Euler's Eyes
2026-10-23
Exercises from chapter VI — Transformations
This is the chapter with the most demonstration exercises. Before opening a solution, decide which family of formulas to use: addition, multiplication, division or transformation into a product.
Addition formulas
Calculate the values of: a) $\cos 15^\circ$; b) $\operatorname{sen} 105^\circ$; c) $\operatorname{tg} 75^\circ$; d) $\sec 285^\circ$.
See the resolution
a) $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \operatorname{sen} 45^\circ \operatorname{sen} 30^\circ = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
b) $\operatorname{sen} 105^\circ = \operatorname{sen}(60^\circ + 45^\circ) = \operatorname{sen} 60^\circ \cos 45^\circ + \operatorname{sen} 45^\circ \cos 60^\circ = \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
c) $\operatorname{tg} 75^\circ = \operatorname{tg}(45^\circ + 30^\circ) = \dfrac{\operatorname{tg} 45^\circ + \operatorname{tg} 30^\circ}{1 - \operatorname{tg} 45^\circ \operatorname{tg} 30^\circ} = \dfrac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \dfrac{3 + \sqrt{3}}{3 - \sqrt{3}}$ $\left(= 2 + \sqrt{3}\right)$
d) $\sec 285^\circ = \sec 75^\circ = \dfrac{1}{\cos 75^\circ} = \dfrac{1}{\cos(45^\circ + 30^\circ)} = \dfrac{4}{\sqrt{6} - \sqrt{2}}$ $\left(= \sqrt{6} + \sqrt{2}\right)$
Calculate $\operatorname{cotg} 165^\circ$, $\sec 255^\circ$ and $\operatorname{cossec} 15^\circ$.
Ver resposta
$\operatorname{cotg} 165^\circ = -(2 + \sqrt{3})$, $\sec 255^\circ = -(\sqrt{6} + \sqrt{2})$, $\operatorname{cossec} 15^\circ = \sqrt{6} + \sqrt{2}$
(FEI-76) Being $\operatorname{tg} A = 2$ and $\operatorname{tg} B = 1$, find $\operatorname{tg}(A - B)$.
Ver resposta
$\dfrac{1}{3}$
(MAPOFEI-75) Calculate the value of the expression $\operatorname{sen} 105^\circ - \cos 75^\circ$.
Ver resposta
$\dfrac{\sqrt{2}}{2}$
Given $\operatorname{sen} x = \dfrac{3}{5}$ and $\cos y = \dfrac{5}{13}$, calculate $\cos(x + y)$, knowing that $0 < x < \dfrac{\pi}{2}$ and $\dfrac{3\pi}{2} < y < 2\pi$.
See the resolution
1º) $\cos x = +\sqrt{1 - \operatorname{sen}^2 x} = +\sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.
2º) $\operatorname{sen} y = -\sqrt{1 - \cos^2 y} = -\sqrt{1 - \frac{25}{169}} = -\frac{12}{13}$.
3º) $\cos(x + y) = \cos x \cos y - \operatorname{sen} x \operatorname{sen} y = \dfrac{4}{5} \cdot \dfrac{5}{13} - \dfrac{3}{5} \cdot \left(-\dfrac{12}{13}\right) = \dfrac{56}{65}$.
Knowing that $\operatorname{tg} a = \dfrac{2}{3}$ and $\operatorname{sen} b = \dfrac{4}{5}$, with $\dfrac{\pi}{2} < b < \pi$, calculate $\operatorname{tg}(a + b)$.
See the resolution
1º) $\cos b = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5}$. 2º) $\operatorname{tg} b = \dfrac{\frac{4}{5}}{-\frac{3}{5}} = -\dfrac{4}{3}$. 3º)
$$\operatorname{tg}(a + b) = \frac{\operatorname{tg} a + \operatorname{tg} b}{1 - \operatorname{tg} a \operatorname{tg} b} = \frac{\frac{2}{3} - \frac{4}{3}}{1 - \frac{2}{3}\left(-\frac{4}{3}\right)} = \frac{-\frac{2}{3}}{\frac{17}{9}} = -\frac{6}{17}.$$Knowing that $\operatorname{sen} x = \dfrac{15}{17}$, $\operatorname{sen} y = -\dfrac{3}{5}$, $0 < x < \dfrac{\pi}{2}$ and $\pi < y < \dfrac{3\pi}{2}$, calculate $\operatorname{sen}(x + y)$, $\cos(x + y)$ and $\operatorname{tg}(x + y)$.
Ver resposta
$\operatorname{sen}(x + y) = -\dfrac{84}{85}$, $\cos(x + y) = \dfrac{13}{85}$, $\operatorname{tg}(x + y) = -\dfrac{84}{13}$
Demonstrate identities:
a) $\operatorname{sen}(a + b) \cdot \operatorname{sen}(a - b) = \cos^2 b - \cos^2 a$;
b) (EPUSP-62) $\begin{vmatrix} 1 & 1 & 1 \\ \operatorname{sen} x & \operatorname{sen} y & \operatorname{sen} z \\ \cos x & \cos y & \cos z \end{vmatrix} = \operatorname{sen}(x - y) + \operatorname{sen}(y - z) + \operatorname{sen}(z - x)$;
c) $\cos^2(a + b) + \cos^2 b - 2\cos(a + b)\cos a \cos b = \operatorname{sen}^2 a$.
See the resolution
a) $1^\circ$ member $= (\operatorname{sen} a \cos b + \operatorname{sen} b \cos a)(\operatorname{sen} a \cos b - \operatorname{sen} b \cos a) = \operatorname{sen}^2 a \cos^2 b - \operatorname{sen}^2 b \cos^2 a$
$= (1 - \cos^2 a)\cos^2 b - (1 - \cos^2 b)\cos^2 a = \cos^2 b - \cos^2 a = 2^\circ$ member.
b) Developing the determinant through the first line:
$$1^\circ \text{ membro} = \begin{vmatrix} \operatorname{sen} y & \operatorname{sen} z \\ \cos y & \cos z \end{vmatrix} - \begin{vmatrix} \operatorname{sen} x & \operatorname{sen} z \\ \cos x & \cos z \end{vmatrix} + \begin{vmatrix} \operatorname{sen} x & \operatorname{sen} y \\ \cos x & \cos y \end{vmatrix}$$$$= \operatorname{sen}(y - z) - \operatorname{sen}(x - z) + \operatorname{sen}(x - y) = \operatorname{sen}(x - y) + \operatorname{sen}(y - z) + \operatorname{sen}(z - x) = 2^\circ \text{ membro}.$$c) $1^\circ$ member $= (\cos a \cos b - \operatorname{sen} a \operatorname{sen} b)^2 + \cos^2 b - 2(\cos a \cos b - \operatorname{sen} a \operatorname{sen} b)\cos a \cos b$
$= \cos^2 a \cos^2 b - 2\operatorname{sen} a \operatorname{sen} b \cos a \cos b + \operatorname{sen}^2 a \operatorname{sen}^2 b + \cos^2 b - 2\cos^2 a \cos^2 b + 2\operatorname{sen} a \operatorname{sen} b \cos a \cos b$
$= \operatorname{sen}^2 a \operatorname{sen}^2 b - \cos^2 a \cos^2 b + \cos^2 b = \operatorname{sen}^2 a(1 - \cos^2 b) - (1 - \operatorname{sen}^2 a)\cos^2 b + \cos^2 b$
$= \operatorname{sen}^2 a - \operatorname{sen}^2 a \cos^2 b - \cos^2 b + \operatorname{sen}^2 a \cos^2 b + \cos^2 b = \operatorname{sen}^2 a = 2^\circ$ member.
