Question 5 Exercise 8.2

Solutions of Question 5 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 exact values for $\sin \theta$, $\cos \theta$ and $\tan \theta$ using the information given: $\sin 2 \theta=\frac{24}{25}, 2 \theta$ in QII

Solution.

Given: $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII.

We have $$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}$$

Since $2\theta$ in QII, therefore $\cos 2\theta$ is negative.

\begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*}

Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\theta}{2}} $$

As $\frac{\pi}{2}< 2\theta < \pi$, so $\frac{\pi}{4}< \theta < \frac{\pi}{2}$, that is, $\theta$ lies in QI and $\sin\theta > 0$. Thus

\begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\theta}{2}} \\ & = \sqrt{\frac{1-\left( -\frac{7}{25} \right)}{2}} \\ & = \sqrt{\frac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{5}} \end{align*} Also $$\cos\theta = \pm\sqrt{1-\sin\theta}$$ As $\theta$ lies in QI, therefore $\cos\theta >0$, thus \begin{align*} \cos\theta & = \sqrt{1-\sin\theta} \\ &=\sqrt{1-\left(\frac{4}{5} \right)^2} \\ & = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} \end{align*} \begin{align*} \implies \boxed{\cos\theta=\frac{3}{5}} \end{align*} Now \begin{align*} \tan\theta & =\frac{\sin\theta}{\cos\theta} \\ & = \frac{4/5}{3/5} \end{align*} \begin{align*} \implies \boxed{\tan\theta=\frac{4}{3}} \end{align*} GOOD

Find exact values for $\sin \theta, \cos \theta$ and $\tan \theta$ using the information given: $\cos 2 \theta=-\frac{7}{25}, 2 \theta$ in QIII

Solution.

Given: \(\cos 2\theta = -\dfrac{7}{25}\) and \(2\theta\) lies in QIII.

We have:

\[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\]

Since \(2\theta\) lies in QIII, we know that \(\sin 2\theta < 0\). Therefore: \begin{align*} \sin 2\theta & = -\sqrt{1 - \left(-\frac{7}{25}\right)^2} \\ & = -\sqrt{1 - \frac{49}{625}} \\ & = -\sqrt{\frac{576}{625}} = -\frac{24}{25}. \end{align*} Also, we have:

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

As \(\pi < 2\theta < \frac{3\pi}{2}\) implies \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\) lies in QII, we know that \(\sin \theta > 0\). Thus: \begin{align*} \sin \theta & = \sqrt{\frac{1 - \left(-\frac{7}{25}\right)}{2}} \\ & = \sqrt{\frac{1 + \frac{7}{25}}{2}} \\ & = \sqrt{\frac{\frac{32}{25}}{2}} \\ & = \sqrt{\frac{16}{25}} = \frac{4}{5}. \end{align*} \begin{align*} \implies \boxed{\sin \theta = \frac{4}{5}}. \end{align*}

Also \[ \cos \theta = \pm \sqrt{1 - \sin^2 \theta}. \]

AS \(\theta\) lies in QII, \(\cos \theta < 0\), so: \begin{align*} \cos \theta & = -\sqrt{1 - \left(\frac{4}{5}\right)^2} \\ & = -\sqrt{1 - \frac{16}{25}} \\ & = -\sqrt{\frac{9}{25}} = -\frac{3}{5}. \end{align*} \begin{align*} \implies \boxed{\cos \theta = -\frac{3}{5}}. \end{align*} Now \begin{align*} \tan \theta & = \frac{\sin \theta}{\cos \theta} \\ & = \frac{\dfrac{4}{5}}{-\dfrac{3}{5}} \\ & = -\frac{4}{3}. \end{align*} \begin{align*} \implies \boxed{\tan \theta = -\frac{4}{3}}. \end{align*}

Find exact values for $\sin \theta, \cos \theta$ and $\tan \theta$ using the information given: $\sin 2 \theta=-\frac{240}{289}, 2 \theta$ in QIII

Solution.

Given: \(\sin 2\theta = -\dfrac{240}{289}\) and \(2\theta\) lies in QIII.

