Question 4 Exercise 8.2

Solutions of Question 4 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ (c) $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\cos \theta=\frac{3}{5}$ where $0<\theta<\frac{\pi}{2}$

Solution.

Given: $\cos\theta=\dfrac{3}{5}$ where $0<\theta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI.

We have $$\sin\theta = \pm \sqrt{1-\cos^2}.$$ Since $\theta$ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}}\\ &= \sqrt{\frac{16}{25}} = \frac{4}{5} \end{align*}

(a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align*}

(b) $\cos 2 \theta$ \begin{align*} \cos 2\theta & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{7}{25}}. \end{align*}

(c) $\tan 2 \theta$ \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta}\\ &= \frac{24/25}{-7/25}\\ \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{24}{7}}. \end{align*}

(d) $\sin \dfrac{\theta}{2}$

We have $$\sin\left(\frac{\theta}{2} \right) = \pm \sqrt{\frac{1-\cos\theta}{2}}$$

As $0<\theta<\dfrac{\pi}{2}$ implies $0<\dfrac{\theta}{2}<\dfrac{\pi}{4}$, that is, $\dfrac{\theta}{2}$ lies in QI, thus $\sin\left(\dfrac{\theta}{2} \right)>0$.

\begin{align*} \sin\left(\frac{\theta}{2} \right) & = \sqrt{\frac{1-\cos\theta}{2}} \\ &= \sqrt{\frac{1-\frac{3}{5}}{2}} \\ &= \sqrt{\frac{\frac{2}{5}}{2}} \\ \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2} \right) = \frac{1}{\sqrt{5}}} \end{align*}

(e) $\cos \dfrac{\theta}{2}$

We have $$\cos\left(\frac{\theta}{2} \right) = \pm \sqrt{\frac{1+\cos\theta}{2}}$$ As $\dfrac{\theta}{2}$ lies in QI, thus $\cos\left(\dfrac{\theta}{2} \right)>0$. \begin{align*} \cos\left(\frac{\theta}{2} \right) & = \sqrt{\frac{1+\cos\theta}{2}} \\ &= \sqrt{\frac{1+\frac{3}{5}}{2}} \\ &= \sqrt{\frac{\frac{8}{5}}{2}} \\ \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2} \right) = \frac{2}{\sqrt{5}}} \end{align*}

(f) $\tan\dfrac{\theta}{2}$

As \begin{align*} \tan\dfrac{\theta}{2} & = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \\ &= \frac{1/sqrt{5}}{2/sqrt{5}} \end{align*} \begin{align*} \implies \boxed{\tan\dfrac{\theta}{2} = \frac{1}{2}} \end{align*}

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ (c) $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\tan \theta=\frac{12}{5}$ where $\pi<\theta<\frac{3 \pi}{2}$

Solution.

Given: \(\tan \theta = \frac{12}{5}\) where \(\pi < \theta < \frac{3\pi}{2}\), i.e., \(\theta\) lies in QIII.

We have:

\begin{align*} \sec \theta &= \pm \sqrt{1+tan^2 \theta}\\ &= \pm \sqrt{1+(\frac{12}{5})^2 }\\ &=\pm \frac{13}{5}\end{align*} \(\theta\) lies in QIII, therefore $\sec \theta <o$
\begin{align*}\sec \theta &=-\frac{13}{5}\\ \cos \theta &=-\frac{5}{13}\end{align*} and

\[\sin \theta = -\sqrt{(1-cos^2 \theta)} = -\sqrt{(1-\frac{25}{169}) } = -\frac{12}{13}\]

(a) \(\sin 2 \theta\) \begin{align*} \sin 2\theta & = 2 \sin \theta \cos \theta \\ & = 2 \left(-\frac{12}{13}\right) \left(-\frac{5}{13}\right) \\ & = \frac{120}{169} \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{120}{169}}. \end{align*}