(MAPOFEI-75) Demonstrate the identity $\operatorname{tg}(45^\circ + x) \cdot \operatorname{cotg}(45^\circ - x) = \dfrac{1 + \operatorname{sen} 2x}{1 - \operatorname{sen} 2x}$.
View a demo
With $t = \operatorname{tg} x$: $\operatorname{tg}(45^\circ + x) = \dfrac{1 + t}{1 - t}$ and $\operatorname{cotg}(45^\circ - x) = \dfrac{1}{\operatorname{tg}(45^\circ - x)} = \dfrac{1 + t}{1 - t}$. The product is
$$\left(\frac{1 + t}{1 - t}\right)^2 = \left(\frac{\cos x + \operatorname{sen} x}{\cos x - \operatorname{sen} x}\right)^2 = \frac{1 + 2\operatorname{sen} x \cos x}{1 - 2\operatorname{sen} x \cos x} = \frac{1 + \operatorname{sen} 2x}{1 - \operatorname{sen} 2x},$$using $\operatorname{sen} 2x = 2\operatorname{sen} x \cos x$, double arc formula from the following topic.
If $a$ and $b$ are acute and positive angles, prove that $\operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b$.
See the resolution
Let $X = \operatorname{sen}(a + b) - \operatorname{sen} a - \operatorname{sen} b = \operatorname{sen} a \cos b + \operatorname{sen} b \cos a - \operatorname{sen} a - \operatorname{sen} b = \operatorname{sen} a(\cos b - 1) + \operatorname{sen} b(\cos a - 1)$.
As $0 < a < \frac{\pi}{2}$, we have $\operatorname{sen} a > 0$ and $\cos a < 1$; likewise, $\operatorname{sen} b > 0$ and $\cos b < 1$. Then
$$X = \underbrace{\operatorname{sen} a}_{> 0} \cdot \underbrace{(\cos b - 1)}_{< 0} + \underbrace{\operatorname{sen} b}_{> 0} \cdot \underbrace{(\cos a - 1)}_{< 0} < 0 \;\Longrightarrow\; \operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b.$$(EPUSP-63) Prove that, if $\dfrac{\pi}{4} < a < \dfrac{\pi}{2}$ and $\dfrac{\pi}{4} < b < \dfrac{\pi}{2}$, then $\operatorname{sen}(a + b) < \operatorname{sen} a + \dfrac{4}{5}\operatorname{sen} b$.
View a demo
$\operatorname{sen}(a + b) = \operatorname{sen} a \cos b + \operatorname{sen} b \cos a$. In the given interval, the cosine is decreasing and positive, therefore $0 < \cos a < \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0{,}707 < \frac{4}{5}$, and the same goes for $\cos b$. Like $\operatorname{sen} a > 0$ and $\operatorname{sen} b > 0$:
$$\operatorname{sen} a \cos b < \operatorname{sen} a \qquad\text{e}\qquad \operatorname{sen} b \cos a < \frac{4}{5}\operatorname{sen} b.$$Adding, $\operatorname{sen}(a + b) < \operatorname{sen} a + \frac{4}{5}\operatorname{sen} b$.
Prove that the interior angles $A$, $B$ and $C$ of a non-right triangle verify the relationship $\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = \operatorname{tg} A \cdot \operatorname{tg} B \cdot \operatorname{tg} C$.
See the resolution
(If $A + B = 90^\circ$, the triangle would be right-angled at $C$; hence $1 - \operatorname{tg} A \operatorname{tg} B \ne 0$.)
Demonstrate the identity $4 \cdot \operatorname{sen}(x + 60^\circ) \cdot \cos(x + 30^\circ) = 3\cos^2 x - \operatorname{sen}^2 x$.
View a demo
The product is of the form $(u + v)(u - v)$ with $u = \frac{\sqrt{3}}{2}\cos x$ and $v = \frac{1}{2}\operatorname{sen} x$:
$$4\left(\frac{3}{4}\cos^2 x - \frac{1}{4}\operatorname{sen}^2 x\right) = 3\cos^2 x - \operatorname{sen}^2 x.$$Study the variation of real functions: a) $f(x) = \operatorname{sen} 2x \cdot \cos x + \operatorname{sen} x \cdot \cos 2x$; b) $g(x) = \dfrac{\sqrt{2}}{2}\cos x + \dfrac{\sqrt{2}}{2}\operatorname{sen} x$; c) $h(x) = \dfrac{1 + \operatorname{tg} x}{1 - \operatorname{tg} x}$.
See the resolution
a) $f(x) = \operatorname{sen}(2x + x) = \operatorname{sen} 3x$, then $D(f) = \mathbb{R}$, $p = \dfrac{2\pi}{3}$ and $\operatorname{Im}(f) = [-1, 1]$.
b) $g(x) = \cos x \cos\frac{\pi}{4} + \operatorname{sen} x \operatorname{sen}\frac{\pi}{4} = \cos\left(x - \frac{\pi}{4}\right)$, then $D(g) = \mathbb{R}$, $p = 2\pi$ and $\operatorname{Im}(g) = [-1, 1]$.
c) $h(x) = \dfrac{\operatorname{tg}\frac{\pi}{4} + \operatorname{tg} x}{1 - \operatorname{tg}\frac{\pi}{4}\operatorname{tg} x} = \operatorname{tg}\left(x + \frac{\pi}{4}\right)$, then $D(h) = \left\{x \in \mathbb{R} \;\middle|\; x \ne \frac{\pi}{4} + k\pi\right\}$, $p = \pi$ and $\operatorname{Im}(h) = \mathbb{R}$. For $x = -\frac{\pi}{4}$, $h = \operatorname{tg} 0 = 0$.
Note: strictly speaking, $h$ is also not defined where $\operatorname{tg} x$ does not exist ($x = \frac{\pi}{2} + k\pi$); at these points the simplified expression would give $\operatorname{tg}\left(\frac{3\pi}{4} + k\pi\right) = -1$.
Study the variation of real functions: a) $f(x) = \cos^2 2x - \operatorname{sen}^2 2x$; b) $g(x) = \sqrt{3} \cdot \cos x - \operatorname{sen} x$; c) $h(x) = \dfrac{\operatorname{sen} x + \cos x}{\cos x - \operatorname{sen} x}$.
