نویسنده : امیر انصاری

پاسخ

1. $$
   r=\sqrt{x^2+y^2}\\
   r= \sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\\
   \sin \theta = \frac{y}{r} = \frac{4}{5}\\
   \cos \theta = \frac{x}{r}=\frac{3}{5}\\
   \tan \theta = \frac{y}{x} = \frac{4}{3}
   $$
   
2. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{(-12)^2+(-5)^2}=\sqrt{144+25}=\sqrt{169}=13\\
   \sin \theta = \frac{y}{r} = -\frac{5}{13}\\
   \cos \theta = \frac{x}{r}= -\frac{12}{13}\\
   \tan \theta = \frac{y}{x} = \frac{-5}{-12}=\frac{5}{12}
   $$
   
3. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{8^2+(-15)^2}=\sqrt{64+225}=\sqrt{289}=17\\
   \sin \theta = \frac{y}{r} = -\frac{15}{17}\\
   \cos \theta = \frac{x}{r} = \frac{8}{17}\\
   \tan \theta = \frac{y}{x} = -\frac{15}{8}
   $$
   
4. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{1^2+(-1)^2}=\sqrt{2}\\
   \sin \theta = \frac{y}{r}=-\frac{1}{\sqrt{2}}\\
   \cos \theta = \frac{x}{r} = \frac{1}{\sqrt{2}}\\
   \tan \theta = \frac{y}{x}=-1
   $$