Exercises from chapter II — Circular functions

We build a table in three steps: 1st) we assign values ​​to $x$; 2nd) we associate the value of $\operatorname{sen} x$ to each $x$; 3rd) we multiply $\operatorname{sen} x$ by $-1$.

With this table we obtain 5 points of the graph, which is the symmetric of the sinusoid in relation to the $x$ axis. It is immediate that $\operatorname{Im}(f) = [-1, 1]$ and $p(f) = 2\pi$.

Same three steps, now multiplying $\operatorname{sen} x$ by 2:

For each $x$, the ordinate is twice the ordinate of the sinusoid. It is immediate that $\operatorname{Im}(f) = [-2, 2]$ and $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-2, 2]$ e $p(f) = 2\pi$.

For a real number $a$: if $a \ge 0$, $|a| = a$; if $a < 0$, $|a| = -a$. Applying the definition:

• $\operatorname{sen} x \ge 0 \Rightarrow |\operatorname{sen} x| = \operatorname{sen} x$: in these sections the graphs of $y = |\operatorname{sen} x|$ and $y = \operatorname{sen} x$ coincide;
• $\operatorname{sen} x < 0 \Rightarrow |\operatorname{sen} x| = -\operatorname{sen} x$: in these sections the graphs are symmetric in relation to the $x$ axis.

It is immediate that $\operatorname{Im}(f) = [0, 1]$ and $p(f) = \pi$.

$\operatorname{Im}(f) = [0, 3]$ e $p(f) = \pi$.

Table in three steps: 1st) we assign values ​​to $t = 2x$; 2nd) we associate the value $\operatorname{sen} 2x$ with each $2x$; 3rd) we calculate $x = \frac{t}{2}$.

For each $x$, the ordinate is the sine of double of $x$. For $\operatorname{sen} t$ to complete a period, $t = 2x$ must go through $[0, 2\pi]$, that is, $x$ must go through $[0, \pi]$. So $p(f) = \pi - 0 = \pi$, and $\operatorname{Im}(f) = [-1, 1]$.

We do $t = \frac{x}{2}$, that is, $x = 2t$:

It is immediate that $\operatorname{Im}(f) = [-1, 1]$ and $p(f) = 4\pi$.

With $t = 3x$, that is, $x = \frac{t}{3}$:

It is immediate that $\operatorname{Im}(f) = [-1, 1]$ and $p(f) = \dfrac{2\pi}{3}$.

$\operatorname{Im}(f) = [-1, 1]$ e $p(f) = 6\pi$.

$\operatorname{Im}(f) = [-3, 3]$ e $p(f) = \dfrac{\pi}{2}$.

For each $x$, the ordinate is the sine of $x$ plus one unit. If each sine undergoes an increase of 1, the sinusoid undergoes a translation of one unit upwards. It is immediate that $\operatorname{Im}(f) = [0, 2]$ and $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-3, -1]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-1, 3]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [1, 3]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-2, 0]$ e $p(f) = \pi$.

$\operatorname{Im}(f) = [-2, 4]$ e $p(f) = 4\pi$.

With $t = x - \frac{\pi}{4}$, that is, $x = t + \frac{\pi}{4}$:

For $\operatorname{sen} t$ to complete a period, $t$ must go through $[0, 2\pi]$, that is, $x$ must go through $\left[\frac{\pi}{4}, \frac{9\pi}{4}\right]$. Then

The graph is the sinusoid shifted from $\frac{\pi}{4}$ to the right.

$\operatorname{Im}(f) = [-1, 1]$ and $p(f) = 2\pi$; a period goes from $-\frac{\pi}{3}$ to $\frac{5\pi}{3}$.

$\operatorname{Im}(f) = [-1, 1]$ and $p(f) = \pi$; a period goes from $\frac{\pi}{6}$ to $\frac{7\pi}{6}$.

$\operatorname{Im}(f) = [-1, 3]$ and $p(f) = 4\pi$; a period goes from $\frac{\pi}{3}$ to $\frac{13\pi}{3}$.