Ver resposta
a) $D(f) = \mathbb{R}$, $p(f) = \dfrac{\pi}{2}$, $\operatorname{Im}(f) = [-1, 1]$ · b) $D(g) = \mathbb{R}$, $p(g) = 2\pi$, $\operatorname{Im}(g) = [-2, 2]$ · c) $D(h) = \left\{x \in \mathbb{R} \mid x \ne \dfrac{\pi}{4} + k\pi\right\}$, $p(h) = \pi$, $\operatorname{Im}(h) = \mathbb{R}$
What is the period of the function $f\colon \mathbb{R} \to \mathbb{R}$ given by
$$f(x) = \operatorname{sen} x \cos 2x \cos 3x - \operatorname{sen} x \operatorname{sen} 2x \operatorname{sen} 3x + \cos x \operatorname{sen} 2x \cos 3x + \cos x \operatorname{sen} 3x \cos 2x\ ?$$Ver resposta
$p(f) = \dfrac{\pi}{3}$ (expression reduces to $f(x) = \operatorname{sen} 6x$)
Multiplication formulas
Being $\operatorname{tg} x = \dfrac{3}{4}$ and $\pi < x < \dfrac{3\pi}{2}$, calculate $\operatorname{sen} 2x$.
See the resolution
In the 3rd quadrant, sine and cosine are negative:
$$\operatorname{sen} x = -\sqrt{\frac{\operatorname{tg}^2 x}{1 + \operatorname{tg}^2 x}} = -\sqrt{\frac{\frac{9}{16}}{1 + \frac{9}{16}}} = -\frac{3}{5} \qquad \cos x = -\sqrt{\frac{1}{1 + \operatorname{tg}^2 x}} = -\sqrt{\frac{1}{1 + \frac{9}{16}}} = -\frac{4}{5}$$$$\operatorname{sen} 2x = 2\operatorname{sen} x \cos x = 2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right) = \frac{24}{25}.$$(FEI-77) Calculate $\operatorname{sen} 2x$ knowing that $\operatorname{tg} x + \operatorname{cotg} x = 3$.
Ver resposta
$\operatorname{sen} 2x = \dfrac{2}{3}$
Being $\operatorname{cotg} x = \dfrac{12}{5}$ and $0 < x < \dfrac{\pi}{2}$, calculate $\cos 2x$.
See the resolution
(MAUÁ-77) Being $\operatorname{sen}\alpha = \dfrac{2}{3}$, with $0 < \alpha < \dfrac{\pi}{2}$: a) calculate $\operatorname{sen}\left(\dfrac{\pi}{2} + 2\alpha\right)$; b) calculate $\cos\left(\dfrac{\pi}{4} + \alpha\right)$.
Ver resposta
a) $\dfrac{1}{9}$ · b) $\dfrac{\sqrt{10} - 2\sqrt{2}}{6}$
Being $\sec x = \dfrac{25}{24}$ and $\dfrac{3\pi}{2} < x < 2\pi$, calculate $\operatorname{tg} 2x$.
See the resolution
If $\cos x = \dfrac{3}{5}$ and $\dfrac{3\pi}{2} < x < 2\pi$, calculate $\operatorname{sen} 3x$.
Ver resposta
$\operatorname{sen} 3x = -\dfrac{44}{125}$
If $\operatorname{sen} x = \dfrac{12}{13}$ and $\dfrac{\pi}{2} < x < \pi$, calculate $\cos 3x$.
Ver resposta
$\cos 3x = \dfrac{2035}{2197}$
If $\sec x = \dfrac{4}{3}$ and $0 < x < \dfrac{\pi}{2}$, calculate $\operatorname{tg} 3x$.
Ver resposta
$\operatorname{tg} 3x = -\dfrac{5\sqrt{7}}{9}$
(MAPOFEI-75) Calculate $\operatorname{sen}^2\dfrac{\pi}{12} - \cos^2\dfrac{\pi}{12} + \operatorname{tg}\dfrac{\pi}{3} + \operatorname{tg}\dfrac{14\pi}{3}$.
Ver resposta
$-\dfrac{\sqrt{3}}{2}$
Prove that: a) $\operatorname{sen} 4a = 4\operatorname{sen} a \cos^3 a - 4\operatorname{sen}^3 a \cos a$; b) $\cos 4a = 8\cos^4 a - 8\cos^2 a + 1$; c) $\operatorname{tg} 4a = \dfrac{4\operatorname{tg} a - 4\operatorname{tg}^3 a}{\operatorname{tg}^4 a - 6\operatorname{tg}^2 a + 1}$.
View a demo
We do $4a = 2 \cdot (2a)$.
a) $\operatorname{sen} 4a = 2\operatorname{sen} 2a \cos 2a = 2(2\operatorname{sen} a \cos a)(\cos^2 a - \operatorname{sen}^2 a) = 4\operatorname{sen} a \cos^3 a - 4\operatorname{sen}^3 a \cos a$.
b) $\cos 4a = 2\cos^2 2a - 1 = 2(2\cos^2 a - 1)^2 - 1 = 2(4\cos^4 a - 4\cos^2 a + 1) - 1 = 8\cos^4 a - 8\cos^2 a + 1$.
c) With $t = \operatorname{tg} a$, $\operatorname{tg} 2a = \dfrac{2t}{1 - t^2}$ and
$$\operatorname{tg} 4a = \frac{2\operatorname{tg} 2a}{1 - \operatorname{tg}^2 2a} = \frac{\frac{4t}{1 - t^2}}{1 - \frac{4t^2}{(1 - t^2)^2}} = \frac{4t(1 - t^2)}{(1 - t^2)^2 - 4t^2} = \frac{4t - 4t^3}{t^4 - 6t^2 + 1}.$$Show by the principle of finite induction that
$$\cos a \cdot \cos 2a \cdot \cos 4a \cdot \ldots \cdot \cos 2^{n-1}a = \frac{\operatorname{sen} 2^n a}{2^n \cdot \operatorname{sen} a} \qquad (\operatorname{sen} a \ne 0).$$View a demo
Base ($n = 1$): the first member is $\cos a$, and the second is $\dfrac{\operatorname{sen} 2a}{2\operatorname{sen} a} = \dfrac{2\operatorname{sen} a \cos a}{2\operatorname{sen} a} = \cos a$.
Step: assume equality valid for $n$. Multiplying both members by $\cos 2^n a$:
$$\cos a \cdots \cos 2^{n-1}a \cdot \cos 2^n a = \frac{\operatorname{sen} 2^n a \cos 2^n a}{2^n \operatorname{sen} a} = \frac{\frac{1}{2}\operatorname{sen} 2^{n+1}a}{2^n \operatorname{sen} a} = \frac{\operatorname{sen} 2^{n+1}a}{2^{n+1}\operatorname{sen} a},$$which is the equality for $n + 1$ (we use $\operatorname{sen} 2\theta = 2\operatorname{sen}\theta\cos\theta$ with $\theta = 2^n a$). According to the principle of induction, it is valid for all $n \ge 1$.
(POLY-61) Sketch the graph of the function $y = 2 \cdot \operatorname{sen}^2 x$ using the graph of $\cos 2x$.