(b) \(\cos 2 \theta\) \begin{align*} \cos 2\theta & = 1 - 2 \sin^2 \theta \\ & = 1 - 2 \left(-\frac{12}{13}\right)^2 \\ & = 1 - \frac{288}{169} \\ & = -\frac{119}{169} \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{119}{169}}. \end{align*}

(c) \(\tan 2 \theta\) \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta} \\ & = \frac{\frac{120}{169}}{-\frac{119}{169}} \\ & = -\frac{120}{119} \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{120}{119}}. \end{align*}

(d) \(\sin \frac{\theta}{2}\)

We have: \[\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}.\] As \(\pi < \theta < \frac{3\pi}{2}\) implies \(\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}\), that is, \(\frac{\theta}{2}\) lies in QII, thus \(\sin\left(\frac{\theta}{2}\right) > 0\). \begin{align*} \sin\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 - \left(-\frac{5}{13}\right)}{2}} \\ & = \sqrt{\frac{1 + \frac{5}{13}}{2}} \\ & = \sqrt{\frac{18}{26}} \\ & = \frac{3}{\sqrt{13}}. \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2}\right) = \frac{3}{\sqrt{13}}}. \end{align*}

(e) \(\cos \frac{\theta}{2}\)

We have: \[\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}.\] As \(\frac{\theta}{2}\) lies in QII, thus \(\cos\left(\frac{\theta}{2}\right) < 0\).

\begin{align*} \cos\left(\frac{\theta}{2}\right) & = -\sqrt{\frac{1 + \left(-\frac{5}{13}\right)}{2}} \\ & = -\sqrt{\frac{1 - \frac{5}{13}}{2}} \\ & = -\sqrt{\frac{8}{26}} \\ & = -\frac{2}{\sqrt{13}} \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2}\right) = -\frac{2}{\sqrt{13}}} \end{align*}

(f) \(\tan \frac{\theta}{2}\)

\begin{align*} \tan \frac{\theta}{2} & = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \\ & = \frac{\frac{3}{\sqrt{13}}}{-\frac{2}{\sqrt{13}}} \\ & = -\frac{3}{2}.\end{align*} \begin{align*} \implies \boxed{\tan \frac{\theta}{2} = -\frac{3}{2}}. \end{align*}

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ (c) $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\sin \theta=-\frac{7}{25}$ where $\frac{3 \pi}{2}<\theta<2 \pi$

Solution.

Given: \(\sin \theta = -\frac{7}{25}\) where \(\frac{3\pi}{2} < \theta < 2\pi\), i.e., \(\theta\) lies in QIV. \begin{align*} \cos \theta &= \sqrt{1 - \sin^2 \theta}\\ \implies \cos \theta &= \sqrt{1 - \left(-\frac{7}{25}\right)^2}\\ & =\sqrt {1 - \frac{49}{625}} = \sqrt{\frac{625 - 49}{625}}\\ &= \sqrt{\frac{576}{625}} = \frac{24}{25}\end{align*}

(a) \(\sin 2\theta\) \begin{align*} \sin 2\theta & = 2 \sin \theta \cos \theta \\ & = 2 \left(-\frac{7}{25}\right) \left(\frac{24}{25}\right) \\ & = -\frac{336}{625}. \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = -\frac{336}{625}}. \end{align*}

(b) \(\cos 2\theta\) \begin{align*} \cos 2\theta & = 1 - 2\sin^2 \theta \\ & = 1 - 2\left(-\frac{7}{25}\right)^2 \\ & = 1 - \frac{98}{625} \\ & = \frac{527}{625}. \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = \frac{527}{625}}. \end{align*}

(c) \(\tan 2\theta\) \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta} \\ & = \frac{-\frac{336}{625}}{\frac{527}{625}} \\ & = -\frac{336}{527}. \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{336}{527}}. \end{align*}

(d) \(\sin \frac{\theta}{2}\)

We have: \[\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}.\]

As \(\frac{3\pi}{2} < \theta < 2\pi\) implies \(\frac{\pi}{2} < \frac{\theta}{2} < \pi\), that is, \(\frac{\theta}{2}\) lies in QII, thus \(\sin\left(\frac{\theta}{2}\right) > 0\).

