Question Number 201427 by mathlove last updated on 06/Dec/23
$$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)}\\{{tanx}−{tany}=\mathrm{1}}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\ $$
Answered by Rasheed.Sindhi last updated on 06/Dec/23
$$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)…\left({i}\right)}\\{{tanx}−{tany}=\mathrm{1}………\left({ii}\right)}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\ $$$$\left({i}\right)\Rightarrow{x}+{y}=\left(\pi/\mathrm{2}\right)−\left({x}−{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}=\pi/\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}=\pi/\mathrm{4} \\ $$$$\left({ii}\right)\Rightarrow\mathrm{tan}\left(\pi/\mathrm{4}\right)\:−\mathrm{tan}\:{y}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\mathrm{1}−\mathrm{tan}\:{y}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\mathrm{tan}\:{y}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\Rightarrow{y}=\mathrm{0} \\ $$$$\mathcal{G}{enerally}\:\left({x},{y}\right)=\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{n}\pi,\:\mathrm{2}{n}\pi\right) \\ $$
Commented by mathlove last updated on 06/Dec/23
$${thanks} \\ $$
Answered by mr W last updated on 06/Dec/23
![sin (x+y)=cos (x−y)=sin ((π/2)−x+y) kπ+(−1)^k (x+y)=(π/2)−x+y ⇒(((2k−1)π)/2)+[1+(−1)^k ]x=[1−(−1)^k ]y case 1: k=2n (((4n−1)π)/2)+2x=0 ⇒x=−nπ+(π/4) tan y=tan x−1=1−1=0 ⇒y=mπ case 2: k=2n+1 (((4n+1)π)/2)=2y ⇒y=nπ+(π/4) tan x=1+tan y=1+1=2 ⇒x=mπ+tan^(−1) 2 summary (with n, m∈Z): { ((x=nπ+(π/4))),((y=mπ)) :} or { ((x=nπ+tan^(−1) 2)),((y=mπ+(π/4))) :}](https://www.tinkutara.com/question/Q201440.png)
$$\mathrm{sin}\:\left({x}+{y}\right)=\mathrm{cos}\:\left({x}−{y}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{x}+{y}\right) \\ $$$${k}\pi+\left(−\mathrm{1}\right)^{{k}} \left({x}+{y}\right)=\frac{\pi}{\mathrm{2}}−{x}+{y} \\ $$$$\Rightarrow\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\pi}{\mathrm{2}}+\left[\mathrm{1}+\left(−\mathrm{1}\right)^{{k}} \right]{x}=\left[\mathrm{1}−\left(−\mathrm{1}\right)^{{k}} \right]{y} \\ $$$${case}\:\mathrm{1}:\:{k}=\mathrm{2}{n} \\ $$$$\frac{\left(\mathrm{4}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}+\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow{x}=−{n}\pi+\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:{y}=\mathrm{tan}\:{x}−\mathrm{1}=\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{y}={m}\pi \\ $$$${case}\:\mathrm{2}:\:{k}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\frac{\left(\mathrm{4}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}=\mathrm{2}{y} \\ $$$$\Rightarrow{y}={n}\pi+\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:{x}=\mathrm{1}+\mathrm{tan}\:{y}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\Rightarrow{x}={m}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$ \\ $$$${summary}\:\left({with}\:{n},\:{m}\in{Z}\right): \\ $$$$\begin{cases}{{x}={n}\pi+\frac{\pi}{\mathrm{4}}}\\{{y}={m}\pi}\end{cases}\:\:{or}\:\begin{cases}{{x}={n}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{2}}\\{{y}={m}\pi+\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$
Commented by mathlove last updated on 06/Dec/23
$${thanks} \\ $$