Question

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

localid="1646481288836" $a.\sin 2\theta b.\cos 2\theta c.\sin \frac{\theta }{2}d.\cos \frac{\theta }{2}e.\tan \left(2\theta \right)f.\tan \frac{\theta }{2}$

localid="1646455477702" $\sin \theta =\frac{3}{5},0<\theta <\frac{\mathrm{\pi }}{2}$

Short answer

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

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

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

Part (d). $\cos \frac{\theta }{2}=\frac{3}{\sqrt{10}}$

Part (e). $\tan 2\theta =\frac{7}{25}$

Part (f).$\tan \frac{\theta }{2}=\frac{24}{7}$

Step-by-step solution

Part (a)  Step 1. Value of sin2θ

We know that, $\sin 2\theta =2\sin \theta \cos \theta$

We know that

localid="1646503586343" $$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ {\cos }^2\theta =1-{\sin }^2\theta \end{gathered}$$

So, substituting $\sin \theta =\frac{3}{5}$, we get:

$$\begin{gathered} {\cos }^2\theta =1-{\left(\frac{3}{5}\right)}^2 \\ \cos \theta =\sqrt{1-\frac{9}{25}} \\ \cos \theta =\sqrt{\frac{16}{25}} \\ \cos \theta =\frac{4}{5} \end{gathered}$$

Substituting localid="1646456605692" $\cos \theta =\frac{4}{5}$in $\sin \left(2\theta \right)$, we get:

$$\begin{gathered} \sin 2\theta =2\times \frac{3}{5}\times \frac{4}{5} \\ =\frac{24}{25} \end{gathered}$$

Part (b) Step 2. Value of cos2θ

We know that,

$$\begin{gathered} {\cos }^{2}2\theta +{\sin }^{2}2\theta =1 \\ \cos 2\theta =\sqrt{1-{\sin }^{2}2\theta } \\ \cos 2\theta =\sqrt{1-{\left(\frac{24}{25}\right)}^2} \\ \cos 2\theta =\sqrt{\frac{625-576}{625}} \\ \cos 2\theta =\sqrt{\frac{49}{625}} \\ \cos 2\theta =\frac{7}{25} \end{gathered}$$

Part (c) Step 1. Value of sinθ2

We know that, $0\le \theta \le \frac{\mathrm{\pi }}{2}$, so

localid="1646483791640" $0\le \frac{\theta }{2}\le \frac{\mathrm{\pi }}{4}$

So,

$\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}$

We know $\cos \theta =\frac{4}{5}$.

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-{\displaystyle \frac{4}{5}}}{2}} \\ =\sqrt{\frac{{\displaystyle \frac{5-4}{5}}}{2}} \\ =\sqrt{\frac{{\displaystyle \frac{1}{5}}}{2}} \\ =\sqrt{\frac{1}{10}} \end{gathered}$$

Part (d) Step 1. Value of cosθ2

We know that,

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

Substituting $\sin \frac{\theta }{2}=\frac{1}{\sqrt{10}}$, we get:

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{1-{\left(\frac{1}{\sqrt{10}}\right)}^2} \\ =\sqrt{1-\frac{1}{10}} \\ =\sqrt{\frac{9}{10}} \\ =\frac{3}{\sqrt{10}} \end{gathered}$$

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,

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

Part (f) Step 1. Value of tanθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\frac{1}{\sqrt{10}} \\ \cos \frac{\theta }{2}=\frac{3}{\sqrt{10}} \end{gathered}$$

Also,

localid="1646495041104" $$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}} \\ \tan \frac{\theta }{2}=\frac{{\displaystyle \frac{1}{\sqrt{10}}}}{{\displaystyle \frac{3}{\sqrt{10}}}} \\ =\frac{1}{\sqrt{10}}\times \frac{\sqrt{10}}{3} \\ =\frac{1}{3} \end{gathered}$$