Fundamental relationships, simplification of expressions, reduction to the first quadrant and parity (chapters III and IV).
Teste 41PUC-75
The value of the expression $25\operatorname{sen}^2 x - 9\operatorname{tg}^2 x$, knowing that $\operatorname{cossec} x = \frac{5}{4}$ and $x$ are from the first quadrant, is:
$\operatorname{sen} x = \frac{4}{5}$, $\cos x = \frac{3}{5}$ e $\operatorname{tg} x = \frac{4}{3}$: $25 \cdot \frac{16}{25} - 9 \cdot \frac{16}{9} = 16 - 16 = 0$.
Teste 42GV-76
If $\operatorname{sen} a = \frac{24}{25}$ and $\sec a$ is negative, then the value of $\sqrt{\dfrac{1 - \cos a}{1 + \cos a}}$ is:
Negative secant means negative cosine: $\cos a = -\frac{7}{25}$. Then $\sqrt{\frac{1 + 7/25}{1 - 7/25}} = \sqrt{\frac{32}{18}} = \frac{4}{3}$.
Teste 43ITA-74
The value of the expression $x = \dfrac{2\operatorname{tg}\theta}{1 - \operatorname{tg}^2\theta}$, when $\cos\theta = -\frac{3}{7}$ and $\operatorname{tg}\theta < 0$, is:
With $\cos\theta = -\frac{3}{7}$ and $\operatorname{tg}\theta < 0$, the sine is positive: $\operatorname{sen}\theta = \frac{2\sqrt{10}}{7}$ and $\operatorname{tg}\theta = -\frac{2\sqrt{10}}{3}$. So $x = \frac{-4\sqrt{10}/3}{1 - 40/9} = \frac{12\sqrt{10}}{31}$, which does not appear in the alternatives.
Teste 44CESCEA-70
If $\operatorname{sen} x = \frac{n - 1}{n}$, then $\dfrac{\operatorname{tg}^2 x + 1}{\operatorname{cotg}^2 x + 1}$ is equal to:
It is known that $\operatorname{sen} x = a \ne 0$ and $\cos x = b \ne 0$. Therefore, $\operatorname{tg} x + \operatorname{cotg} x$ is equal to:
$\operatorname{tg} x + \operatorname{cotg} x = \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} = \frac{1}{ab}$, because $a^2 + b^2 = \operatorname{sen}^2 x + \cos^2 x = 1$.
Teste 46MACK-73
The roots of the equation $2x^2 - px - 1 = 0$ are $\operatorname{sen}\theta$ and $\cos\theta$, where $\theta$ is a real number. The value of $p$ is:
Sum of roots: $\operatorname{sen}\theta + \cos\theta = \frac{p}{2}$; product: $\operatorname{sen}\theta\cos\theta = -\frac{1}{2}$. Then $\frac{p^2}{4} = 1 + 2\left(-\frac{1}{2}\right) = 0$, then $p = 0$.
Teste 47ITA-71
Let $x \in \left]0, \frac{\pi}{2}\right[$. Which statement below is true?
With $t = \operatorname{tg} x > 0$: $t + \frac{1}{t} \ge 2$, because $t + \frac{1}{t} - 2 = \frac{(t - 1)^2}{t} \ge 0$ (equality only in $x = \frac{\pi}{4}$).
Teste 48CESCEA-72
Mark the statement false:
$\operatorname{sen}^4 x + \cos^4 x = 1 - \frac{1}{2}\operatorname{sen}^2 2x$, which is only valid 1 when $\operatorname{sen} 2x = 0$; is not all $\mathbb{R}$. The rest are true: in (b), $3\operatorname{sen}^2 + 2\cos^2 = 2 + \operatorname{sen}^2 \le 3$; in (d), $\sec^2 x = \operatorname{tg}^2 x + 1$ always.
Teste 49CESCEM-70
If $\theta \ne \frac{\pi}{2} + 2k\pi$, $k$ integer, then $\dfrac{\cos^2\theta}{1 - \operatorname{sen}\theta}$ is equal to:
$\frac{\cos^2\theta}{1 - \operatorname{sen}\theta} = \frac{(1 - \operatorname{sen}\theta)(1 + \operatorname{sen}\theta)}{1 - \operatorname{sen}\theta} = 1 + \operatorname{sen}\theta$. (The printed statement has $\theta = \frac{\pi}{2} + 2k\pi$, precisely the values in which the expression does not exist; the correct sign is $\ne$.)
Teste 50PUC-70
The expression $\dfrac{\operatorname{cossec} x - \operatorname{sen} x}{\sec x - \cos x}$ is identically equal to:
The roots of the equation $x^2 - (2\operatorname{tg} a)x - 1 = 0$ are:
$x = \frac{2\operatorname{tg} a \pm \sqrt{4\operatorname{tg}^2 a + 4}}{2} = \operatorname{tg} a \pm \sqrt{\sec^2 a} = \operatorname{tg} a \pm \sec a$.
Teste 54ITA-73
Eliminating $\theta$ in equations $x\operatorname{sen}\theta + y\cos\theta = 2a\operatorname{sen}\theta$ and $x\cos\theta - y\operatorname{sen}\theta = a\cos\theta$, with $a > 0$, we have:
Solving the system in $x$ and $y$: $x = a(1 + \operatorname{sen}^2\theta)$ and $y = a\operatorname{sen}\theta\cos\theta$. None of the relations offered is valid for all $\theta$: alternative (c), for example, is valid for $\theta = 0$ ($x = a$, $y = 0$), but fails for $\theta = \frac{\pi}{2}$ ($x = 2a$, $y = 0$).
Teste 55MACK-76
The value of $k$ for which $(\cos x + \operatorname{sen} x)^2 + k\operatorname{sen} x\cos x - 1 = 0$ is an identity is:
$(\cos x + \operatorname{sen} x)^2 = 1 + 2\operatorname{sen} x\cos x$; the expression is valid $(2 + k)\operatorname{sen} x\cos x$, which is identically null only for $k = -2$.
Teste 56CESGRANRIO-76
In the figure, the radius $\overline{OA}$ of the circle is worth 6, the segment $\overline{OB}$ is worth 3 and the segment $\overline{CB}$ is perpendicular to $\overline{OA}$. The measure, in radians, of the angle $\theta = C\hat{O}B$ is:
In the right triangle $OBC$, $OC = 6$ (radius) and $OB = 3$: $\cos\theta = \frac{3}{6} = \frac{1}{2}$, therefore $\theta = \frac{\pi}{3}$.
Teste 57CESCEA-75
In the figure, the circle has its center at the origin and radius 1; the ray that forms $45^\circ$ with the axis $X$ cuts the circle at $M$ and the tangent line to the circle at the point $(1, 0)$ at $N$. The length of segment $\overline{MN}$ is:
$N$ is on the tangent axis, with $AN = \operatorname{tg} 45^\circ = 1$; then $ON = \sqrt{2}$. Such as $OM = 1$ (radius), $MN = \sqrt{2} - 1$.
Teste 58GV-74
The expression $\sqrt{\cos\pi + \log_2 16 - e^{\operatorname{sen} 2\pi}}$ has the same numerical value as:
Knowing that $x + y = \frac{\pi}{3}$ and $x - y = \frac{\pi}{2}$, then $\operatorname{sen} x + \operatorname{sen} y$ is equal to:
$\operatorname{sen} x + \operatorname{sen} y = 2\operatorname{sen}\frac{x + y}{2}\cos\frac{x - y}{2} = 2\operatorname{sen}\frac{\pi}{6}\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$.
Teste 60CESCEM-75
The sine of one of the acute angles of a rhombus is equal to $\frac{1}{2}$; Therefore, the tangent of the largest internal angle is:
The acute angle measures $30^\circ$ and the largest internal angle, $150^\circ$: $\operatorname{tg} 150^\circ = -\frac{\sqrt{3}}{3}$.