See the resolution
From the identity $\cos 2x = 1 - 2\operatorname{sen}^2 x$, we have $y = 2\operatorname{sen}^2 x \Longrightarrow y = 1 - \cos 2x$.
| $x$ | $0$ | $\frac{\pi}{4}$ | $\frac{\pi}{2}$ | $\frac{3\pi}{4}$ | $\pi$ |
|---|---|---|---|---|---|
| $2x$ | $0$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ | $2\pi$ |
| $\cos 2x$ | $1$ | $0$ | $-1$ | $0$ | $1$ |
| $y$ | $0$ | $1$ | $2$ | $1$ | $0$ |
Study the variation of real functions: a) $f(x) = \cos^4 x - \operatorname{sen}^4 x$; b) $g(x) = 8\operatorname{sen}^2 x \cos^2 x$; c) $h(x) = \cos^4 x + \operatorname{sen}^4 x$.
Ver resposta
a) $p(f) = \pi$, $D(f) = \mathbb{R}$, $\operatorname{Im}(f) = [-1, 1]$ · b) $p(g) = \dfrac{\pi}{2}$, $D(g) = \mathbb{R}$, $\operatorname{Im}(g) = [0, 2]$ · c) $p(h) = \dfrac{\pi}{2}$, $D(h) = \mathbb{R}$, $\operatorname{Im}(h) = \left[\dfrac{1}{2}, 1\right]$
What is the period of real functions: a) $f(x) = \operatorname{sen} x \cdot \cos x$; b) $g(x) = \dfrac{1 - \operatorname{tg}^2 2x}{1 + \operatorname{tg}^2 2x}$; c) $h(x) = \cos^6 x + \operatorname{sen}^6 x$?
Ver resposta
$p(f) = \pi$, $p(g) = \dfrac{\pi}{2}$, $p(h) = \dfrac{\pi}{2}$
Division formulas
Calculate the circular functions of $\dfrac{\pi}{8}$.
See the resolution
As $\frac{\pi}{8}$ is in the 1st quadrant, all signs are positive:
$$\operatorname{sen}\frac{\pi}{8} = \sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2} \qquad \cos\frac{\pi}{8} = \sqrt{\frac{1 + \cos\frac{\pi}{4}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$$$\operatorname{tg}\frac{\pi}{8} = \sqrt{\frac{1 - \cos\frac{\pi}{4}}{1 + \cos\frac{\pi}{4}}} = \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} = \sqrt{2} - 1.$$(The other three are the inverses: $\operatorname{cotg}\frac{\pi}{8} = \sqrt{2} + 1$, $\sec\frac{\pi}{8} = \frac{2}{\sqrt{2 + \sqrt{2}}}$, $\operatorname{cossec}\frac{\pi}{8} = \frac{2}{\sqrt{2 - \sqrt{2}}}$.)
If $\operatorname{sen} x = \dfrac{24}{25}$ and $\dfrac{\pi}{2} < x < \pi$, calculate the circular functions of $\dfrac{x}{2}$.
See the resolution
$\cos x = -\sqrt{1 - \frac{576}{625}} = -\frac{7}{25}$. As $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$, all signs are positive:
$$\operatorname{sen}\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \qquad \cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \qquad \operatorname{tg}\frac{x}{2} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{16}{9}} = \frac{4}{3}.$$If $\operatorname{tg} x = \dfrac{5}{12}$, calculate $\operatorname{sen}\dfrac{x}{2}$.
See the resolution
then there are 4 possibilities for $\operatorname{sen}\frac{x}{2}$: $\;+\dfrac{1}{\sqrt{26}}$, $\;-\dfrac{1}{\sqrt{26}}$, $\;+\dfrac{5}{\sqrt{26}}$ or $\;-\dfrac{5}{\sqrt{26}}$.
(FAUUSP-69) Knowing that $x$ is an arc of the first quadrant and $\cos x = \dfrac{1}{3}$, determine $\operatorname{sen}\dfrac{x}{2}$ and $\operatorname{tg}\dfrac{x}{2}$.
Ver resposta
$\operatorname{sen}\dfrac{x}{2} = \dfrac{\sqrt{3}}{3}$, $\operatorname{tg}\dfrac{x}{2} = \dfrac{\sqrt{2}}{2}$
Knowing that $\cos x = \dfrac{24}{25}$ and $0 < x < \dfrac{\pi}{2}$, calculate $\operatorname{sen}\dfrac{x}{4}$, $\cos\dfrac{x}{4}$ and $\operatorname{tg}\dfrac{x}{4}$.
Ver resposta
$\operatorname{sen}\dfrac{x}{4} = \sqrt{\dfrac{10 - 7\sqrt{2}}{20}}$, $\cos\dfrac{x}{4} = \sqrt{\dfrac{10 + 7\sqrt{2}}{20}}$, $\operatorname{tg}\dfrac{x}{4} = \sqrt{\dfrac{10 - 7\sqrt{2}}{10 + 7\sqrt{2}}}$
Being $\sec x = 4$ and $\dfrac{3\pi}{2} < x < 2\pi$, calculate $\operatorname{tg}\left(\dfrac{\pi + x}{2}\right)$.
Ver resposta
$\operatorname{tg}\left(\dfrac{\pi + x}{2}\right) = +\sqrt{\dfrac{5}{3}}$
Study the variation of the function $f\colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \sqrt{\dfrac{1 - \cos 2x}{2}}$.
See the resolution
From $\cos 2x = 1 - 2\operatorname{sen}^2 x$ comes $\dfrac{1 - \cos 2x}{2} = \operatorname{sen}^2 x$; therefore $f(x) = \sqrt{\operatorname{sen}^2 x} = |\operatorname{sen} x|$. We have already seen that $D(f) = \mathbb{R}$, $p = \pi$ and $\operatorname{Im}(f) = [0, 1]$.
Study the variation of the function $f\colon \mathbb{R} - \left\{x \;\middle|\; x = \dfrac{\pi}{2} + k\pi\right\} \to \mathbb{R}$ given by $f(x) = (1 - \cos 2x)^{\frac{1}{2}} \cdot (1 + \cos 2x)^{-\frac{1}{2}}$.
Ver resposta
$f(x) = |\operatorname{tg} x|$, $\operatorname{Im}(f) = \mathbb{R}_+$, $D(f) = \left\{x \in \mathbb{R} \mid x \ne \dfrac{\pi}{2} + k\pi\right\}$, $p(f) = \pi$
What is the period of the function $f\colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = (1 + \cos 4x)^{\frac{1}{2}}$?
