Question

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

$\tan \theta =\frac{4}{3},\mathrm{\pi }<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2}$

Short answer

Part (a). $\sin \left(2\theta \right)=\frac{24}{25}$

Part (b). $\cos 2\theta =-\frac{7}{25}$

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

part (d). $\cos \frac{\theta }{2}=\frac{1}{\sqrt{5}}$

part (e). role="math" localid="1646510106260" $\tan 2\theta =-\frac{24}{7}$

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

Step-by-step solution

Part (a) Step 1.  value of sin2θ

Angle is the third quadrant.

$\sin \theta$and $\cos \theta$are negative.

so,

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

We have:

$$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ =2\times \left(-\frac{3}{5}\right)\times \left(-\frac{4}{5}\right) \\ =\frac{24}{25} \end{gathered}$$

Part(b) step 1. Value of cos2θ

we know that,

$$\begin{gathered} \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ ={\left(-\frac{3}{5}\right)}^2-{\left(-\frac{4}{5}\right)}^2 \\ =\frac{9}{25}-\frac{16}{25} \\ =\frac{9-16}{25} \\ =-\frac{7}{25} \end{gathered}$$

part(c) step 1. value of  sinθ2

We know that,

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

part(d) Step 2. value of cosθ2

We know that,

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

Part (e) step 1. value of tan2θ

We know that,

$$\begin{gathered} \sin 2\theta =\frac{24}{25} \\ \cos 2\theta =-\frac{7}{25} \end{gathered}$$

Also,

$$\begin{gathered} \tan \left(2\theta \right)=\frac{\sin 2\theta }{\cos 2\theta } \\ =\frac{{\displaystyle \frac{24}{25}}}{{\displaystyle \frac{-7}{25}}} \\ =\frac{-7}{24} \end{gathered}$$

Part (f) Step 6. value of tanθ2

We know that,

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

Also,

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