Arc sine, arc cosine and arc tangent (chapter VII).
Teste 158MACK-74
Let $f$, $g$ and $h$ be functions of $A$ in $A$, where $A = [-1, 1]$, defined as follows: $f(x) = \operatorname{sen} x$, $g(x) = \operatorname{sen}\pi x$, $h(x) = \frac{\pi}{2}x$. We can state that:
$f$ (sine in $[-1, 1]$) is injective, with image $[-\operatorname{sen} 1, \operatorname{sen} 1]$; $g(x) = \operatorname{sen}\pi x$ has image $[-1, 1]$ (surjective), but is not injective ($g(0) = g(1) = 0$); $h$ is injective and its image $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ doesn’t even fit into $A$. Only $g$ is surjective. The printed template indicates (e), but none of the three is both an injector and a surjector of $A$ in $A$; the true statement is (d).
Teste 159MACK-73
The domain of the function defined by $y = \operatorname{arc\,sen}\sqrt{2x - 3}$ is:
The root requires $2x - 3 \ge 0$ and the arcsine requires $\sqrt{2x - 3} \le 1$: $0 \le 2x - 3 \le 1$, that is, $\frac{3}{2} \le x \le 2$.
Teste 160MACK-77
The value of $\operatorname{arc\,sen}\left(\cos\frac{33\pi}{5}\right)$ is:
$\cos\frac{33\pi}{5} = \cos\frac{3\pi}{5} = \operatorname{sen}\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) = \operatorname{sen}\left(-\frac{\pi}{10}\right)$, and $-\frac{\pi}{10}$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Teste 161MACK-74
The value of $\operatorname{tg}\left(2\operatorname{arc\,sen}\frac{\sqrt{3}}{2}\right)$ is:
$\operatorname{arc\,sen}\frac{\sqrt{3}}{2} = \frac{\pi}{3}$ e $\operatorname{tg}\frac{2\pi}{3} = -\sqrt{3}$.
Teste 162PUC-71
Since the determinations of the arcs are between $0$ and $\frac{\pi}{2}$, the value of the expression $y = \operatorname{sen}\left(\operatorname{arc\,sen}\dfrac{1}{1 + a^2} + \operatorname{arc\,cos}\dfrac{1}{1 + a^2}\right)$ is:
For all $u \in [-1, 1]$, $\operatorname{arc\,sen} u + \operatorname{arc\,cos} u = \frac{\pi}{2}$; then $y = \operatorname{sen}\frac{\pi}{2} = 1$.
Teste 163ITA-72
For all $\alpha$ and $\beta$, with $|\beta| < 1$, the expression $\operatorname{tg}(\operatorname{arc\,tg}\alpha + \operatorname{arc\,sen}\beta)$ is equal to:
With $\operatorname{tg} A = \alpha$ and $\operatorname{tg} B = \frac{\beta}{\sqrt{1 - \beta^2}}$: $\operatorname{tg}(A + B) = \frac{\alpha\sqrt{1 - \beta^2} + \beta}{\sqrt{1 - \beta^2} - \alpha\beta}$, which does not appear. The printed template indicates (a), but it fails, for example, for $\alpha = 1$, $\beta = 0$: $\operatorname{tg}\frac{\pi}{4} = 1$, while (a) gives $-1$.
Teste 164PUC-70
$2\operatorname{arc\,tg}\frac{1}{3} + \operatorname{arc\,tg}\frac{1}{7}$ is equal to:
It is the exercise 269: $\operatorname{tg}\left(2\operatorname{arc\,tg}\frac{1}{3}\right) = \frac{3}{4}$ and $\operatorname{tg}\left(2\operatorname{arc\,tg}\frac{1}{3} + \operatorname{arc\,tg}\frac{1}{7}\right) = \frac{3/4 + 1/7}{1 - 3/28} = 1$, with the sum between 0 and $\frac{\pi}{2}$.