Ver resposta
$D(f) = \mathbb{R}$, $\operatorname{Im}(f) = [0, \sqrt{2}]$ and $p(f) = \dfrac{\pi}{2}$, as $f(x) = \sqrt{2\cos^2 2x} = \sqrt{2}\,|\cos 2x|$. (The printed edition template shows $\dfrac{\pi}{4}$, which does not match: $f\left(x + \frac{\pi}{4}\right) = \sqrt{2}\,|\operatorname{sen} 2x|$.)
a) For every real $\alpha \ne \dfrac{k\pi}{2}$ ($k \in \mathbb{Z}$), prove that $\dfrac{1}{\operatorname{sen}\alpha} = \operatorname{cotg}\dfrac{\alpha}{2} - \operatorname{cotg}\alpha$.
b) Show, using the previous result, that
$$\frac{1}{\operatorname{sen} a} + \frac{1}{\operatorname{sen} 2a} + \frac{1}{\operatorname{sen} 4a} + \cdots + \frac{1}{\operatorname{sen} 2^n a} = \operatorname{cotg}\frac{a}{2} - \operatorname{cotg} 2^n a.$$View a demo
a) With $\operatorname{sen}\alpha = 2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}$ and $\cos\alpha = 2\cos^2\frac{\alpha}{2} - 1$:
$$\operatorname{cotg}\frac{\alpha}{2} - \operatorname{cotg}\alpha = \frac{\cos\frac{\alpha}{2}}{\operatorname{sen}\frac{\alpha}{2}} - \frac{2\cos^2\frac{\alpha}{2} - 1}{2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{2\cos^2\frac{\alpha}{2} - 2\cos^2\frac{\alpha}{2} + 1}{2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{1}{\operatorname{sen}\alpha}.$$b) Applying (a) with $\alpha = a, 2a, 4a, \ldots, 2^n a$, each installment becomes a difference:
$$\left(\operatorname{cotg}\frac{a}{2} - \operatorname{cotg} a\right) + \left(\operatorname{cotg} a - \operatorname{cotg} 2a\right) + \cdots + \left(\operatorname{cotg} 2^{n-1}a - \operatorname{cotg} 2^n a\right).$$The intermediate terms cancel out (“telescopic sum”), and we are left with $\operatorname{cotg}\frac{a}{2} - \operatorname{cotg} 2^n a$.
Transformation into product
Transform into product: a) $y = \operatorname{sen} 5x + \operatorname{sen} 3x$; b) $y = \cos 3x + \cos x$; c) $y = \operatorname{sen} 7a + \operatorname{sen} 5a - \operatorname{sen} 3a - \operatorname{sen} a$; d) $y = \cos 9a + \cos 5a - \cos 3a - \cos a$; e) $y = \operatorname{sen} a + \operatorname{sen} b + \operatorname{sen} c - \operatorname{sen}(a + b + c)$.
See the resolution
a) $y = 2\operatorname{sen}\dfrac{5x + 3x}{2}\cos\dfrac{5x - 3x}{2} = 2\operatorname{sen} 4x \cos x$
b) $y = 2\cos\dfrac{3x + x}{2}\cos\dfrac{3x - x}{2} = 2\cos 2x \cos x$
c) $y = (\operatorname{sen} 7a + \operatorname{sen} 5a) - (\operatorname{sen} 3a + \operatorname{sen} a) = 2\operatorname{sen} 6a \cos a - 2\operatorname{sen} 2a \cos a = 2\cos a(\operatorname{sen} 6a - \operatorname{sen} 2a)$
$= 2\cos a(2\operatorname{sen} 2a \cos 4a) = 4\cos a \operatorname{sen} 2a \cos 4a$
d) $y = (\cos 9a + \cos 5a) - (\cos 3a + \cos a) = 2\cos 7a \cos 2a - 2\cos 2a \cos a = 2\cos 2a(\cos 7a - \cos a)$
$= -4\cos 2a \operatorname{sen} 4a \operatorname{sen} 3a$
e) $y = (\operatorname{sen} a + \operatorname{sen} b) - [\operatorname{sen}(a + b + c) - \operatorname{sen} c] = 2\operatorname{sen}\dfrac{a + b}{2}\cos\dfrac{a - b}{2} - 2\operatorname{sen}\dfrac{a + b}{2}\cos\dfrac{a + b + 2c}{2}$
$= -2\operatorname{sen}\dfrac{a + b}{2}\left[\cos\dfrac{a + b + 2c}{2} - \cos\dfrac{a - b}{2}\right] = -2\operatorname{sen}\dfrac{a + b}{2}\left(-2\operatorname{sen}\dfrac{a + c}{2}\operatorname{sen}\dfrac{b + c}{2}\right)$
$= 4\operatorname{sen}\dfrac{a + b}{2}\operatorname{sen}\dfrac{a + c}{2}\operatorname{sen}\dfrac{b + c}{2}$
Transform into product: a) $y = 1 + \operatorname{sen} 2x$; b) $y = 1 + \cos x$; c) $y = 1 + \cos a + \cos 2a$; d) $y = \operatorname{sen} a + 2\operatorname{sen} 3a + \operatorname{sen} 5a$.
See the resolution
a) $y = \operatorname{sen}\dfrac{\pi}{2} + \operatorname{sen} 2x = 2\operatorname{sen}\left(\dfrac{\pi}{4} + x\right)\cos\left(\dfrac{\pi}{4} - x\right) = 2\operatorname{sen}^2\left(\dfrac{\pi}{4} + x\right)$ (because $\cos\left(\frac{\pi}{4} - x\right) = \operatorname{sen}\left(\frac{\pi}{4} + x\right)$)
b) $y = \cos 0 + \cos x = 2\cos\dfrac{x}{2}\cos\left(-\dfrac{x}{2}\right) = 2\cos^2\dfrac{x}{2}$
c) $y = (\cos 2a + \cos 0) + \cos a = 2\cos^2 a + \cos a = 2\cos a\left(\cos a + \dfrac{1}{2}\right) = 2\cos a\left[\cos a + \cos\dfrac{\pi}{3}\right]$
$= 4\cos a \cos\left(\dfrac{a}{2} + \dfrac{\pi}{6}\right)\cos\left(\dfrac{a}{2} - \dfrac{\pi}{6}\right)$
d) $y = (\operatorname{sen} 5a + \operatorname{sen} a) + 2\operatorname{sen} 3a = 2\operatorname{sen} 3a \cos 2a + 2\operatorname{sen} 3a = 2\operatorname{sen} 3a[\cos 2a + \cos 0]$
$= 2\operatorname{sen} 3a[2\cos a \cos a] = 4\operatorname{sen} 3a \cos^2 a$
Transform into product: a) $y = \operatorname{sen} x + \cos x$; b) $y = \cos 2x - \operatorname{sen} 2x$; c) $y = \cos^2 3x - \cos^2 x$; d) $y = \operatorname{sen}^2 5x - \operatorname{sen}^2 x$; e) $y = \dfrac{\operatorname{sen} a + \operatorname{sen} b}{\cos a + \cos b}$.
