Question 2, Review Exercise

Solutions of Question 2 of Review Exercise 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.

Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$ where $\theta$ is obtuse and $\phi$ is acute. Find the values of $\sin (\theta-\phi)$.

Solution. Given: $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-\left(\frac{5}{13}\right)^2\\ & =1-\frac{25}{169} \\ \implies \cos^2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\phi$ lies in I Q. This implies $\cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\phi)&=\sin \theta \cos \phi-\cos \theta \sin \phi\\ &=\left(\frac{3}{5}\right) \left(\frac{12}{13}\right)-\left(-\frac{4}{5}\right) \left(\frac{5}{13}\right)\\ &=\frac{36}{65}+\frac{20}{65}\\ \implies \sin(\theta -\phi)&=\frac{56}{65} \end{align*}

Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$ where $\theta$ is obtuse and $\phi$ is acute. Find the values of $\tan (\theta-\phi)$.

Solution.

Given $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \tan \theta & = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} \\ \implies \tan \theta & = -\frac{3}{4} \end{align*} Take \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-(\frac{5}{13})^2\\ & =1-\frac{25}{169} \\ \implies \cos^2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\phi$ lies in IQ. This implies $\cos\phi > 0$, thus $$\cos \phi=\frac{12}{13}$$ Now \begin{align*} \tan \phi & = \frac{\sin \phi}{\cos \phi} = \frac{\frac{5}{13}}{\frac{12}{13}}\\ \implies \tan \phi & = \frac{5}{12} \end{align*} Finally, we have \begin{align*} \tan(\theta - \phi) &= \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} \\ &= \frac{-\frac{3}{4} - \frac{5}{12}}{1 + \left(-\frac{3}{4}\right) \cdot( \frac{5}{12})} \\ &= \frac{-14}{12}\times \frac{16}{11} \\ &= -\frac{56}{33} \end{align*}

Given that $\sin \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$ where $\theta$ is obtuse and $\phi$ is acute. Find the values of: $\tan (\theta+\phi)$

Solution.

Given $\sin \theta=\dfrac{3}{5}$ and $\sin \phi=\dfrac{5}{13}$, where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \tan \theta & = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} \\ \implies \tan \theta & = -\frac{3}{4} \end{align*} Take \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-(\frac{5}{13})^2\\ & =1-\frac{25}{169} \\ \implies \cos^2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\phi$ lies in IQ. This implies $\cos\phi > 0$, thus $$\cos \phi=\frac{12}{13}$$ Now \begin{align*} \tan \phi & = \frac{\sin \phi}{\cos \phi} = \frac{\frac{5}{13}}{\frac{12}{13}}\\ \implies \tan \phi & = \frac{5}{12} \end{align*} Finally, we have \begin{align*} \tan(\theta + \phi) &= \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} \\ &= \frac{-\frac{3}{4} + \frac{5}{12}}{1 - \left(-\frac{3}{4}\right) \cdot \frac{5}{12}} \\ &= \frac{-\frac{4}{12}}{\frac{63}{48}} \\ &= -\frac{16}{63} \end{align*}

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