====== 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.
=====Question 5(i)=====
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
=====Question 5(ii)=====
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*}
=====Question 5(iii)=====
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*}
=====Question 5(iv)=====
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*}
====Go to ====
[[math-11-nbf:sol:unit08:ex8-2-p2|< Question 4 ]]
[[math-11-nbf:sol:unit08:ex8-2-p4|Question 6 >]]