See the resolution
a) $y = \operatorname{sen} x + \operatorname{sen}\left(\dfrac{\pi}{2} - x\right) = 2\operatorname{sen}\dfrac{\pi}{4}\cos\left(x - \dfrac{\pi}{4}\right) = \sqrt{2}\cos\left(x - \dfrac{\pi}{4}\right)$
b) $y = \cos 2x - \cos\left(\dfrac{\pi}{2} - 2x\right) = -2\operatorname{sen}\dfrac{\pi}{4}\operatorname{sen}\left(2x - \dfrac{\pi}{4}\right) = -\sqrt{2}\operatorname{sen}\left(2x - \dfrac{\pi}{4}\right)$
c) $y = (\cos 3x + \cos x)(\cos 3x - \cos x) = (2\cos 2x \cos x)(-2\operatorname{sen} 2x \operatorname{sen} x)$
$= -(2\operatorname{sen} 2x \cos 2x)(2\operatorname{sen} x \cos x) = -\operatorname{sen} 4x \operatorname{sen} 2x$
d) $y = (\operatorname{sen} 5x + \operatorname{sen} x)(\operatorname{sen} 5x - \operatorname{sen} x) = (2\operatorname{sen} 3x \cos 2x)(2\operatorname{sen} 2x \cos 3x)$
$= (2\operatorname{sen} 3x \cos 3x)(2\operatorname{sen} 2x \cos 2x) = \operatorname{sen} 6x \operatorname{sen} 4x$
e) $y = \dfrac{2\operatorname{sen}\frac{a + b}{2}\cos\frac{a - b}{2}}{2\cos\frac{a + b}{2}\cos\frac{a - b}{2}} = \operatorname{tg}\dfrac{a + b}{2}$
Transform into product:
a) $y = \operatorname{sen}(a + b + c) - \operatorname{sen}(a - b + c)$; b) $y = \cos(a + 2b) + \cos a$; c) $y = \operatorname{sen} a + \operatorname{sen}(a + r) + \operatorname{sen}(a + 2r) + \operatorname{sen}(a + 3r)$; d) $y = \cos(a + 3b) + \cos(a + 2b) + \cos(a + b) + \cos a$; e) $y = \cos^2 p - \cos^2 q$; f) $y = \operatorname{sen}^2 p - \operatorname{sen}^2 q$; g) $y = \cos^2 p - \operatorname{sen}^2 q$; h) $y = \dfrac{\operatorname{sen} 2a + \operatorname{sen} 2b}{\cos 2a - \cos 2b}$; i) $y = \dfrac{1 + \operatorname{sen} a}{1 - \operatorname{sen} a}$.
Ver resposta
a) $2\operatorname{sen} b\cos(a + c)$ · b) $2\cos(a + b)\cos b$ · c) $4\cos r\cos\dfrac{r}{2}\operatorname{sen}\dfrac{2a + 3r}{2}$ · d) $4\cos b\cos\dfrac{b}{2}\cos\dfrac{2a + 3b}{2}$ · e) $\operatorname{sen}(p + q)\operatorname{sen}(q - p)$ · f) $\operatorname{sen}(p + q)\operatorname{sen}(p - q)$ · g) $\cos(p + q)\cos(p - q)$ · h) $\operatorname{cotg}(b - a)$ · i) $\operatorname{tg}\left(\dfrac{\pi}{4} + \dfrac{a}{2}\right)\operatorname{cotg}\left(\dfrac{\pi}{4} - \dfrac{a}{2}\right)$
(MAPOFEI-76) Transform the product $\cos 2x \cdot \cos 4x$ into an equivalent sum.
Ver resposta
$\dfrac{1}{2}(\cos 6x + \cos 2x)$
(FEFAAP-77) Prove that $(\operatorname{sen} A + \cos A)^4 = 4\cos^4\left(A - \dfrac{\pi}{4}\right)$.
View a demo
By exercise 173a, $\operatorname{sen} A + \cos A = \sqrt{2}\cos\left(A - \frac{\pi}{4}\right)$. Raising to the fourth power: $(\sqrt{2})^4\cos^4\left(A - \frac{\pi}{4}\right) = 4\cos^4\left(A - \frac{\pi}{4}\right)$.
Calculate the numerical value of the expression $y = \operatorname{sen}\dfrac{13\pi}{12} \cdot \cos\dfrac{11\pi}{12}$.
See the resolution
Doing $\dfrac{p + q}{2} = \dfrac{13\pi}{12}$ and $\dfrac{p - q}{2} = \dfrac{11\pi}{12}$, we obtain $p = \dfrac{24\pi}{12} = 2\pi$ and $q = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$. Therefore
$$y = \frac{1}{2}\left(2\operatorname{sen}\frac{13\pi}{12}\cos\frac{11\pi}{12}\right) = \frac{1}{2}\left(\operatorname{sen} 2\pi + \operatorname{sen}\frac{\pi}{6}\right) = \frac{1}{2}\left(0 + \frac{1}{2}\right) = \frac{1}{4}.$$Calculate the numerical value of the expressions: a) $y = \cos\dfrac{7\pi}{8} \cdot \cos\dfrac{\pi}{8}$; b) $y = \operatorname{sen}\dfrac{13\pi}{12} \cdot \operatorname{sen}\dfrac{7\pi}{12}$; c) $y = \operatorname{sen}\dfrac{5\pi}{24} \cdot \cos\dfrac{\pi}{24}$.
Ver resposta
a) $-\dfrac{1}{4}(2 + \sqrt{2})$ · b) $-\dfrac{1}{4}$ · c) $\dfrac{\sqrt{2} + 1}{4}$
Prove that $\cos 40^\circ \cdot \cos 80^\circ \cdot \cos 160^\circ = -\dfrac{1}{8}$.
See the resolution
because $\cos 200^\circ = \cos 160^\circ$ (supplementary arcs: $200^\circ = 360^\circ - 160^\circ$).
Prove that $\operatorname{tg} 81^\circ - \operatorname{tg} 63^\circ - \operatorname{tg} 27^\circ + \operatorname{tg} 9^\circ = 4$.
View a demo
Group $(\operatorname{tg} 81^\circ + \operatorname{tg} 9^\circ) - (\operatorname{tg} 63^\circ + \operatorname{tg} 27^\circ)$ and use $\operatorname{tg} p + \operatorname{tg} q = \dfrac{\operatorname{sen}(p + q)}{\cos p \cos q}$ with $p + q = 90^\circ$:
$$\operatorname{tg} 81^\circ + \operatorname{tg} 9^\circ = \frac{1}{\cos 81^\circ \cos 9^\circ} = \frac{2}{\cos 90^\circ + \cos 72^\circ} = \frac{2}{\cos 72^\circ} \qquad \operatorname{tg} 63^\circ + \operatorname{tg} 27^\circ = \frac{2}{\cos 36^\circ}.$$The expression is worth $\dfrac{2}{\cos 72^\circ} - \dfrac{2}{\cos 36^\circ} = \dfrac{2(\cos 36^\circ - \cos 72^\circ)}{\cos 36^\circ \cos 72^\circ}$. Now:
- $\cos 36^\circ \cos 72^\circ = \dfrac{1}{4}$: multiplying by $\operatorname{sen} 36^\circ$, $\operatorname{sen} 36^\circ \cos 36^\circ \cos 72^\circ = \frac{1}{2}\operatorname{sen} 72^\circ \cos 72^\circ = \frac{1}{4}\operatorname{sen} 144^\circ = \frac{1}{4}\operatorname{sen} 36^\circ$;
- $\cos 36^\circ - \cos 72^\circ = 2\operatorname{sen} 54^\circ \operatorname{sen} 18^\circ = 2\cos 36^\circ \cos 72^\circ = \dfrac{1}{2}$.
