Question, Exercise 10.1
Solutions of Question 4 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Question 4(i)
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \right)$.
Solution
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant,
$\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.
We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$
As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant,
\begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\
&=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\
&=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\
\Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align}
Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$
As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant,
\begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\
\Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align}
Now
\begin{align}\sin (\alpha-\beta )&=\sin \alpha\cos \beta+\cos \alpha\sin \beta \\
&=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)\\
&=\dfrac{48}{65}-\frac{15}{65}\\
\Rightarrow \quad \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align}
Question 4(ii)
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\cos \left( \alpha +\beta \right)$.
Solution
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant,
$\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.
We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$ As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\ \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\cos (\alpha+\beta )&=\cos \alpha\cos \beta-\sin \alpha\sin \beta \\ &=\left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)\\ &=\dfrac{36}{65}-\frac{20}{65}\\ \Rightarrow \quad \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align}
Question 4(iii)
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\tan \left( \alpha +\beta \right)$ .
Solution
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant,
$\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.
We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$ As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant, \begin{align} &=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\ \Rightarrow \quad \sin \beta &=\dfrac{5}{13}.\\ \tan \alpha &= \dfrac{\sin \alpha}{\cos \alpha}\\ &=\dfrac{\dfrac{-4}{5}}{\dfrac{-3}{5}}\\ \tan \alpha &=\dfrac{4}{3}\\ \tan \beta &= \dfrac{\sin \beta}{\cos \beta}\\ &=\dfrac{\dfrac{5}{13}}{\dfrac{-12}{13}}\\ \tan \beta &=\dfrac{-5}{12}.\end{align}
Now \begin{align}\tan \left( \alpha +\beta \right)&=\dfrac{\tan \alpha+ \tan \beta}{1-\tan \alpha \tan \beta} \\ &=\dfrac{\dfrac{4}{3}+ \dfrac{-5}{12}}{1-(\dfrac{4}{3}) (\dfrac{-5}{12})}\\ &=\dfrac{\dfrac{4}{3}- \dfrac{5}{12}}{1+ \dfrac{20}{36}}=\dfrac{\dfrac{11}{12}}{ \dfrac{56}{36}}\\ \Rightarrow \quad \tan \left( \alpha+\beta \right)&=\dfrac{33}{56}.\end{align}