In the interactive book: complete chapter V, with all demonstrations and exercises.

Make a square

Draw a square with side 1 and draw a diagonal. It divides two right angles in half: two triangles with angles of $45°$, $45°$ and $90°$ appear. According to Pythagoras’ theorem, the diagonal measures $\sqrt{1^2+1^2}=\sqrt2$.

For an acute angle in a right triangle, sine is the opposite side divided by the hypotenuse; cosine is the adjacent side divided by the hypotenuse. These ratios coincide with the cycle coordinates: just reduce the hypotenuse to the measure 1, without changing the angles.

Demo: the value of 45 degrees

In the square triangle, the legs measure 1 and the hypotenuse $\sqrt2$. Logo:

$$\operatorname{sen}45°=\cos45°=\frac1{\sqrt2}=\frac{\sqrt2}{2}.$$

We multiply numerator and denominator by $\sqrt2$ to obtain the last form. The tangent is $1/1=1$.

Demonstration: the values ​​of 30 and 60 degrees

Now make equilateral triangle next to 2. Its three angles are equal and add up to $180°$, therefore they measure $60°$. Plot the height from a vertex. By symmetry, it divides the base into two parts measuring 1 and the upper angle into two parts measuring $30°$.

In the right triangle formed, the hypotenuse is 2, the smaller side is 1 and the height is $h$. Pythagoras provides:

$$h^2+1^2=2^2\quad\Rightarrow\quad h=\sqrt3.$$

For the angle of $30°$, the opposite side measures 1; for $60°$, measure $\sqrt3$. We get the table:

Angle Sine Cosine Tangent
$30°=\pi/6$ $1/2$ $\sqrt3/2$ $\sqrt3/3$
$45°=\pi/4$ $\sqrt2/2$ $\sqrt2/2$ 1
$60°=\pi/3$ $\sqrt3/2$ $1/2$ $\sqrt3$

The table is the result of two drawings. If you forget an entry, redo the triangle.

Regular polygons also have sines

A regular polygon has all sides and all angles equal. Inscribe one of them, with $n\ge3$ sides, in a circle of radius $R$. Connect two neighboring vertices to the center. The isosceles triangle formed has a central angle of $360°/n$ and a base length of $\ell$.

The height that starts from the center cuts the base and the central angle in half. In one half, the hypotenuse is worth $R$, the side opposite the angle $180°/n$ is worth $\ell/2$. Therefore:

$$\operatorname{sen}\frac{\pi}{n}=\frac{\ell}{2R}, \qquad \ell=2R\operatorname{sen}\frac{\pi}{n}.$$

For a hexagon, $n=6$: $\ell=2R\operatorname{sen}30°=R$. For a square, $\ell=2R\operatorname{sen}45°=R\sqrt2$. This relationship explains how polygon measurements help calculate trigonometric values.

Check the scale

If equilateral triangle has side 10 instead of 2, all measurements will be multiplied by 5. The reasons will remain the same. Sine is not a length: it is a ratio without unity.

In everyday life

  • Hobs and tiles. Hexagons and equilateral triangles cover the floor without leaving holes because their angles ($120^\circ$ and $60^\circ$) divide $360^\circ$ exactly.
  • Origami. The most common folds are $45^\circ$ (folding the square diagonally) and $30^\circ$ and $60^\circ$ (dividing a sheet into three).
  • Squares. The pair of squares in the school kit holds the angles of $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$.
The notable angles on the case and toolbox: the two school squares and the hex nut, a regular hexagon.

Exercises

Try solving it in your notebook before opening the solution. Exercises 1 and 2 are input; 3 and 4, for practice; 5 and 6, for deepening.

1. What is the diagonal of a square with side 6 cm?

See resolution 1

$d^2=6^2+6^2=72$, then $d=6\sqrt2$ cm.

2. How tall is one equilateral triangle by 8 cm?

See resolution 2

The base half is 4 and the hypotenuse is 8. $h=\sqrt{64-16}=4\sqrt3$ cm.

3. Calculate $2\operatorname{sen}30°+\cos60°$.

See resolution 3

$2\cdot1/2+1/2=3/2$.

4. A regular hexagon is inscribed in a circle of radius 5 cm. What is its perimeter?

See resolution 4

Each side measures 5 cm. The perimeter is $6\cdot5=30$ cm.

5. A square is inscribed in a circle of radius 3 cm. How long is your side?

See resolution 5

$\ell=2\cdot3\operatorname{sen}45°=3\sqrt2$ cm.

6. In the circle of radius 1, an inscribed equilateral triangle has side $\sqrt3$. Use this to find $\operatorname{sen}60°$.

See resolution 6

For $n=3$, $\operatorname{sen}(\pi/3)=\ell/(2R)=\sqrt3/2$.

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