Therefore, the expression is worth $\dfrac{2 \cdot \frac{1}{2}}{\frac{1}{4}} = 4$.
Show that, if $A$, $B$ and $C$ are internal angles of a triangle, the relations hold:
a) $\operatorname{sen} A + \operatorname{sen} B + \operatorname{sen} C = 4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$; b) $\cos A + \cos B + \cos C = 1 + 4\operatorname{sen}\dfrac{A}{2}\operatorname{sen}\dfrac{B}{2}\operatorname{sen}\dfrac{C}{2}$; c) $\operatorname{sen} 2A + \operatorname{sen} 2B + \operatorname{sen} 2C = 4\operatorname{sen} A \operatorname{sen} B \operatorname{sen} C$; d) $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A \cos B \cos C$.
See the resolution
Preliminaries. I) $A + B + C = \pi \Rightarrow B + C = \pi - A$, then $\operatorname{sen}(B + C) = \operatorname{sen} A$ and $\cos(B + C) = -\cos A$. II) $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$, then $\operatorname{sen}\dfrac{B + C}{2} = \cos\dfrac{A}{2}$ and $\cos\dfrac{B + C}{2} = \operatorname{sen}\dfrac{A}{2}$.
a) $\operatorname{sen} A + \operatorname{sen} B + \operatorname{sen} C = \operatorname{sen} A + 2\operatorname{sen}\dfrac{B + C}{2}\cos\dfrac{B - C}{2} = 2\operatorname{sen}\dfrac{A}{2}\cos\dfrac{A}{2} + 2\cos\dfrac{A}{2}\cos\dfrac{B - C}{2}$
$= 2\cos\dfrac{A}{2}\left[\cos\dfrac{B + C}{2} + \cos\dfrac{B - C}{2}\right] = 2\cos\dfrac{A}{2}\left[2\cos\dfrac{B}{2}\cos\dfrac{C}{2}\right] = 4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$.
b) $\cos A + \cos B + \cos C = \cos A + 2\cos\dfrac{B + C}{2}\cos\dfrac{B - C}{2} = \left(1 - 2\operatorname{sen}^2\dfrac{A}{2}\right) + 2\operatorname{sen}\dfrac{A}{2}\cos\dfrac{B - C}{2}$
$= 1 - 2\operatorname{sen}\dfrac{A}{2}\left[\cos\dfrac{B + C}{2} - \cos\dfrac{B - C}{2}\right] = 1 - 2\operatorname{sen}\dfrac{A}{2}\left[-2\operatorname{sen}\dfrac{B}{2}\operatorname{sen}\dfrac{C}{2}\right] = 1 + 4\operatorname{sen}\dfrac{A}{2}\operatorname{sen}\dfrac{B}{2}\operatorname{sen}\dfrac{C}{2}$.
c) $\operatorname{sen} 2A + \operatorname{sen} 2B + \operatorname{sen} 2C = \operatorname{sen} 2A + 2\operatorname{sen}(B + C)\cos(B - C) = 2\operatorname{sen} A \cos A + 2\operatorname{sen} A \cos(B - C)$
$= 2\operatorname{sen} A[-\cos(B + C) + \cos(B - C)] = -2\operatorname{sen} A(-2\operatorname{sen} B \operatorname{sen} C) = 4\operatorname{sen} A \operatorname{sen} B \operatorname{sen} C$.
d) As $\cos^2\alpha = \dfrac{\cos 2\alpha + 1}{2}$:
$\cos^2 A + \dfrac{\cos 2B + 1}{2} + \dfrac{\cos 2C + 1}{2} = 1 + \cos^2 A + \cos(B + C)\cos(B - C) = 1 + \cos^2 A - \cos A \cos(B - C)$
$= 1 - \cos A[\cos(B - C) - \cos A] = 1 - \cos A[\cos(B - C) + \cos(B + C)] = 1 - 2\cos A \cos B \cos C$.
Show that, if $A$, $B$, $C$ are internal angles of a triangle, the relations hold:
a) $\operatorname{sen} B + \operatorname{sen} C - \operatorname{sen} A = 4\operatorname{sen}\dfrac{B}{2}\operatorname{sen}\dfrac{C}{2}\cos\dfrac{A}{2}$; b) $\cos B + \cos C - \cos A = -1 + 4\cos\dfrac{B}{2}\cos\dfrac{C}{2}\operatorname{sen}\dfrac{A}{2}$; c) $\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$; d) $\operatorname{sen}^2 A + \operatorname{sen}^2 B + \operatorname{sen}^2 C = 2(1 + \cos A \cos B \cos C)$; e) $\dfrac{1}{\operatorname{tg} A \operatorname{tg} B} + \dfrac{1}{\operatorname{tg} B \operatorname{tg} C} + \dfrac{1}{\operatorname{tg} C \operatorname{tg} A} = 1$ $\left(A, B, C \ne \dfrac{\pi}{2}\right)$.
View a demo
We use the preliminaries of exercise 181.
a) $2\operatorname{sen}\dfrac{B + C}{2}\cos\dfrac{B - C}{2} - 2\operatorname{sen}\dfrac{A}{2}\cos\dfrac{A}{2} = 2\cos\dfrac{A}{2}\left[\cos\dfrac{B - C}{2} - \cos\dfrac{B + C}{2}\right] = 2\cos\dfrac{A}{2} \cdot 2\operatorname{sen}\dfrac{B}{2}\operatorname{sen}\dfrac{C}{2}$.
b) $2\cos\dfrac{B + C}{2}\cos\dfrac{B - C}{2} - \left(1 - 2\operatorname{sen}^2\dfrac{A}{2}\right) = -1 + 2\operatorname{sen}\dfrac{A}{2}\left[\cos\dfrac{B - C}{2} + \cos\dfrac{B + C}{2}\right] = -1 + 4\operatorname{sen}\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$.
c) $2\cos(A + B)\cos(A - B) + 2\cos^2 C - 1 = -2\cos C\cos(A - B) + 2\cos^2 C - 1 = -1 - 2\cos C[\cos(A - B) + \cos(A + B)] = -1 - 4\cos A \cos B \cos C$.
d) $\operatorname{sen}^2 A + \operatorname{sen}^2 B + \operatorname{sen}^2 C = 3 - (\cos^2 A + \cos^2 B + \cos^2 C) = 3 - (1 - 2\cos A \cos B \cos C) = 2(1 + \cos A \cos B \cos C)$, for exercise 181d.
e) For the year 142, $\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = \operatorname{tg} A \operatorname{tg} B \operatorname{tg} C$. Dividing both members by $\operatorname{tg} A \operatorname{tg} B \operatorname{tg} C$ (not null), we obtain the requested relationship.
Prove that, if $a + b + c = \dfrac{\pi}{2}$, then: a) $\operatorname{tg} a \operatorname{tg} b + \operatorname{tg} b \operatorname{tg} c + \operatorname{tg} c \operatorname{tg} a = 1$; b) $\cos^2 a + \cos^2 b + \cos^2 c - 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 2$.
