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

Adding turns is not adding heights

A turn of $30°$ followed by another of $30°$ results in $60°$. But $\operatorname{sen}30°+\operatorname{sen}30°=1$, while $\operatorname{sen}60°=\sqrt3/2$. We need to find out how the coordinates change when rotating.

This is an advanced step. If the letter beads are still new, follow the example of $75°$ first and return to the demonstration with an adult.

Demonstration of addition formulas

After a rotation of angle $a$, the horizontal vector of length 1 points to $(\cos a,\operatorname{sen} a)$. The vertical vector, perpendicular to it and also of length 1, points to $(-\operatorname{sen} a,\cos a)$. It is the first rotated vector plus $90°$.

Before this rotation, the angle point $b$ is the sum of a horizontal displacement of size $\cos b$ and a vertical displacement of size $\operatorname{sen} b$. Rotation preserves lengths and sum of displacements. After rotating $a$, its position is:

$$\begin{aligned} (\cos(a+b),\operatorname{sen}(a+b)) &=\cos b(\cos a,\operatorname{sen} a)\\ &\quad+\operatorname{sen} b(-\operatorname{sen} a,\cos a). \end{aligned}$$

Comparing the coordinates, we have:

$$\cos(a+b)=\cos a\cos b-\operatorname{sen} a\operatorname{sen} b,$$

$$\operatorname{sen}(a+b)=\operatorname{sen} a\cos b+\cos a\operatorname{sen} b.$$

Replacing $b$ with $-b$ and using parity gives:

$$\cos(a-b)=\cos a\cos b+\operatorname{sen} a\operatorname{sen} b,$$

$$\operatorname{sen}(a-b)=\operatorname{sen} a\cos b-\cos a\operatorname{sen} b.$$

Dividing the sine formula by the cosine and then by $\cos a\cos b$, we obtain:

$$\tan(a\pm b)=\frac{\tan a\pm\tan b}{1\mp\tan a\tan b}.$$

This last expression requires that the tangents of $a$ and $b$ exist and that the final denominator is not zero.

Example: a new value

As $75°=45°+30°$:

$$\begin{aligned} \operatorname{sen}75°&=\frac{\sqrt2}{2}\frac{\sqrt3}{2} +\frac{\sqrt2}{2}\frac12\\ &=\frac{\sqrt6+\sqrt2}{4}. \end{aligned}$$

Double and triple

Do $b=a$ in addition formulas:

$$\operatorname{sen}2a=2\operatorname{sen} a\cos a,$$

$$\cos2a=\cos^2a-\operatorname{sen}^2a=1-2\operatorname{sen}^2a=2\cos^2a-1.$$

The first form of the cosine flips the others when using $\operatorname{sen}^2a+\cos^2a=1$. For tangent, $\tan2a=2\tan a/(1-\tan^2a)$, when terms exist.

Applying addition to $2a+a$ and substituting the previous formulas produces $\operatorname{sen}3a=3\operatorname{sen} a-4\operatorname{sen}^3a$ and $\cos3a=4\cos^3a-3\cos a$. It is possible to multiply an angle by repeating the same idea.

Divide the angle in half

In the expressions for $\cos2a$, replace $a$ with $\theta/2$ and isolate the squares:

$$\operatorname{sen}^2\frac\theta2=\frac{1-\cos\theta}{2},\qquad \cos^2\frac\theta2=\frac{1+\cos\theta}{2}.$$

When extracting the root, the sign depends on the quadrant of $\theta/2$. If $0<\theta<180°$, both roots are positive. Dividing sine by cosine, we also get:

$$\tan\frac\theta2=\frac{\operatorname{sen}\theta}{1+\cos\theta}.$$

This form requires $1+\cos\theta\ne0$. The equivalent form $(1-\cos\theta)/\operatorname{sen}\theta$ still requires $\operatorname{sen}\theta\ne0$.

Transform sums into products

Add or subtract addition and subtraction formulas. Then write $u=a+b$ and $v=a-b$, so that $a=(u+v)/2$ and $b=(u-v)/2$:

$$\operatorname{sen} u+\operatorname{sen} v=2\operatorname{sen}\frac{u+v}{2}\cos\frac{u-v}{2},$$

$$\operatorname{sen} u-\operatorname{sen} v=2\cos\frac{u+v}{2}\operatorname{sen}\frac{u-v}{2},$$

$$\cos u+\cos v=2\cos\frac{u+v}{2}\cos\frac{u-v}{2},$$

$$\cos u-\cos v=-2\operatorname{sen}\frac{u+v}{2}\operatorname{sen}\frac{u-v}{2}.$$

For example, $\operatorname{sen}75°+\operatorname{sen}15°=2\operatorname{sen}45°\cos30°=\sqrt6/2$. Changing shape can make a bill shorter.

In everyday life

  • Rotate photos. The cell phone rotates a photo by calculating, for each point, the formulas for adding arcs.
  • Tune instruments. Two almost identical notes produce a “pulse” in volume — the transformation from sum to product explains this beat, which musicians use to tune.
  • Throws. The range of a throw depends on $\operatorname{sen} 2\theta$ (double arc), and therefore the angle of $45^\circ$ is the one that goes the furthest without air resistance.
The beat between two nearby notes: sum transformed into product.

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 $\cos75°$ using $45°+30°$.

See resolution 1

$\cos75°=(\sqrt6-\sqrt2)/4$, using the sum cosine formula.

2. If $\operatorname{sen} a=3/5$ and $\cos a=4/5$, calculate $\operatorname{sen}2a$ and $\cos2a$.

See resolution 2

$\operatorname{sen}2a=2(3/5)(4/5)=24/25$; $\cos2a=(16-9)/25=7/25$.

3. Deduct $\operatorname{sen}15°$ using subtraction.

See resolution 3

$\operatorname{sen}(45°-30°)=(\sqrt6-\sqrt2)/4$.

4. Find $\cos22{,}5°$ by half formula.

See resolution 4

The angle is in the first quadrant. We choose the positive root: $\sqrt{(1+\sqrt2/2)/2}=\sqrt{2+\sqrt2}/2$.

5. Transform $\cos80°+\cos20°$ into product.

See resolution 5

The average is $50°$ and half the difference is $30°$. Result: $2\cos50°\cos30°=\sqrt3\cos50°$.

6. Use the half tangent formula to find $\tan30°$.

See resolution 6

With $\theta=60°$: $(\sqrt3/2)/(1+1/2)=\sqrt3/3$.

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