We have:

\[\cos 2\theta = \pm \sqrt{1 - \sin^2 2\theta}.\]

Since \(2\theta\) lies in QIII, we know that \(\cos 2\theta < 0\). Therefore: \begin{align*} \cos 2\theta & = -\sqrt{1 - \left(-\frac{240}{289}\right)^2} \\ & = -\sqrt{1 - \frac{57600}{83521}} \\ & = -\sqrt{\frac{25921}{83521}} = -\frac{161}{289}. \end{align*} Also, we have,

\[\sin\theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}\]

As \(\pi < 2\theta < \frac{3\pi}{2}\) implies \(\frac{\pi}{2} < \theta < \pi\), i.e., \(\theta\) lies in QII, we know that \(\sin \theta > 0\). Thus: \begin{align*} \sin\theta & = \sqrt{\frac{1 - \left(-\frac{161}{289}\right)}{2}} \\ & = \sqrt{\frac{1 + \frac{161}{289}}{2}} \\ & = \sqrt{\frac{\frac{450}{289}}{2}} = \sqrt{\frac{225}{289}} &= \frac{15}{17}. \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{15}{17}}. \end{align*} Also

\[\cos \theta = \pm \sqrt{1 - \sin^2 \theta}\]

As \(\theta\) lies in QII, \(\cos \theta < 0\), so: \begin{align*} \cos \theta & = -\sqrt{1 - \left(\frac{15}{17}\right)^2} \\ & = -\sqrt{1 - \frac{225}{289}} \\ & = -\sqrt{\frac{64}{289}} = -\frac{8}{17}. \end{align*} \begin{align*} \implies \boxed{\cos \theta = -\frac{8}{17}}. \end{align*} Now \begin{align*} \tan \theta & = \frac{\sin \theta}{\cos \theta} \\ & = \frac{\frac{15}{17}}{-\frac{8}{17}} \\ & = -\frac{15}{8}. \end{align*} \begin{align*} \implies \boxed{\tan \theta = -\frac{15}{8}}. \end{align*}

Find exact values for $\sin \theta, \cos \theta$ and $\tan \theta$ using the information given: $\cos 2 \theta=\frac{120}{169}, 2 \theta$ in QIV

Solution.

Given: \(\cos 2\theta = \frac{120}{169}\) and \(2\theta\) lies in QIV.

We have:

\[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\]

Since \(2\theta\) lies in QIV, we know that \(\sin 2\theta < 0\). Therefore: \begin{align*} \sin 2\theta & = -\sqrt{1 - \left(\frac{120}{169}\right)^2} \\ & = -\sqrt{1 - \frac{14400}{28561}} \\ & = -\sqrt{\frac{14161}{28561}} = -\frac{119}{169}. \end{align*} Also, we have:
\[\sin \theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}\]

As \( \frac{3\pi}{2} < 2\theta < 2\pi \) implies \( \frac{3\pi}{4} < \theta < \pi \), i.e., \(\theta\) lies in QII, we know that \(\sin \theta > 0\). Thus: \begin{align*} \sin \theta & = \sqrt{\frac{1 - \frac{120}{169}}{2}} \\ & = \sqrt{\frac{\frac{49}{169}}{2}} \\ & = \sqrt{\frac{49}{338}} = \frac{7}{\sqrt{338}}. \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{7}{\sqrt{338}}}. \end{align*}

\[\cos \theta = \pm \sqrt{1 - \sin^2 \theta}\]

As \(\theta\) lies in QII, we know that \(\cos \theta < 0\). Therefore: \begin{align*} \cos \theta & = -\sqrt{1 - \left(\frac{7}{\sqrt{338}}\right)^2} \\ & = -\sqrt{1 - \frac{49}{338}} \\ & = -\sqrt{\frac{289}{338}} = -\frac{17}{\sqrt{338}}. \end{align*} \begin{align*} \implies \boxed{\cos\theta = -\frac{17}{\sqrt{338}}}. \end{align*} \begin{align*} \tan \theta & = \frac{\sin \theta}{\cos \theta} \\ & = \frac{\frac{7}{\sqrt{338}}}{-\frac{17}{\sqrt{338}}} \\ & = -\frac{7}{17}. \end{align*} \begin{align*} \implies \boxed{\tan\theta = -\frac{7}{17}}. \end{align*}

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