Question

daca t=sin2u, sa se calculeze in functie de t expresia E=tg patrat de u + ctg patrat de u

Answer 1

[tex]t=sin2u\\ t=2\cdot sin\ u\cdot cos\ u\\ sin\ u\cdot cos\ u=\frac{t}{2}|()^2\\ sin^2 u\cdot cos^2u=\frac{t^2}{4}(1)\\ ...................................\\ E=tg^2u+ctg^2u\\ \\ E=\frac{sin^2u}{cos^2u}+\frac{cos^2u}{sin^2u}\\ \\ E=\frac{sin^4u+cos^4u}{sin^2u\cdot cos^2u} \\ \\ E=\frac{(sin^2u+cos^2u)^2-2sin^2u\cdot cos^2u}{sin^2u\cdot cos^2u}\\ \\ E=\frac{1-2\cdot \frac{t^2}{4}}{\frac{t^2}{4}}\\ \\ E=(1-\frac{t^2}{2})\cdot \frac{4}{t^2}\\ \\ Asadar:\boxed{E=\frac{4}{t^2}-2} [/tex]

Answer 2

Scriem ecuatiile pentru sin2u si cos2u
$\sin{2u}=2\sin{u}\cos{u}\Rightarrow\sin^{2}{2u}=4\sin^{2}{u}\cos^{2}{u}$
$\cos{2u}=2\cos^{2}{u}-1\Rightarrow(\cos(2u)+1)=2\cos^{2}{u}\Rightarrow (\cos(2u)+1)^{2}=4\cos^{4}{u}$
Impartim cele 2 ecuatii acum ridicate la patrat
$\frac{\sin^{2}{2u}}{(\cos(2u)+1)^{2}}=\frac{4\sin^{2}{u}\cos^{2}{u}}{4\cos^{4}{u}}=\frac{\sin^{2}{u}}{\cos^{2}{u}}=tg^{2}{u}$
Putem sa exprimam si cos2u cu ajutorul lui t
$\sin^{2}{2u}+\cos^{2}{2u}=1\Rightarrow \cos^{2}{2u}=1-\sin^{2}{2u}=1-t^{2}\Rightarrow \cos{2u}=\sqrt{1-t^{2}}$
Deci obtinem ca
$tg^{2}{u}=\frac{t^{2}}{(\sqrt{1-t^{2}}+1)^{2}}$
Expresia pentru cos2u mai poate fi scrisa si asa
$\cos{2u}=1-2\sin^{2}{u}\Rightarrow(1-\cos(2u))=2\sin^{2}{u}\Rightarrow(\cos(2u)-1)^{2}=4\sin^{4}{u}<span>$
Acum facem din nou raportul intre ele
$\frac{\sin^{2}{2u}}{(\cos(2u)-1)^{2}}=\frac{4\sin^{2}{u}\cos^{2}{u}}{4\sin^{4}{u}}=\frac{\cos^{2}{u}}{\sin^{2}{u}}=ctg^{2}{u}$
Deci rezulta ca
$ctg^{2}{u}=\frac{t^{2}}{(\sqrt{1-t^{2}}-1)^{2}}<span>$
Deci expresia E poate fi scrisa ca
$E=\frac{t^{2}}{(\sqrt{1-t^{2}}+1)^{2}}+<span>\frac{t^{2}}{(\sqrt{1-t^{2}}-1)^{2}}</span>$