Question

Use the information given about the angle , to find the exact value of

$\tan \theta =\frac{1}{2},\pi <\theta <\frac{3\pi }{2}$

Short answer

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

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

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

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

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

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

Step-by-step solution

Part (a) Step 1. Value of sin2θ

Angle is in the third quadrant.

So, sine and cosine are negative.

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

we know that,

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

Part(b) step 1. Value of  cos2θ

we know that,

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

Part (c) Step 1. Value of sinθ2

We know that,

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

Part (d) Step 1. Value of cosθ2

We know that,

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

Part (e) Step 1. Value of tan2θ

We 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:

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

We know that,

role="math" localid="1646544830476" $$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}} \\ \tan \frac{\theta }{2}=\frac{\sqrt{\frac{2+\sqrt{5}}{2\sqrt{5}}}}{\sqrt{\frac{\sqrt{5}-2}{2\sqrt{5}}}} \\ \tan \frac{\theta }{2}=\frac{\sqrt{2+\sqrt{5}}}{\sqrt{\sqrt{5}-2}} \end{gathered}$$