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Question Number 205219 by hardmath last updated on 13/Mar/24
{ ((sinx + cosx = 1)),((sinx − cosy = 1)) :} ⇒ x = ?
$$\begin{cases}{\mathrm{sinx}\:+\:\mathrm{cosx}\:=\:\mathrm{1}}\\{\mathrm{sinx}\:−\:\mathrm{cosy}\:=\:\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 13/Mar/24
{ ((sinx + cosx = 1...i)),((sinx − cosy = 1...ii)) :} ;x = ? cos x+cos y=0 cos y=−cos x ii ⇒sin x+cos x=1 sin^2 x+2 sin x cos x+cos^2 x=1 1+2 sin x cos x=1 sin x cos x=0 sin x=0 ∣ cos x=0 x=nπ ∣ x=(π/2)+nπ
$$\begin{cases}{\mathrm{sinx}\:+\:\mathrm{cosx}\:=\:\mathrm{1}…{i}}\\{\mathrm{sinx}\:−\:\mathrm{cosy}\:=\:\mathrm{1}…{ii}}\end{cases}\:\:\:;\mathrm{x}\:=\:? \\ $$$$\mathrm{cos}\:{x}+\mathrm{cos}\:{y}=\mathrm{0} \\ $$$$\mathrm{cos}\:{y}=−\mathrm{cos}\:{x} \\ $$$${ii}\:\Rightarrow\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}+\mathrm{cos}^{\mathrm{2}} {x}=\mathrm{1}\: \\ $$$$\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{1}\: \\ $$$$\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}=\mathrm{0}\:\mid\:\mathrm{cos}\:{x}=\mathrm{0} \\ $$$${x}={n}\pi\:\:\:\mid\:\:{x}=\frac{\pi}{\mathrm{2}}+{n}\pi \\ $$
Commented by mr W last updated on 13/Mar/24
through squaring you added also the roots for eqn. sin x+cos x=−1. i think only x=2nπ or 2nπ+(π/2) are suitable.
$${through}\:{squaring}\:{you}\:{added}\:{also} \\ $$$${the}\:{roots}\:{for}\:{eqn}.\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=−\mathrm{1}. \\ $$$${i}\:{think}\:{only}\:{x}=\mathrm{2}{n}\pi\:{or}\:\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\:{are} \\ $$$${suitable}. \\ $$
Commented by Rasheed.Sindhi last updated on 13/Mar/24
Very right sir! Thanks.
$${Very}\:{right}\:\boldsymbol{{sir}}!\:\mathcal{T}{hanks}. \\ $$
Answered by MM42 last updated on 13/Mar/24
(√2)sin(x+(π/4))=1⇒ { ((x=2kπ)),((x=2kπ+(π/2))) :} if x=2kπ⇒cosy=−1⇒y=(2k+1)π if x=2kπ+(π/2)⇒cosy=0⇒y=kπ+(π/2) ⇒Ans={(x,y) ∣ x=2kπ & y=(2kπ+1)π or x=2kπ+(π/2) & y=kπ+(π/2)}
$$\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{1}\Rightarrow\begin{cases}{{x}=\mathrm{2}{k}\pi}\\{{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$${if}\:{x}=\mathrm{2}{k}\pi\Rightarrow{cosy}=−\mathrm{1}\Rightarrow{y}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi \\ $$$${if}\:{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\Rightarrow{cosy}=\mathrm{0}\Rightarrow{y}={k}\pi+\frac{\pi}{\mathrm{2}}\: \\ $$$$\Rightarrow{Ans}=\left\{\left({x},{y}\right)\:\mid\:{x}=\mathrm{2}{k}\pi\:\&\:{y}=\left(\mathrm{2}{k}\pi+\mathrm{1}\right)\pi\:\:{or}\:{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\:\&\:{y}={k}\pi+\frac{\pi}{\mathrm{2}}\right\} \\ $$$$ \\ $$

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