\begin{align*} \sin\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 - \frac{24}{25}}{2}} \\ & = \sqrt{\frac{\frac{1}{25}}{2}} \\ & = \frac{1}{\sqrt{50}} \\ & = \frac{1}{5\sqrt{2}}. \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2}\right) = \frac{1}{5\sqrt{2}}}. \end{align*}

(e) \(\cos \frac{\theta}{2}\)

We have: \[\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}.\]

As \(\frac{\theta}{2}\) lies in QII, thus \(\cos\left(\frac{\theta}{2}\right) < 0\). \begin{align*} \cos\left(\frac{\theta}{2}\right) & = -\sqrt{\frac{1 + \frac{24}{25}}{2}} \\ & = -\sqrt{\frac{49}{50}} \\ & = -\frac{7}{\sqrt{50}} \\ & = -\frac{7}{5\sqrt{2}}. \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2}\right) = -\frac{7}{5\sqrt{2}}}. \end{align*}

(f) \(\tan \frac{\theta}{2}\)

As \begin{align*} \tan \frac{\theta}{2} & = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \\ & = \frac{\frac{1}{5\sqrt{2}}}{-\frac{7}{5\sqrt{2}}} \\ & = -\frac{1}{7} \end{align*} \begin{align*} \implies \boxed{\tan \frac{\theta}{2} = -\frac{1}{7}}. \end{align*}

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ (c) $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\sec \theta=\sqrt{5}$ where $\frac{3 \pi}{2}<\theta<2 \pi$

Solution.

Given: \(\sec \theta = \sqrt{5}\) where \(\frac{3\pi}{2} < \theta < 2\pi\), i.e., \(\theta\) lies in QIV. \begin{align*} \cos \theta &= \frac{1}{\sec \theta} \\ \implies \cos \theta &= \frac{1}{\sqrt{5}}\end{align*} \begin{align*} \sin \theta & = \sqrt{1 - \cos^2 \theta} \\ & = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} \\ & = \sqrt{1 - \frac{1}{5}} \\ & = \sqrt{\frac{4}{5}}\\ & = \frac{2}{\sqrt{5}} \end{align*} Since \(\theta\) lies in QIV, where sine is negative, we have:

\[\sin \theta = -\frac{2}{\sqrt{5}}\]

(a) \(\sin 2\theta\) \begin{align*} \sin 2\theta & = 2 \sin \theta \cos \theta \\ & = 2 \left(-\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{5}}\right) \\ & = -\frac{4}{5}. \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = -\frac{4}{5}}. \end{align*}

(b) \(\cos 2\theta\) \begin{align*} \cos 2\theta & = 2\cos^2 \theta - 1 \\ & = 2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 \\ & = 2\left(\frac{1}{5}\right) - 1 \\ & = -\frac{3}{5}. \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{3}{5}}. \end{align*}

(c) \(\tan 2\theta\) \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta} \\ & = \frac{-\frac{4}{5}}{-\frac{3}{5}} \\ & = \frac{4}{3}. \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = \frac{4}{3}}. \end{align*}

(d) \(\sin \frac{\theta}{2}\)

We have:

\[\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}\]

As \(\frac{3\pi}{2} < \theta < 2\pi\) implies \(\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}\), that is, \(\frac{\theta}{2}\) lies in QI, thus \(\sin\left(\frac{\theta}{2}\right) > 0\).

\begin{align*} \sin\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 - \frac{1}{\sqrt{5}}}{2}} \\ & = \sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}}. \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}}}. \end{align*}

(e) \(\cos \frac{\theta}{2}\)

We have:

\[\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}\]

As \(\frac{\theta}{2}\) lies in QI, thus \(\cos\left(\frac{\theta}{2}\right) > 0\). \begin{align*} \cos\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 + \frac{1}{\sqrt{5}}}{2}} \\ & = \sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}}. \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}}}. \end{align*}

(f) \(\tan \frac{\theta}{2}\)

As \begin{align*} \tan \frac{\theta}{2} & = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \\ & = \frac{\sqrt{\frac{\sqrt{5} - 1}{2\sqrt{5}}}}{\sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}}} \\ & = \sqrt{\frac{\sqrt{5} - 1}{\sqrt{5} + 1}}. \end{align*} \begin{align*} \implies \boxed{\tan \frac{\theta}{2} = \sqrt{\frac{\sqrt{5} - 1}{\sqrt{5} + 1}}}. \end{align*}

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ © $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\csc \theta=4$ where $\frac{\pi}{2}<\theta<\pi$

Solution.

