Question

Determinati sinx, știind ca​

Answer 1

Răspuns:

${ \sin(x) }^{2} + { \cos(x) }^{2} = 1 \\ { \sin(x) }^{2} + { \frac{4}{5} }^{2} = 1 \\ { \sin(x) }^{2} = 1 - \frac{16}{25} \\ { \sin(x) }^{2} = \frac{25 - 16}{25} \\ { \sin(x) }^{2} = \frac{11}{25} \\ \sin(x) = \sqrt{ \frac{11}{25} } \\ \sin(x ) = \frac{ \sqrt{11} }{5}$

Answer 2

 

$\displaystyle\bf\\x\in\left(\frac{3\pi}{2},~2\pi \right)\\\\x~este~in~cadranul~~4\\\\In~cadranul~4~cosinusul~este~pozitiv~iar~sinusul~este~negativ.\\\\cos\,x=\frac{4}{5}\\\\sin\,x=\pm\sqrt{1-cos^2x}\\\\\textbf{Formula are }\pm\textbf{ in fata radicalului, dar noi alegem semnul}\\\textbf{in functie ce cadranul in care suntem.}$

.

$\displaystyle\bf\\sin\,x=\pm\sqrt{1-cos^2x}\\\\sin\,x=\pm\sqrt{1-\left(\frac{4}{5}\right)^\b2}}\\\\\\sin\,x=\pm\sqrt{1-\frac{16}{25}}\\\\\\sin\,x=\pm\sqrt{\frac{25-16}{25}}\\\\\\sin\,x=\pm\sqrt{\frac{9}{25}}\\\\\\\textbf{Alegem semnul minus deoarece}\\\textbf{sinusul este negativ in cadranul 4}\\\\\boxed{\bf~sin\,x=-\frac{3}{5}}$