Question 9, Exercise 8.1

Solutions of Question 9 of Exercise 8.1 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 $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$

Solution.

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ As $\alpha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Also $$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$ As $\beta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}} \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \sin (\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= -\frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\ &= -\frac{7}{5\sqrt{2}}. \end{align*} \begin{align*} \sin (\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) - \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= -\frac{3}{5\sqrt{2}} + \frac{4}{5\sqrt{2}} \\ &= \frac{1}{5\sqrt{2}}. \end{align*}

Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\frac{1}{\sqrt{2}}$ and $\cos \beta=-\frac{3}{5}$ find: $\cos (\alpha \pm \beta)$

Solution.

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ As $\alpha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Also $$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$ As $\beta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}} \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} \begin{align*} \cos (\alpha + \beta) &= \left( -\frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= \frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\ &= -\frac{1}{5\sqrt{2}}. \end{align*} \begin{align*} \cos (\alpha - \beta) &= \left( -\frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= \frac{3}{5\sqrt{2}} + \frac{4}{5\sqrt{2}} \\ &= \frac{7}{5\sqrt{2}}. \end{align*}

Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\frac{1}{\sqrt{2}}$ and $\cos \beta=-\frac{3}{5}$ find: $\tan (\alpha \pm \beta)$

Solution.

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ As $\alpha$ lies in QII and $\cos$ is -ive in QII, \begin{align*} \cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}\\ \implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}. \end{align*} Now \begin{align*} tan \alpha&=\frac{\sin \alpha}{\cos \alpha}\\ &=\frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}}\\ tan \alpha&= -1 \end{align*} As $\beta$ lies in QII and $\sin$ is +ive in QII, \begin{align*} \sin \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}} \\ & =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} \end{align*} Now \begin{align*} tan \beta&=\frac{\sin \beta}{\cos \beta}\\ &=\frac{\dfrac{4}{5}}{-\dfrac{3}{5}}\\ tan \beta &= -\dfrac{4}{3} \end{align*} \begin{align*} \tan(\alpha + \beta) &= \frac{-1 + \left( -\frac{4}{3} \right)}{1 - \left( -1 \right) \left( -\frac{4}{3} \right)} \\ &= \frac{-1 - \frac{4}{3}}{1 - \frac{4}{3}} \\ &= \frac{-\frac{7}{3}}{\frac{-1}{3}} \\ &= 7. \end{align*} \begin{align*} \tan(\alpha - \beta) &= \frac{-1 - \left( -\frac{4}{3} \right)}{1 + \left( -1 \right) \left( -\frac{4}{3} \right)} \\ &= \frac{-1 + \frac{4}{3}}{1 + \frac{4}{3}} \\ &= \frac{\frac{1}{3}}{\frac{7}{3}} \\ &= \frac{1}{7}. \end{align*}