Euler's Eyes
2026-10-19
Symmetries: one table serves the entire round
In the interactive book: complete chapter IV, with all demonstrations and exercises.
The wheel mirror
Draw a point in the first quadrant and its reflection in the second. The height is the same, but the horizontal position changes sign. Symmetry allows us to calculate values for an entire revolution knowing only the first quadrant.
If $\alpha$ is between $0°$ and $90°$, the corresponding point is $(\cos\alpha,\operatorname{sen}\alpha)$. Reflect it on the axes and in the center:
| Angle | Coordinates | Sine | Cosine |
|---|---|---|---|
| $180°-\alpha$ | $(-\cos\alpha,\operatorname{sen}\alpha)$ | $\operatorname{sen}\alpha$ | $-\cos\alpha$ |
| $180°+\alpha$ | $(-\cos\alpha,-\operatorname{sen}\alpha)$ | $-\operatorname{sen}\alpha$ | $-\cos\alpha$ |
| $360°-\alpha$ | $(\cos\alpha,-\operatorname{sen}\alpha)$ | $-\operatorname{sen}\alpha$ | $\cos\alpha$ |
Demo: reading a reflection
The $150°=180°-30°$ point is the reflection of the $30°$ point on the vertical axis. A reflection on this axis changes $(x,y)$ to $(-x,y)$. Therefore, without needing to memorize two new measurements:
$$\operatorname{sen}150°=\operatorname{sen}30°=\frac12,$$$$\cos150°=-\cos30°=-\frac{\sqrt3}{2}.$$
The tangent is the quotient of these measurements. So, in $150°$, it is $-1/\sqrt3=-\sqrt3/3$. The same reasoning justifies each line in the table.
The sign map
In the first quadrant, sine and cosine are positive. In the second, positive sine and negative cosine. In the third, both negative. In the fourth, negative sine and positive cosine. The tangent is positive when the two signs are the same and negative when they are different.
Do not apply this rule to axes as if they were quadrants: there a coordinate is zero and some functions cease to exist.
Negative spins, even and odd functions
The angle points $\theta$ and $-\theta$ are reflections on the horizontal axis. Therefore:
$$\cos(-\theta)=\cos\theta,\qquad \operatorname{sen}(-\theta)=-\operatorname{sen}\theta.$$A function with $f(-x)=f(x)$ is par; one with $f(-x)=-f(x)$ is odd. Cosine and secant are pairs. Sine, tangent, cotangent and cosecant are odd, always in their respective domains.
Swap height for width
In a right triangle, the two acute angles add up to $90°$. The side opposite to one is adjacent to the other. Like this:
$$\operatorname{sen}(90°-\alpha)=\cos\alpha,$$$$\cos(90°-\alpha)=\operatorname{sen}\alpha.$$
This is the reason for the name cosine: the sine of the complement. For tangents, $\tan(90°-\alpha)=\cot\alpha$, when expressions exist. In radians, replace $90°$, $180°$, and $360°$ with $\pi/2$, $\pi$, and $2\pi$.
A three-move method
To study $765°$, remove two turns: $765°-720°=45°$. To study $225°$, write $180°+45°$. First identify the quadrant, then the reference angle and finally the sign of the function. This method allows you to check the result using the drawing.
In everyday life
- Rose of the winds. Northeast, northwest, southwest and southeast are reflections of each other: the direction $45^\circ$ is known, the other three only exchange signs.
- Calculator. Before calculating $\cos 300^\circ$, the machine uses symmetry: $\cos 300^\circ = \cos 60^\circ = \frac{1}{2}$.
- Clock hands. The position of the hand at 7 h is the reflection of the position at 1 h in relation to the center, with the two sign coordinates swapped.
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. Calculate $\operatorname{sen}210°$.
See resolution 1
$210°=180°+30°$. In the third quadrant the sine is negative: $-1/2$.
2. Calculate $\cos330°$.
See resolution 2
$330°=360°-30°$. The cosine is positive: $\sqrt3/2$.
3. Calculate $\tan135°$.
See resolution 3
$135°=180°-45°$. The tangent is negative and its modulus is 1; result $-1$.
4. Simplify $\operatorname{sen}(-\theta)+\operatorname{sen}\theta$.
See resolution 4
The sine is odd; the sum is $-\operatorname{sen}\theta+\operatorname{sen}\theta=0$.
5. Calculate $\cos765°$.
See resolution 5
We remove $720°$, leaving $45°$. The result is $\sqrt2/2$.
6. If $\operatorname{sen}20°=a$, write $\cos70°$ using $a$.
See resolution 6
As $70°=90°-20°$, we have $\cos70°=\operatorname{sen}20°=a$.