Teste 165LINS-67
Admitting the variation of $\operatorname{arc\,sen} x$ in the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the solution to the equation $\operatorname{arc\,sen} x = 2\operatorname{arc\,sen}\frac{1}{2}$ is:
$2\operatorname{arc\,sen}\frac{1}{2} = \frac{\pi}{3}$, and $\operatorname{arc\,sen} x = \frac{\pi}{3} \iff x = \frac{\sqrt{3}}{2}$, which does not appear in the alternatives.
Teste 166MACK-75
The solution set of $\operatorname{arc\,sen} 2x - 3\operatorname{arc\,sen} x = 0$ has:
With $\theta = \operatorname{arc\,sen} x$: you need $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $2x = \operatorname{sen} 3\theta = 3x - 4x^3$, that is, $x(1 - 4x^2) = 0$. The three roots $0$ and $\pm\frac{1}{2}$ are fine ($\theta = \pm\frac{\pi}{6}$ gives $3\theta = \pm\frac{\pi}{2} = \operatorname{arc\,sen}(\pm 1)$).
Teste 167VALEPARAIBANA-72
Solving the equation $\operatorname{arc\,tg}\left(\dfrac{1 + e^x}{2}\right) + \operatorname{arc\,tg}\left(\dfrac{1 - e^x}{2}\right) = \dfrac{\pi}{4}$, we obtain:
Being $A = \frac{1 + e^x}{2}$ and $B = \frac{1 - e^x}{2}$, $A + B = 1$ and $\operatorname{tg}(\operatorname{arc\,tg} A + \operatorname{arc\,tg} B) = \frac{1}{1 - AB} = 1 \Rightarrow AB = 0 \Rightarrow e^{2x} = 1 \Rightarrow x = 0$. Check: $\operatorname{arc\,tg} 1 + \operatorname{arc\,tg} 0 = \frac{\pi}{4}$.
Teste 168MACK-76
Being $f(x) = \operatorname{arc\,sen} x$ and $g(x) = 1 + \operatorname{cotg}^2 x$, the value of $(g \circ f)\left(\frac{1}{3}\right)$ is:
$g(f(x)) = 1 + \operatorname{cotg}^2(\operatorname{arc\,sen} x) = \operatorname{cossec}^2(\operatorname{arc\,sen} x) = \frac{1}{x^2}$; for $x = \frac{1}{3}$, it is worth 9.
Teste 169ITA-71
Let us consider the equation $[\log_e(\operatorname{sen} x)]^2 - \log_e(\operatorname{sen} x) - 6 = 0$. The solution(s) of the equation is(are) given by:
With $L = \log_e(\operatorname{sen} x)$: $L^2 - L - 6 = 0 \Rightarrow L = 3$ or $L = -2$. Like $\operatorname{sen} x \le 1$, $L \le 0$; there remains $\operatorname{sen} x = e^{-2}$, that is, $x = \operatorname{arc\,sen}\frac{1}{e^2}$ (minus the arcs with the same sine).
Teste 170ITA-75
Let $S = \log_3(\operatorname{tg} x_1) + \log_3(\operatorname{tg} x_2) + \log_3(\operatorname{tg} x_3) + \cdots$, where $x_1 = \frac{\pi}{3}$ and $x_{n+1} = \operatorname{arc\,tg}\sqrt{\operatorname{tg} x_n}$, $n = 1, 2, 3, \ldots$ Under these conditions, we can ensure that:
$\operatorname{tg} x_1 = \sqrt{3} = 3^{1/2}$ and each tangent is the square root of the previous one: $\operatorname{tg} x_n = 3^{1/2^n}$. Then $S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$.
Teste 171ITA-72
Let us consider the function $S(x) = \sum\limits_{n=1}^{\infty}(\operatorname{sen} x)^n$, where $0 < x < \frac{\pi}{2}$. For what values of $x$ do we have $10 \le S(x) \le 20$?
With $s = \operatorname{sen} x \in\, ]0, 1[$, the geometric series sums to $S = \frac{s}{1 - s}$. $10 \le \frac{s}{1 - s} \le 20 \iff \frac{10}{11} \le s \le \frac{20}{21}$, and as the arcsine is increasing, we obtain (c).