Given: \(\csc \theta = 4\) where \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\) lies in QII. \begin{align*}\sin \theta &= \frac{1}{\csc \theta}\\ & = \frac{1}{4}\end{align*} \begin{align*} \cos \theta & =\sqrt {1 - \sin^2 \theta} \\ & = \sqrt{1 - \left(\frac{1}{4}\right)^2 }\\ & = \sqrt{1 - \frac{1}{16}} \\ & = \frac{15}{16}\\ &=\frac{\sqrt{15}}{4} \end{align*} Since \(\theta\) lies in QII, where cosine is negative, we have: $$\cos \theta = -\frac{\sqrt{15}}{4}$$

(a) \(\sin 2\theta\) \begin{align*} \sin 2\theta & = 2 \sin \theta \cos \theta \\ & = 2 \left(\frac{1}{4}\right) \left(-\frac{\sqrt{15}}{4}\right) \\ & = -\frac{\sqrt{15}}{8}. \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = -\frac{\sqrt{15}}{8}}. \end{align*}

(b) \(\cos 2\theta\) \begin{align*} \cos 2\theta & = 2\cos^2 \theta - 1 \\ & = 2\left(-\frac{\sqrt{15}}{4}\right)^2 - 1 \\ & = 2\left(\frac{15}{16}\right) - 1 \\ & = \frac{30}{16} - 1 \\ & = \frac{30}{16} - \frac{16}{16} \\ & = \frac{14}{16} \\ & = \frac{7}{8}. \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = \frac{7}{8}}. \end{align*}

(c) \(\tan 2\theta\) \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta} \\ & = \frac{-\frac{\sqrt{15}}{8}}{\frac{7}{8}} \\ & = -\frac{\sqrt{15}}{7}. \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = -\frac{\sqrt{15}}{7}}. \end{align*}

(d) \(\sin \frac{\theta}{2}\)

We have:

\[\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}.\]

As \(\frac{\pi}{2} < \theta < \pi\) implies \(0 < \frac{\theta}{2} < \frac{\pi}{2}\), that is, \(\frac{\theta}{2}\) lies in QI, thus \(\sin\left(\frac{\theta}{2}\right) > 0\). \begin{align*} \sin\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 - \left(-\frac{\sqrt{15}}{4}\right)}{2}} \\ & = \sqrt{\frac{1 + \frac{\sqrt{15}}{4}}{2}} \\ & = \sqrt{\frac{\frac{4}{4} + \frac{\sqrt{15}}{4}}{2}} \\ & = \sqrt{\frac{4 + \sqrt{15}}{8}}. \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}}}. \end{align*}

(e) \(\cos \frac{\theta}{2}\)

We have:

\[\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}\]

As \(\frac{\theta}{2}\) lies in QI, thus \(\cos\left(\frac{\theta}{2}\right) > 0\). \begin{align*} \cos\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 + \left(-\frac{\sqrt{15}}{4}\right)}{2}} \\ & = \sqrt{\frac{1 - \frac{\sqrt{15}}{4}}{2}} \\ & = \sqrt{\frac{\frac{4}{4} - \frac{\sqrt{15}}{4}}{2}} \\ & = \sqrt{\frac{4 - \sqrt{15}}{8}}. \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}}}. \end{align*}

(f) \(\tan \frac{\theta}{2}\)

As \begin{align*} \tan \frac{\theta}{2} & = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \\ & = \frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} \\ & = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}. \end{align*} \begin{align*} \implies \boxed{\tan \frac{\theta}{2} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}}. \end{align*}

Find (a) $\sin 2 \theta$ (b) $\cos 2 \theta$ (c) $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\cot \theta=-\frac{1}{2}$ where $\frac{\pi}{2}<\theta<\pi$

Solution.

