Rownanie (granica+trygon)

[nabuch]

Witam!
Takie oto zadanko. I mam z nim problem.
Bede wdzieczny za kazda pomoc.
\(\displaystyle{ \lim_{t\to } (\sqrt{(t+sinx)(t+cosx)}-t)}= - \frac{1}{2}}\)

[soku11]

\(\displaystyle{ \lim_{t\to } (\sqrt{(t+sinx)(t+cosx)}-t)}= - \frac{1}{2}\\
L:\\
\lim_{t\to } ft( \frac{(t+sinx)(t+cosx)-t^{2}}{\sqrt{(t+sinx)(t+cosx)}+t}\right)=
\lim_{t\to } ft( \frac{t^{2} +tcosx+tsinx+sinxcosx- t^{2}}{\sqrt{(t+sinx) (t+cosx)}+t}\right)=\\
\lim_{t\to } ft(\frac{tcosx+tsinx+sinxcosx}{\sqrt{t^{2} +tcosx+tsinx+sinxcosx} +t}\right)=
\lim_{t\to } ft(\frac{t(cosx+sinx+\frac{sinxcosx}{t})} {t(\sqrt{1 +\frac{cosx}{t}+\frac{sinx}{t}+\frac{sinxcosx}{t^{2}}} +1}\right)=
\frac{cosx+sinx}{2}\\
\\
\frac{cosx+sinx}{2}=-\frac{1}{2}\\
cosx+sinx=-1\\
sin(\frac{\pi}{2}-x)+sinx=-1\\}\)

Z tym sobie juz dasz rade POZDRO

[janek1337]

moglbys ta ostatnia linijke tez rozwinac?

[soku11]

\(\displaystyle{ sin\left(\frac{\pi}{2}-x\right)+sinx=-1\\
2in\frac{\pi}{4}cos\left(\frac{\pi}{4}-x\right)=-1\\
\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=-1\\
cos\left(-(x-\frac{\pi}{4})\right)=-\frac{1}{\sqrt{2}}\\
cos\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\
x-\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi\ \ \ \ x-\frac{\pi}{4}=-\frac{3\pi}{4}+2k\pi\\
x=\frac{\pi}{4}+2k\pi\ \ \ \ x=-\frac{1\pi}{2}+2k\pi\\}\)

POZDRO