Let’s do $cx + d = t$. When $x$ goes through $\mathbb{R}$, $t$ also goes through $\mathbb{R}$ (the affine function $t = cx + d$ is surjective). As a result, $\operatorname{sen} t$ traverses the range $[-1, 1]$, $b \cdot \operatorname{sen} t$ traverses $[-b, b]$, and $y = a + b \cdot \operatorname{sen} t$ traverses $[a - b, a + b]$, which is the image of $f$.

For $f$ to complete a period, $t$ must range from $0$ to $2\pi$:

therefore

$\operatorname{Im}(f) = [-1, 3]$ and $p(f) = \pi$; a period goes from $\frac{\pi}{6}$ to $\frac{7\pi}{6}$.

To exist $x$ satisfying equality, we must have

$\operatorname{Im}(f) = [-1, 1]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-2, 2]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-3, 3]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [0, 1]$ e $p(f) = \pi$.

$\operatorname{Im}(f) = [-1, 1]$ e $p(f) = \pi$.

$\operatorname{Im}(f) = [-1, 1]$ e $p(f) = 4\pi$.

$\operatorname{Im}(f) = [0, 2]$ e $p(f) = 2\pi$.

$\operatorname{Im}(f) = [-1, 3]$ e $p(f) = \dfrac{2\pi}{3}$.

$\operatorname{Im}(f) = [-1, 1]$ and $p(f) = 2\pi$; a period goes from $\frac{\pi}{4}$ to $\frac{9\pi}{4}$.

$\operatorname{Im}(f) = [-2, 2]$ and $p(f) = 2\pi$; a period goes from $\frac{\pi}{3}$ to $\frac{7\pi}{3}$.

$\operatorname{Im}(f) = [-3, 1]$ and $p(f) = \dfrac{2\pi}{3}$; a period goes from $\frac{\pi}{12}$ to $\frac{3\pi}{4}$.

Examining the cycle, we notice that $|\operatorname{sen} 135^\circ| = |\cos 135^\circ|$ and that

because in this arc the point $P$ is further from the $u$ axis than the $v$ axis.

As $\operatorname{sen} 107^\circ > 0$, $\cos 107^\circ < 0$ and $|\operatorname{sen} 107^\circ| > |\cos 107^\circ|$, it follows

For each $x$, this function associates a $y$ which is the sum of the sine with the cosine of $x$. We place the sinusoid and cosineoid in the same diagram and, for each $x$, we add the ordinates of the points found on each curve.

We will see later that $\operatorname{Im}(f) = \left[-\sqrt{2}, \sqrt{2}\right]$ and $p(f) = 2\pi$.

Let $P$ be the image of $x$, inside the arc $\overset{\frown}{AB}$, and let $P_1$ and $P_2$ be its projections on the axes: $OP_2 = \cos x$ and $P_2P = OP_1 = \operatorname{sen} x$, both positive. In the triangle $OP_2P$ (rectangle in $P_2$), the triangular inequality gives

The inequality is strict because $O$, $P_2$, and $P$ are not aligned when $0 < x < \frac{\pi}{2}$.

Let’s do $2x = t$. There is $\operatorname{tg} t$ if and only if $t \ne \frac{\pi}{2} + k\pi$ ($k \in \mathbb{Z}$). Then

Let’s do $x - \frac{\pi}{4} = t$. We have:

For $\operatorname{tg} t$ to describe a complete period we must have

then $p(f) = \dfrac{3\pi}{4} - \left(-\dfrac{\pi}{4}\right) = \pi$.

As the function associates the value $\operatorname{tg}\left(x - \frac{\pi}{4}\right)$ to each $x$, the graph is (by analogy with the functions already seen) the tangentoid shifted from $\frac{\pi}{4}$ to the right.

$D(f) = \left\{x \in \mathbb{R} \;\middle|\; x \ne \dfrac{\pi}{6} + \dfrac{k\pi}{2},\ k \in \mathbb{Z}\right\}$ and $p(f) = \dfrac{\pi}{2}$; a branch goes from $-\frac{\pi}{3}$ to $\frac{\pi}{6}$ and cuts the $x$ axis at $-\frac{\pi}{12}$.