Question

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

$$\begin{gathered} a.\sin 2\theta b.\cos 2\theta c.\sin \frac{\theta }{2}d.\cos \frac{\theta }{2}e.\tan \left(2\theta \right)f.\tan \frac{\theta }{2} \\ csc\theta =-\sqrt{5},\cos \theta <0 \end{gathered}$$

Short answer

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

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

Part (c). $\sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}}$

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,

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

Now,

role="math" localid="1646566268717" $$\begin{gathered} \cos \theta =\sqrt{1-{\sin }^2\theta } \\ =\sqrt{1-{\left(-\frac{1}{\sqrt{5}}\right)}^2} \\ =\sqrt{1-\frac{1}{5}} \\ =\sqrt{\frac{4}{5}} \\ =-\frac{2}{\sqrt{5}}[\cos \theta <0] \end{gathered}$$

Also

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

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 3. Value of sinθ2

We know that,

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

Part (d) Step 1. Value of cosθ2

We know that,

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

Part (e) Step 1. Value of  tan2θ

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 know that,

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