left(\frac{1}{3}\right)}^2-{\left(\frac{2\sqrt{2}}{3}\right)}^2 \\ =\frac{1}{9}-\frac{8}{9} \\ =-\frac{7}{9} \end{gathered}$$

Part (c) Step 1. Value of sinθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-\left(\frac{1}{3}\right)}{2}} \\ =\frac{1}{\sqrt{3}} \end{gathered}$$

Part (d) Step 1. Value of cosθ2

We know that,

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1+{\displaystyle \frac{1}{3}}}{2}} \\ =\sqrt{\frac{2}{3}} \end{gathered}$$

Part (e) Step 1. value of tan2θ

We have:

$$\begin{gathered} \sin 2\theta =\frac{4\sqrt{2}}{9} \\ \cos 2\theta =-\frac{7}{9} \end{gathered}$$

We know that,

$$\begin{gathered} \tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta } \\ =\frac{{\displaystyle \frac{4\sqrt{2}}{9}}}{-{\displaystyle \frac{7}{9}}} \\ =-\frac{4\sqrt{2}}{7} \end{gathered}$$

Part (f) step 1. Value of tanθ2

We know that,

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos \frac{\theta }{2}} \\ \tan \frac{\theta }{2}=\frac{{\displaystyle \frac{1}{\sqrt{3}}}}{{\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}} \\ \tan \frac{\theta }{2}=\frac{1}{\sqrt{2}} \end{gathered}$$