Question

Cum se demonstreaza formulele urmatoare? (dau coronita)

$sin^{2} x = \frac{tgx}{1+tg^{2}x } \\ \\ cos^{2} x = \frac{1}{1+tg^{2}x }$

Si cand folosim plus sau minus, in functie de cine trebuie sa vedem ca sa folosim un semn?

Answer 1

   
[tex]\displaystyle\\ \text{la prima formulatrebuia sa scrii: } \\\\ \sin^{2} x = \frac{\tex{tg}^2x}{1+\text{tg}^{2}x } ~~\text{ in loc de: }~~ \sin^{2} x = \frac{\text{tg }x}{1+\text{tg}^{2}x }\\\\ [/tex]


[tex]\displaystyle\\ \text{Rezolvare:}\\\\ \text{Aratam ca: }~~~\sin^{2} x = \frac{\tex{tg}^2x}{1+\text{tg}^{2}x }\\\\ \frac{\tex{tg}^2x}{1+\text{tg}^{2}x} =\frac{ \frac{\sin^2x}{\cos^2x}}{1+ \frac{\sin^2x}{\cos^2x}}=\frac{\frac{\sin^2x}{\cos^2x}}{ \frac{\cos^2x +\sin^2x}{\cos^2x}} =\frac{ \frac{\sin^2x}{\cos^2x}}{ \frac{1}{\cos^2x}} =\frac{\sin^2x}{\cos^2x}}\cdot \frac{\cos^2x}{1}=\sin^2x [/tex]


[tex]\displaystyle\\ \text{Aratam ca: }~~~\cos^{2} x = \frac{1}{1+\text{tg}^{2}x }\\\\ \frac{1}{1+\text{tg}^{2}x }= \frac{1}{1+ \frac{\sin^2x}{\cos^2x}}= \frac{1}{ \frac{\cos^2x+\sin^2x}{\cos^2x}}= \frac{\cos^2x}{\cos^2x+\sin^2x}=\cos^2x [/tex]