Question

Use the information given about the angle $0<\theta <2\mathrm{\pi }$, to find the exact value of

$$\begin{gathered} a.\sin 2\theta b.\cos 2\theta c.\sin \frac{\theta }{2}d.\cos \frac{\theta }{2}e.\tan \left(2\theta \right)f.\tan \frac{\theta }{2} \\ cot\theta =-2,sec\theta <0 \end{gathered}$$

Short answer

Part (a). $\sin 2\theta =\frac{-4}{5}$

part (b). $\cos 2\theta =\frac{3}{5}$

part (c). role="math" localid="1646651000338" $\sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}}$

part(d). $\cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}-2}{2\sqrt{5}}}$

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

part (f).$\tan \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}+2}{\sqrt{5}-2}}$

Step-by-step solution

Part (a) Step 1. Value of sin2θ

We know that,

$$\begin{gathered} \tan \theta =\frac{1}{cot\theta } \\ \tan \theta =\frac{-1}{2} \end{gathered}$$

$\theta$is in the second quadrant.

So,

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

Also,

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

part (b) Step 1. Value of cos2θ

We know that,

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

Part (c) Step 1. Value of sinθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-\left(-{\displaystyle \frac{2}{\sqrt{5}}}\right)}{2}} \\ =\sqrt{\frac{1+{\displaystyle \frac{2}{\sqrt{5}}}}{2}} \\ =\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}} \end{gathered}$$

Part (d) Step 1. Value of cosθ2

We know that,

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1+\left(-{\displaystyle \frac{\sqrt{2}}{5}}\right)}{2}} \\ =\sqrt{\frac{\sqrt{5}-2}{2\sqrt{5}}} \end{gathered}$$

Part (e) Part 1. Value of tan2θ

We have :

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

We know that,

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

Part (f) Step 1. Value of tanθ2

We have:

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}} \\ \cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}-2}{2\sqrt{5}}} \end{gathered}$$

We know that,

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}} \\ =\frac{{\displaystyle \frac{\sqrt{\sqrt{5}+2}}{2\sqrt{5}}}}{\frac{\sqrt{\sqrt{5}-2}}{2\sqrt{5}}} \\ =\frac{\sqrt{5}+2}{\sqrt{5}-2} \end{gathered}$$