Given: \(\cos \theta = \dfrac{3}{5}\) where \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\) lies in QII. $$ \implies \cos \theta = -\dfrac{3}{5}$$ \begin{align*} \sin \theta & = \sqrt{1 - \left(-\frac{3}{5}\right)^2} \\ & = \sqrt{1 - \frac{9}{25}} \\ & = \sqrt{\frac{16}{25}} = \frac{4}{5}. \end{align*}

(a) \(\sin 2 \theta\) \begin{align*} \sin 2\theta & = 2 \sin \theta \cos \theta \\ & = 2 \left(\frac{4}{5}\right) \left(-\frac{3}{5}\right) \\ & = -\frac{24}{25}. \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = -\frac{24}{25}}. \end{align*}

(b) \(\cos 2 \theta\) \begin{align*} \cos 2\theta & = 1 - 2 \sin^2 \theta \\ & = 1 - 2 \left(\frac{4}{5}\right)^2 \\ & = 1 - 2 \left(\frac{16}{25}\right) \\ & = 1 - \frac{32}{25} \\ & = \frac{25}{25} - \frac{32}{25} \\ & = -\frac{7}{25}. \end{align*} \begin{align*} \implies \boxed{\cos 2\theta = -\frac{7}{25}}. \end{align*}

(c) \(\tan 2 \theta\) \begin{align*} \tan 2\theta & = \frac{\sin 2\theta}{\cos 2\theta} \\ & = \frac{-\dfrac{24}{25}}{-\dfrac{7}{25}} \\ & = \frac{24}{7}. \end{align*} \begin{align*} \implies \boxed{\tan 2\theta = \frac{24}{7}}. \end{align*}

(d) \(\sin \frac{\theta}{2}\)

We have:

\[\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}.\]

As \(\frac{\pi}{2} < \theta < \pi\), we know \(\frac{\theta}{2}\) lies in QI, so \(\sin \left(\frac{\theta}{2}\right) > 0\): \begin{align*} \sin\left(\frac{\theta}{2}\right) & = \sqrt{\frac{1 - \left(-\frac{3}{5}\right)}{2}} \\ & = \sqrt{\frac{1 + \dfrac{3}{5}}{2}} \\ & = \sqrt{\frac{\dfrac{8}{5}}{2}} \\ & = \sqrt{\frac{8}{10}} \\ & = \frac{2}{\sqrt{5}}. \end{align*} \begin{align*} \implies \boxed{\sin\left(\frac{\theta}{2}\right) = \frac{2}{\sqrt{5}}}. \end{align*}

(e) \(\cos \frac{\theta}{2}\)

We have:

\[\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}.\]

As \(\dfrac{\theta}{2}\) lies in QI, we know \(\cos \left(\frac{\theta}{2}\right) > 0\): \begin{align*} \cos\left(\frac{\theta}{2}\right) & = \sqrt{\dfrac{1 + \left(-\dfrac{3}{5}\right)}{2}} \\ & = \sqrt{\dfrac{1 - \dfrac{3}{5}}{2}} \\ & = \sqrt{\dfrac{\frac{2}{5}}{2}} \\ & = \sqrt{\frac{2}{10}} \\ & = \frac{1}{\sqrt{5}}. \end{align*} \begin{align*} \implies \boxed{\cos\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{5}}}. \end{align*}

(f) \(\tan \frac{\theta}{2}\)

As: \begin{align*} \tan \dfrac{\theta}{2} & = \dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}} \\ & = \frac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}} \\ & = 2. \end{align*} \begin{align*} \implies \boxed{\tan \frac{\theta}{2} = 2}. \end{align*}

< Question 1, 2 & 3 Question 5 >