Question

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

Short answer

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

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

part (c). $\sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-1}{2\sqrt{10}}}$

part (d). $\cos \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}+1}{2\sqrt{10}}}$

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

part (f).$\tan \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}}$

Step-by-step solution

Part (a) Step 1. Value of sin2θ

Using ,

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

We know that,

role="math" localid="1646682420032" $$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ \sin 2\theta =2\times \frac{-3}{\sqrt{10}}\times \frac{1}{\sqrt{10}} \\ \sin 2\theta =\frac{-6}{10} \\ \sin 2\theta =\frac{-3}{5} \end{gathered}$$

Part (b) step 1. Value of cos2θ

We know that,

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

Part (c) Step 1. Value of sinθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1-{\displaystyle \frac{1}{\sqrt{10}}}}{2}} \\ \sin \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-1}{2\sqrt{10}}} \end{gathered}$$

Part (d) Step 1. Value of cosθ2

We know that,

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1+{\displaystyle \frac{1}{\sqrt{10}}}}{2}} \\ =\sqrt{\frac{\sqrt{10}+1}{2\sqrt{10}}} \end{gathered}$$

Part (e) Step 1. Value of tan2θ

We know that,

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

Part (f) Step 1. Value of tanθ2

We know that,

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}} \\ \tan \frac{\theta }{2}=\frac{\sqrt{\frac{\sqrt{10}-1}{2\sqrt{10}}}}{\sqrt{\frac{\sqrt{10}+1}{2\sqrt{10}}}} \\ \tan \frac{\theta }{2}=\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}} \end{gathered}$$