Tests — Transformations

It is known that $\operatorname{tg} 75^\circ = 2 + \sqrt{3}$ and $\operatorname{tg} 60^\circ = \sqrt{3}$. The value of $\operatorname{tg} 15^\circ$ is:

If $0 < a < \frac{\pi}{2}$ and $0 < b < \frac{\pi}{2}$, then:

For every real $x$, the relationship always holds:

The expression $N = \operatorname{sen}\alpha\cos\alpha\cos 2\alpha\cos 4\alpha\cos 8\alpha\cos 16\alpha\cos 32\alpha$ is equivalent to:

The shortest period of the function $f(x) = \operatorname{sen} x\cos x$ is:

Let the functions $f_1$ and $f_2$ of domain $\mathbb{R}$ be defined by $f_1(x) = \operatorname{sen} x + \cos x$ and $f_2(x) = 3\operatorname{sen} x\cos x$. Since $I_1$ and $I_2$ are the image sets of $f_1$ and $f_2$, respectively, we have that:

It is known that $\operatorname{sen} 2x = 2\operatorname{sen} x\cos x$. Therefore, $\operatorname{sen} 4x$ is equal to:

Since $x$ is an arc of the fourth quadrant and $\operatorname{sen} x = -\frac{1}{2}$, the value of $\operatorname{sen} 4x$ is:

If $\cos 2x = 2\cos^2 x - 1$, then the value of $\cos 4x$ is:

Knowing that $\cos 2x = \frac{2}{3}$, the value of $\operatorname{tg}^2 x$ is:

Since $x$ is an arc of the first quadrant and $\operatorname{sen} x = a$, the expression $2\cos 2x + \operatorname{sen}^2 2x$ is equal to:

Knowing that $\operatorname{sen} a = \frac{3}{5}$ and $\cos a = \frac{4}{5}$, then $\operatorname{sen} 2a + \cos 2a$ is equal to:

Let $f$ and $g$ be functions defined by $f(x) = \cos 2x$ and $g(x) = \operatorname{sen}^2 x - 1$. So $f(x) + g(x)$ is:

The period of the function $f$ defined by $f(x) = \operatorname{sen}^4 x$ is:

The period of function $f(x) = \operatorname{sen}^2 3x - \cos 4x$ is:

The expression $\dfrac{\operatorname{tg} x}{1 + \operatorname{tg} x} + \dfrac{\operatorname{tg} x}{1 - \operatorname{tg} x}$ is identical to:

Let $f(x) = \operatorname{tg}\left(x + \frac{\pi}{4}\right) \cdot \operatorname{tg}\left(x - \frac{\pi}{4}\right)$. We can state that:

Let $w = \operatorname{tg}\alpha + \operatorname{cotg}\alpha$, with $0 < \alpha < \frac{\pi}{2}$. Then:

If $\operatorname{tg} x = t$, then $\cos 2x + \operatorname{sen} 2x$ is equivalent to:

Another form for the expression $\dfrac{3\operatorname{sen} 2x}{1 - \cos 2x}$ is:

$\left[\dfrac{1 - \operatorname{tg} x}{1 + \operatorname{tg} x}\right]^2$ is worth:

If $\operatorname{tg} x = m$ and $\operatorname{tg} 2x = 3m$, with $m > 0$, the acute angle $x$ measures:

Let $D = \left\{x \in \mathbb{R} \;\middle|\; x \ne \log\frac{n\pi}{2},\ n = 1, 2, 3, \ldots\right\}$ (natural logarithm). With respect to the function $f\colon D \to \mathbb{R}$ defined by $f(x) = \dfrac{\operatorname{sen}(3e^x)}{\operatorname{sen} e^x} - \dfrac{\cos(3e^x)}{\cos e^x}$, we can state that:

With $u$ being the measure in radians of an angle and $v = \frac{\pi}{4} - u$, the expression $S = \dfrac{\operatorname{sen} u + \cos u}{\sqrt{2}\operatorname{sen} u\cos u}$, as a function of $x = \cos v$, is:

For $0 \le t \le 2\pi$, the expression $\frac{1}{2}\sqrt{2(1 - \cos t)}$ is equal to:

Let $0 < \alpha < \frac{\pi}{2}$. From the figure below (circle radius 1 and center at the origin) it can be directly concluded that:

If $\operatorname{cotg}\frac{a}{2} = \sqrt{3}$, then:

If $\operatorname{tg}\frac{a}{2} = \frac{1}{2}$, then $\operatorname{tg} a$ is valid:

Simplifying the expression $\dfrac{1}{1 + \sec x} \cdot \sqrt{\dfrac{1 + \cos x}{1 - \cos x}}$, we obtain:

The expression $\operatorname{tg}\frac{x}{2} + \operatorname{cotg}\frac{x}{2}$, for $0 < x < \frac{\pi}{2}$, is equivalent to:

If $0 < a_n < \frac{\pi}{2}$ and $\cos a_n = \frac{n}{n + 1}$, then $\cos\frac{a_n}{2}$ is valid:

The equality $\operatorname{tg} x = a\operatorname{cotg} x + b\operatorname{cotg} 2x$ is valid for every real $x$ such as $x \ne \frac{k\pi}{2}$. Then $a$ and $b$ are valid, respectively:

Knowing that $\operatorname{sen} x = \frac{m - n}{m + n}$, with $n > 0$ and $m > 0$, we can say that $\operatorname{tg}\left(\frac{\pi}{4} - \frac{x}{2}\right)$ is equal to:

Transforming the expression $\cos 70^\circ - \operatorname{sen} 60^\circ$ into a product, we obtain:

Select the true statement:

The expression $\operatorname{sen} x - \cos x$ is identical to:

The expression $\operatorname{sen}(135^\circ + x) + \operatorname{sen}(135^\circ - x)$ is equal to:

$\operatorname{sen}\alpha + 2\operatorname{sen} 2\alpha + \operatorname{sen} 3\alpha$ is equal to:

Let $P = \operatorname{sen}^2 ax - \operatorname{sen}^2 bx$. We then have to:

The smallest value that $y$ can assume in the equality $y = \cos x + \cos 2x$ is:

Being $\operatorname{sen} x - \operatorname{sen} y = 2\operatorname{sen}\frac{x - y}{2}\cos\frac{x + y}{2}$ and remembering that $|\operatorname{sen} z| \le |z|$, $|\cos t| \le 1$ and $|ab| = |a||b|$, we can state that, for any real numbers $x$ and $y$:

The expression $\dfrac{\operatorname{sen}(a - x) + \operatorname{sen}(2a - 3x)}{\cos(a - x) + \cos(2a - 3x)}$ is the same as:

With $u$ being the measure in radians of an angle and $v = \frac{\pi}{4} - u$, the expression $S = \dfrac{\operatorname{sen} u + \cos u}{\sqrt{2}\operatorname{sen} u\cos u}$, as a function of $x = \cos v$, is:

$\cos\frac{\pi}{12} \cdot \cos\frac{8\pi}{12}$ is worth:

Simplifying $4\operatorname{sen}\alpha\operatorname{sen}(60^\circ - \alpha)\operatorname{sen}(60^\circ + \alpha)$, we obtain:

For what values ​​of $t$ does the system $x + y = \pi$, $\operatorname{sen} x + \operatorname{sen} y = \log_{10} t^2$ admit a solution?

The graph from $y = \operatorname{sen} x - \cos x$, to $0 \le x \le \pi$, is: