Question

Am nevoie urgenta de E2 va rog!

Answer 1

$\displaystyle x\in \Bigg(0,\frac{\pi}{2} \Bigg),~y\in \Bigg(\frac{\pi}{2} ,\pi\Bigg),~sinx=\frac{4}{5} ,~cosy=-\frac{3}{5} \\\\\\ \boxed{sin(\alpha-\beta)=sin\alpha cos\beta-sin\beta cos \alpha};~~~~~~~~~\boxed{sin^2\alpha+cos^2\alpha=1}\\\\ \\ sin^2x+cos^2x=1 \Rightarrow cosx=\pm\sqrt{1-sin^2x} \\\\ x\in \Bigg(0,\frac{\pi}{2}\Bigg) \rightarrow primul~cadran \Rightarrow cosx= \sqrt{1-sin^2x}$

$\displaystyle cosx=\sqrt{1-\Bigg(\frac{4}{5} \Bigg)^2} =\sqrt{1-\frac{16}{25} } =\sqrt{\frac{25-16}{25} } =\sqrt{\frac{9}{25} } =\frac{\sqrt{9} }{\sqrt{25} }=\frac{3}{5}$

$\displaystyle sin^2y+cos^2y=1 \Rightarrow siny=\pm \sqrt{1-cos^2y} \\\\\\ y \in \Bigg(\frac{\pi}{2} ,\pi\Bigg)\rightarrow al~doilea~cadran\Rightarrow siny=\sqrt{1-cos^2y} \\\\\\ siny=\sqrt{1-\Bigg(-\frac{3}{5} \Bigg)^2} =\sqrt{1-\frac{9}{25} } =\sqrt{\frac{25-9}{25} }=\sqrt{\frac{16}{25} } =\frac{\sqrt{16} }{\sqrt{25} }=\frac{4}{5}$

$\displaystyle sin(x-y)=\frac{4}{5} \cdot \Bigg(-\frac{3}{5} \Bigg) -\frac{4}{5} \cdot \frac{3}{5} =-\frac{12}{25} -\frac{12}{25} =-\frac{24}{25}$

$\displaystyle \boxed{cos(\alpha-\beta)=cos\alpha cos\beta+sin\alpha sin\beta}\\\\ cos(x-y)=\frac{3}{5} \cdot \Bigg(-\frac{3}{5} \Bigg)+\frac{4}{5} \cdot \frac{4}{5} =-\frac{9}{25} +\frac{16}{25} =\frac{7}{25}$

$\displaystyle \boxed{cos(\alpha+\beta)=cos\alpha cos\beta -sin\alpha sin\beta}\\\\ cos(x+y)=\frac{3}{5} \cdot \Bigg(-\frac{3}{5} \Bigg)-\frac{4}{5} \cdot \frac{4}{5} =-\frac{9}{25} -\frac{16}{25} =-\frac{25}{25} =-1$

$\displaystyle \boxed{sin2\alpha=2 sin\alpha cos\alpha}\\\\ sin2y=2 \cdot \frac{4}{5} \cdot \Bigg(-\frac{3}{5}\Bigg) =\frac{8}{5} \cdot \Bigg(-\frac{3}{5} \Bigg)=-\frac{24}{25}$

$\displaystyle \boxed{cos2 \alpha=cos^2\alpha-sin^2\alpha}\\\\ cos2x=\Bigg(\frac{3}{5}\Bigg)^2-\Bigg(\frac{4}{5} \Bigg)^2=\frac{3^2}{5^2} -\frac{4^2}{5^2} =\frac{9}{25}-\frac{16}{25} = -\frac{7}{25}$