I want the explanation in English

We use an isosceles right-angled triangle.
Consider a right-angled triangle with two equal sides, each of length 1. The angles will be $45^\circ$, $45^\circ$, and $90^\circ$.
Using the Pythagorean Theorem, the length of the hypotenuse ($h$) is:

$$h = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$$

Now we can calculate the trigonometric values for $45^\circ$:

$$\sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\cos 45^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\tan 45^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{1} = 1$$

We use an equilateral triangle with side lengths of 2. All angles are $60^\circ$.
Draw an altitude from one vertex to the midpoint of the opposite side. This altitude bisects the top angle and the base, creating two $30^\circ-60^\circ-90^\circ$ right-angled triangles.
In each right-angled triangle:

• The hypotenuse is 2.
• The side opposite the $30^\circ$ angle (or adjacent to the $60^\circ$ angle) is 1 (half of the equilateral triangle's side).
• Using the Pythagorean Theorem, the length of the side opposite the $60^\circ$ angle (or adjacent to the $30^\circ$ angle) is $h$:

$$h^2 + 1^2 = 2^2$$ $$h^2 + 1 = 4$$ $$h^2 = 3$$ $$h = \sqrt{3}$$

For the $30^\circ$ angle:

• Opposite side is 1.
• Adjacent side is $\sqrt{3}$.
• Hypotenuse is 2.

$$\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}$$ $$\cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}$$ $$\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}}$$

For the $60^\circ$ angle:

• Opposite side is $\sqrt{3}$.
• Adjacent side is 1.
• Hypotenuse is 2.

$$\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}$$ $$\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}$$ $$\tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

These derivations show how the standard trigonometric values for these special angles are obtained using basic geometry and the Pythagorean Theorem.