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

Two measures that combine

Mark a point with coordinates $(3/5,4/5)$ in the first quadrant. Its horizontal and vertical displacements form the legs of a right-angled triangle. The distance to the center is 1. This is a point in the cycle because $(3/5)^2+(4/5)^2=9/25+16/25=1$.

Demonstration: the fundamental identity

The Pythagorean theorem says that, in a right-angled triangle, the square of the hypotenuse is the sum of the squares of the legs. For a point in the cycle, the lengths of the legs are $|\cos\theta|$ and $|\operatorname{sen}\theta|$, and the hypotenuse measures 1. Squaring eliminates the signs. Therefore, in any quadrant:

$$\operatorname{sen}^2\theta+\cos^2\theta=1.$$

The writing $\operatorname{sen}^2\theta$ means $(\operatorname{sen}\theta)^2$. At the axis points, the equality also applies directly: one coordinate is zero and the other is 1 or −1.

We call this equality identity: it applies to every angle of its domain. $\operatorname{sen}\theta=1/2$ is an equation that only applies to certain angles.

Two new identities

If $\cos\theta\ne0$, divide each identity term by $\cos^2\theta$:

$$\frac{\operatorname{sen}^2\theta}{\cos^2\theta}+1 =\frac1{\cos^2\theta}.$$

By definitions, $1+\tan^2\theta=\sec^2\theta$. If $\operatorname{sen}\theta\ne0$, division by $\operatorname{sen}^2\theta$ gives $1+\cot^2\theta=\csc^2\theta$.

The conditions follow the formula. An equality with fractions can only be used where all fractions exist.

Example: the sign that the square hides

We know that $\operatorname{sen}\theta=3/5$. Then:

$$\cos^2\theta=1-\frac9{25}=\frac{16}{25}.$$

Thus, $\cos\theta=4/5$ or $-4/5$. If the point is in the second quadrant, we choose $-4/5$ as it is left of center. In this case, $\tan\theta=(3/5)/(-4/5)=-3/4$.

How to demonstrate an identity

Start on one side and use already known equalities until you reach the other. It is not enough to test one angle: a claim about all angles requires a general argument.

For example, for $\cos\theta\ne0$:

$$\begin{aligned} (1-\operatorname{sen}\theta)(1+\operatorname{sen}\theta) &=1-\operatorname{sen}^2\theta\\ &=\cos^2\theta. \end{aligned}$$

The equality obtained is valid even when the cosine is zero, as no division was made. If we divide both sides by $\cos^2\theta$ to get 1, we now require $\cos\theta\ne0$. Simplifying an expression does not return points that were prohibited at the beginning.

Bug lab

Someone wrote $\operatorname{sen}\theta+\cos\theta=1$, because the sum of the squares is 1. Try $\theta=45°$: the sum is $\sqrt2$, different from 1. A single counterexample demolishes a universal claim. Keeping this difference helps in all mathematics.

In everyday life

  • Cart on the ramp. The weight $P$ of a cart on an inclined ramp $\alpha$ is divided into $P\operatorname{sen}\alpha$, which makes it go down, and $P\cos\alpha$, which presses it against the ramp; and $(P\operatorname{sen}\alpha)^2 + (P\cos\alpha)^2 = P^2$.
  • Screen size. Width and height of a screen are the cosine and sine of the diagonal, and therefore $\text{largura}^2 + \text{altura}^2 = \text{diagonal}^2$.
  • Energy bills. Industries pay a fine when the “power factor” $\cos\varphi$ is low; the corresponding sine leaves the fundamental relation.
Width² + height² = diagonal²: the fundamental relationship on a TV screen.

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. If $\cos\theta=5/13$ and $\theta$ is in the first quadrant, find the sine.

See resolution 1

$\operatorname{sen}^2\theta=1-25/169=144/169$. As it is positive, $\operatorname{sen}\theta=12/13$.

2. If $\tan\theta=2$, how much is $\sec^2\theta$ worth?

See resolution 2

$1+2^2=5$. This determines the square of the secant, without determining its sign.

3. Demonstrate $\operatorname{sen}\theta\,\csc\theta=1$. What is the condition?

See resolution 3

By definition, $\operatorname{sen}\theta\cdot(1/\operatorname{sen}\theta)=1$ as long as $\operatorname{sen}\theta\ne0$.

4. Simplify $(1-\cos^2\theta)/\operatorname{sen}\theta$.

See resolution 4

The numerator is $\operatorname{sen}^2\theta$. The result is $\operatorname{sen}\theta$, with the original condition $\operatorname{sen}\theta\ne0$.

5. If $\operatorname{sen}\theta=-8/17$ and the point is in the third quadrant, find the cosine.

See resolution 5

$\cos^2\theta=225/289$ and the cosine is negative: $-15/17$.

6. Is equality $\tan\theta=\operatorname{sen}\theta$ an identity?

See resolution 6

No: in $45°$, the tangent is 1 and the sine is $\sqrt2/2$. This counterexample is enough.

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