Question

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

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

Short answer

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

part (b). $\cos 2\theta =-\frac{1}{2}$

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

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

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

part (f).$\tan \frac{\theta }{2}=\frac{1}{\sqrt{3}}$

Step-by-step solution

Part (a) step 1. Value of sin2θ

We know that,

$$\begin{gathered} \cos \theta =\frac{1}{sec\theta } \\ \cos \theta =\frac{1}{2} \end{gathered}$$

We know that,

$$\begin{gathered} \sin \theta =\sqrt{1-{\cos }^2\theta } \\ \sin \theta =\sqrt{1-{\left(\frac{1}{2}\right)}^2} \\ \sin \theta =-\frac{\sqrt{3}}{2}[csc<0] \end{gathered}$$

Also,

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

Part (b)Step 1. Value of cos2θ

We know that,

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

Part (c) Step 3. Value of sinθ2

We know that,

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

Part (d) Step 1. Value of cosθ2

We know that,

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

Part (e) Step 1. Value of tan2θ

We have:

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

We know that,

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

Part (f) Step 1. 

We have:

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

We know that,

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