Teste 61FAAP-75
Once the formula $\operatorname{sen}^2 x + \operatorname{sen}^2 2x + \operatorname{sen}^2 3x + \cdots + \operatorname{sen}^2 nx = \dfrac{n}{2} - \dfrac{\operatorname{sen}(nx)\cos[(n + 1)x]}{2\operatorname{sen} x}$ is known, valid for all $x \in \mathbb{R}$ such that $\operatorname{sen} x \ne 0$, the sum $\operatorname{sen}^2\frac{\pi}{3} + \operatorname{sen}^2\frac{2\pi}{3} + \operatorname{sen}^2\frac{3\pi}{3} + \cdots + \operatorname{sen}^2\frac{9\pi}{3}$ is worth:
With $n = 9$ and $x = \frac{\pi}{3}$: $\operatorname{sen}(9x) = \operatorname{sen} 3\pi = 0$, and the sum is $\frac{9}{2}$. (Checking: the squared sines are worth $\frac{3}{4}$, $\frac{3}{4}$, $0$, repeated three times: $3 \cdot \frac{3}{2} = \frac{9}{2}$.)
Teste 62CESCEM-74
Given the angle $\alpha = 1782^\circ$, then:
$1782^\circ = 4 \cdot 360^\circ + 342^\circ$, and $342^\circ = 360^\circ - 18^\circ$ is in the 4th quadrant: sine and tangent change signs, cosine does not.
Teste 63FEI-66
If $\cos x = \frac{3}{5}$, then $\operatorname{sen}\left(x + \frac{\pi}{2}\right)$ is equal to:
$\operatorname{sen}\left(x + \frac{\pi}{2}\right) = \cos x = \frac{3}{5}$.
Teste 64MACK-76
If $x = \frac{\pi}{5}$, the value of $2\cos\pi\operatorname{sen}(\pi - x)\operatorname{sen}\left(\frac{3\pi}{2} + x\right)$ is equal to:
Numerator: $-\operatorname{sen} x - \cos x + \cos x + 3\operatorname{sen} x = 2\operatorname{sen} x$. Denominator: $-\cos x - \cos x - \operatorname{sen} x + \operatorname{sen} x = -2\cos x$. Quotient: $-\operatorname{tg} x$.
Teste 66MACK-75
The sum of the 12 first terms of the sequence $\cos\alpha$, $\cos(\alpha + \pi)$, $\cos(\alpha + 2\pi)$, … is:
The terms are $\cos\alpha, -\cos\alpha, \cos\alpha, -\cos\alpha, \ldots$; twelve terms cancel out in pairs.
Teste 67FFCLUSP-67
$\log\operatorname{tg} 1^\circ + \log\operatorname{tg} 2^\circ + \log\operatorname{tg} 3^\circ + \cdots + \log\operatorname{tg} 89^\circ$ is equal to:
The sum of the logarithms is the logarithm of the product $\operatorname{tg} 1^\circ \cdot \operatorname{tg} 2^\circ \cdots \operatorname{tg} 89^\circ$. Like $\operatorname{tg} k^\circ \cdot \operatorname{tg}(90^\circ - k^\circ) = 1$ and $\operatorname{tg} 45^\circ = 1$, the product is 1 and the logarithm, 0.
Teste 68PUC-77
Which of the functions below is an even function?
$f(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = f(x)$. The other four are odd.
Teste 69CESCEM-71
We say that a real function is even if $f(x) = f(-x)$ and that it is odd if $f(x) = -f(-x)$. Of the following statements, indicate which is false:
If $f$ and $g$ are odd, $(fg)(-x) = (-f(x))(-g(x)) = f(x)g(x)$: the product is even. Statements (b), (c) and (d) are true.
Instructions for tests 70 to 74 (CESCEA-71). In each of them the graph of a function defined in $\mathbb{R}$ is given. Consider the denominations: I — odd function; II — non-limited function; III — periodic function; IV — even function; V — identity function.
Teste 70CESCEA-71
Which alternative describes the graph below?
The graph is symmetrical in relation to the $y$ axis (even function) and grows without limit: II and IV. It is function $|x|$.
Teste 71CESCEA-71
Which alternative describes the graph below?
It is the tangent: odd, unlimited and periodic (I, II and III).
Teste 72CESCEA-71
Which alternative describes the graph below?
It is the straight line $y = x$: the identity function, which is odd and not limited (I, II and V).
Teste 73CESCEA-71
Which alternative describes the graph below?
The graph is symmetrical about the axis $y$ (even) and descends without limit near $x = 0$ (not limited): II and IV.
Teste 74CESCEA-71
Which alternative describes the graph below?
It is a triangular wave: it repeats every 4 units (periodic) and is symmetrical in relation to the $y$ axis (pair): III and IV.
