نویسنده : امیر انصاری

پاسخ

1. $$
   P(x,0)\\
   r=\sqrt{x^2+y^2} = \sqrt{x^2+0^2} = \sqrt{x^2} = x\\
   \sin 0^{\circ} = \frac{y}{r} = \frac{0}{x}=0\\
   \cos 0^{\circ} = \frac{x}{r} = \frac{x}{x}=1\\
   \tan 0^{\circ} = \frac{y}{x} = \frac{0}{x} = 0
   $$
   
2. $$
   P(0,y)\\
   r=\sqrt{x^2+y^2} = \sqrt{0+y^2}=\sqrt{y^2}=y\\
   \sin 90^{\circ} = \frac{y}{r} = \frac{y}{y}=1\\
   \cos 90^{\circ} = \frac{x}{r} = \frac{0}{y} = 0\\
   \tan 90^{\circ} = \frac{y}{x} = \frac{y}{0} = \text{undefined} (\text{ تعریف نشده })
   $$
   
3. $$
   P(-x,0)\\
   r=\sqrt{x^2+y^2}=\sqrt{(-x)^2+0}=\sqrt{x^2}=x\\
   \sin 180^{\circ} = \frac{y}{r}=\frac{0}{x}=0\\
   \cos 180^{\circ} = \frac{x}{r}=\frac{-x}{x}=-1\\
   \tan 180^{\circ} = \frac{y}{x}=\frac{0}{-x}=0
   $$
   
4. $$
   P(0,-y)\\
   r=\sqrt{x^2+y^2}=\sqrt{0+(-y)^2}=\sqrt{y^2}=y\\
   \sin 270^{\circ} = \frac{y}{r} = \frac{-y}{y}=-1\\
   \cos 270^{\circ} = \frac{x}{r}= \frac{0}{y}=0\\
   \tan 270^{\circ} = \frac{y}{x} = \frac{-y}{0}=\text{undefined} (\text{ تعریف نشده })
   $$
   
5. $$
   P(x,0)\\
   r=\sqrt{x^2+y^2} = \sqrt{x^2+0^2} = \sqrt{x^2} = x\\
   \sin 0^{\circ} = \frac{y}{r} = \frac{0}{x}=0\\
   \cos 0^{\circ} = \frac{x}{r} = \frac{x}{x}=1\\
   \tan 0^{\circ} = \frac{y}{x} = \frac{0}{x} = 0
   $$