Question

Use the information given about the angle $0<\theta <2\pi$, to find the exact value of$$\begin{gathered} a.\sin 2\theta b.\cos 2\theta c.\sin \frac{\theta }{2}d.\cos \frac{\theta }{2}e.\tan \left(2\theta \right)f.\tan \frac{\theta }{2} \\ cot\theta =3,\cos \theta <0 \end{gathered}$$

Short answer

$\cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-3}{2\sqrt{10}}}$

Part (a) . $\sin 2\theta =\frac{3}{5}$

part (b). $\cos 2\theta =\frac{4}{5}$

part (c). $\sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}+3}{2\sqrt{10}}}$

part (d). $\cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-3}{2\sqrt{10}}}$

part (e). $\tan 2\theta =\frac{3}{4}$

part (f).$\tan \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}+3}{\sqrt{10}-3}}$

Step-by-step solution

part (a) Step 1. value of sin2θ

We know that,

$$\begin{gathered} \tan \left(\theta \right)=\frac{1}{cot\theta } \\ \tan \theta =\frac{1}{3} \end{gathered}$$

And,

$$\begin{gathered} \sin \theta =-\frac{1}{\sqrt{10}} \\ \cos \theta =-\frac{3}{\sqrt{10}} \end{gathered}$$

We know that,

$$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ =2·\frac{-1}{\sqrt{10}}·\frac{-3}{\sqrt{10}} \\ =\frac{6}{10} \\ =\frac{3}{5} \end{gathered}$$

Part (b) Step 1. Value of cos2θ

We know that,

$$\begin{gathered} \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ =={\left(-\frac{3}{\sqrt{10}}\right)}^2-{\left(-\frac{1}{\sqrt{10}}\right)}^2 \\ =\frac{4}{5} \end{gathered}$$

Part (c) step 1. value of sinθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-\left(-\frac{3}{\sqrt{10}}\right)}{2}} \\ =\sqrt{\frac{\sqrt{10}+3}{2\sqrt{10}}} \end{gathered}$$

part (d) step 1.  value of cosθ2

We know that,

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1+\left(-\frac{3}{\sqrt{10}}\right)}{2}} \\ \cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-3}{2\sqrt{10}}} \end{gathered}$$

Part (a) Step 1. Value of tan2θ

We know that,

$$\begin{gathered} \tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta } \\ =\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}} \\ =\frac{3}{4} \end{gathered}$$

Part (f) step 1. Value of tanθ2

We know that,

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}} \\ =\frac{\sqrt{\frac{\sqrt{10}+3}{2\sqrt{10}}}}{\sqrt{\frac{\sqrt{10}-3}{2\sqrt{10}}}} \\ =\sqrt{\frac{\sqrt{10}+3}{\sqrt{10}-3}} \end{gathered}$$