Solve this

Question:

If $\cos \mathrm{x}=\frac{3}{5}$ and $\cos \mathrm{y}=\frac{-24}{25}$, where $\frac{3 \pi}{2}<\mathrm{x}<2 \pi$ and $\pi<\mathrm{y}<\frac{3 \pi}{2}$, find the values of

(i) $\sin (x+y)$

(ii) $\cos (x-y)$

(iii) $\tan (x+y)$

Solution:

Given $\cos x=\frac{3}{5}$ and $\cos y=\frac{-24}{25}$

We will first find out value of sinx and siny

$\sin x=\sqrt{\left(1-\cos ^{2} x\right)} \Rightarrow \sqrt{\left(1-\left(\frac{3}{5}\right)^{2}\right)}=\sqrt{\left(\frac{25-9}{25}\right)} \Rightarrow \sqrt{\left(\frac{16}{25}\right)}=\frac{4}{5}$

siny $=\sqrt{\left(1-\cos ^{2} y\right)} \Rightarrow \sqrt{\left(1-\left(\frac{-24}{25}\right)^{2}\right)}=\sqrt{\left(\frac{625-576}{625}\right)} \Rightarrow \sqrt{\left(\frac{49}{625}\right)}=\frac{7}{25}$

(i) $\sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y$

$=\frac{4}{5} \cdot \frac{-24}{25}+\frac{3}{5} \cdot \frac{7}{25} \Rightarrow \frac{-96+21}{125}=\frac{-75}{125}$

$=\frac{-3}{5}$

(ii) $\cos (x-y)=\cos x \cdot \cos y+\sin x \cdot \sin y$

$=\frac{3}{5} \cdot \frac{-24}{25}+\frac{4}{5} \cdot \frac{7}{25} \Rightarrow \frac{-72+28}{125}=\frac{-44}{125}$

(iii) Here first we will calculate value of tanx and tany,

$\tan x=\frac{\sin x}{\cos x} \Rightarrow \frac{4 / 5}{3 / 5}=\frac{4}{3}$ and tany $=\frac{\sin y}{\cos y} \Rightarrow \frac{7 / 25}{-24 / 25}=\frac{7}{-24}$

$\tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\operatorname{tany}}{1-\tan \mathrm{x} \cdot \operatorname{tany}} \Rightarrow \frac{\frac{4}{3}+\frac{-7}{24}}{1+\frac{4}{3} \cdot \frac{-7}{24}}=\frac{\frac{32-7}{24}}{\frac{72-28}{72}} \Rightarrow \frac{\frac{25}{24}}{\frac{44}{72}}=\frac{75}{44}$