View a demo
a) $a + b = \frac{\pi}{2} - c \Rightarrow \operatorname{tg}(a + b) = \operatorname{cotg} c \Rightarrow \dfrac{\operatorname{tg} a + \operatorname{tg} b}{1 - \operatorname{tg} a \operatorname{tg} b} = \dfrac{1}{\operatorname{tg} c} \Rightarrow \operatorname{tg} c(\operatorname{tg} a + \operatorname{tg} b) = 1 - \operatorname{tg} a \operatorname{tg} b$, which is the requested relationship (assuming defined tangents).
b) For the year 186, $\operatorname{sen}^2 a + \operatorname{sen}^2 b + \operatorname{sen}^2 c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$. Replacing each $\operatorname{sen}^2$ with $1 - \cos^2$: $3 - (\cos^2 a + \cos^2 b + \cos^2 c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$, which is equivalent to the requested relationship.
Prove that, if $(\operatorname{sen} 2A, \operatorname{sen} 2B, \operatorname{sen} 2C)$ is an arithmetic progression, then the same occurs with $(\operatorname{tg}(A + B), \operatorname{tg}(C + A), \operatorname{tg}(B + C))$.
See the resolution
By hypothesis, $\operatorname{sen} 2B - \operatorname{sen} 2A = \operatorname{sen} 2C - \operatorname{sen} 2B$, that is,
$$2\operatorname{sen}(B - A)\cos(B + A) = 2\operatorname{sen}(C - B)\cos(C + B).$$Writing $B - A = (B + C) - (A + C)$ and $C - B = (C + A) - (A + B)$ and developing by the sine of the difference:
$$[\operatorname{sen}(B + C)\cos(A + C) - \operatorname{sen}(A + C)\cos(B + C)]\cos(B + A) = [\operatorname{sen}(C + A)\cos(A + B) - \operatorname{sen}(A + B)\cos(C + A)]\cos(C + B).$$Dividing by $\cos(A + B)\cos(B + C)\cos(C + A)$:
$$\frac{\operatorname{sen}(B + C)}{\cos(B + C)} - \frac{\operatorname{sen}(C + A)}{\cos(C + A)} = \frac{\operatorname{sen}(C + A)}{\cos(C + A)} - \frac{\operatorname{sen}(A + B)}{\cos(A + B)},$$that is, $\operatorname{tg}(B + C) - \operatorname{tg}(C + A) = \operatorname{tg}(C + A) - \operatorname{tg}(A + B)$: the three numbers are in arithmetic progression.
Prove that, if $(\operatorname{sen}(a + b - c), \operatorname{sen}(a - b + c), \operatorname{sen}(-a + b + c))$ is an arithmetic progression, then the same occurs with $(\operatorname{tg} a, \operatorname{tg} b, \operatorname{tg} c)$.
View a demo
Let them be $\alpha = a + b - c$, $\beta = a - b + c$ and $\gamma = -a + b + c$. The hypothesis is $(\operatorname{sen}\beta - \operatorname{sen}\alpha) + (\operatorname{sen}\beta - \operatorname{sen}\gamma) = 0$. Transforming into product:
$$\operatorname{sen}\beta - \operatorname{sen}\alpha = 2\operatorname{sen}(c - b)\cos a \qquad \operatorname{sen}\beta - \operatorname{sen}\gamma = 2\operatorname{sen}(a - b)\cos c.$$Logo $\operatorname{sen}(c - b)\cos a = \operatorname{sen}(b - a)\cos c$. Dividing by $\cos a \cos b \cos c$ and using $\operatorname{tg} p - \operatorname{tg} q = \dfrac{\operatorname{sen}(p - q)}{\cos p \cos q}$:
$$\frac{\operatorname{sen}(c - b)}{\cos b \cos c} = \frac{\operatorname{sen}(b - a)}{\cos a \cos b} \;\Longrightarrow\; \operatorname{tg} c - \operatorname{tg} b = \operatorname{tg} b - \operatorname{tg} a.$$Prove that if $a + b + c = \dfrac{\pi}{2}$, then $\operatorname{sen}^2 a + \operatorname{sen}^2 b + \operatorname{sen}^2 c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$.
See the resolution
As $b + c = \frac{\pi}{2} - a$, we have $\cos(b + c) = \operatorname{sen} a$. Then
$1^\circ$ member $= \operatorname{sen}^2 a + \dfrac{1 - \cos 2b}{2} + \dfrac{1 - \cos 2c}{2} + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 + \operatorname{sen}^2 a - \dfrac{\cos 2b + \cos 2c}{2} + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c$
$= 1 + \operatorname{sen}^2 a - \cos(b + c)\cos(b - c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 + \operatorname{sen}^2 a - \operatorname{sen} a \cos(b - c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c$
$= 1 + \operatorname{sen} a[\cos(b + c) - \cos(b - c)] + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 - 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$.
Study the variation of the function $f\colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \cos x - \operatorname{sen} x$.
See the resolution
Therefore $D(f) = \mathbb{R}$, $p = 2\pi$ and $\operatorname{Im}(f) = \left[-\sqrt{2}, \sqrt{2}\right]$ — which confirms the graph from exercise 62.
Study the variation of the function $f\colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \operatorname{sen} 2x + \cos 2x$.
Ver resposta
$f(x) = \sqrt{2}\operatorname{sen}\left(2x + \dfrac{\pi}{4}\right)$: $D(f) = \mathbb{R}$, $p(f) = \pi$, $\operatorname{Im}(f) = \left[-\sqrt{2}, \sqrt{2}\right]$
What is the period of function $f(x) = \dfrac{1 + \operatorname{tg} x}{1 - \operatorname{tg} x}$?
Ver resposta
$p(f) = \pi$
Prove that if the angles of a triangle $ABC$ verify the relationship $\cos A + \cos B = \operatorname{sen} C$, then the triangle is right-angled.
See the resolution
(because $\cos\frac{A + B}{2} = \operatorname{sen}\frac{C}{2}$). As $\operatorname{sen}\frac{C}{2} \ne 0$, comes $\cos\dfrac{A - B}{2} = \cos\dfrac{C}{2}$, where
$$A - B = C \;\Rightarrow\; A = B + C = \frac{\pi}{2} \qquad\text{ou}\qquad A - B = -C \;\Rightarrow\; B = A + C = \frac{\pi}{2}.$$Prove that if the angles of a triangle $ABC$ verify the relationship $\operatorname{sen} 4A + \operatorname{sen} 4B + \operatorname{sen} 4C = 0$, then the triangle is right-angled.
View a demo
As $2A + 2B = 2\pi - 2C$, we have $\operatorname{sen}(2A + 2B) = -\operatorname{sen} 2C$. Then
$$\operatorname{sen} 4A + \operatorname{sen} 4B + \operatorname{sen} 4C = 2\operatorname{sen}(2A + 2B)\cos(2A - 2B) + 2\operatorname{sen} 2C \cos 2C = 2\operatorname{sen} 2C[\cos 2C - \cos(2A - 2B)].$$Such as $2C = 2\pi - 2A - 2B$, $\cos 2C = \cos(2A + 2B)$, and $\cos(2A + 2B) - \cos(2A - 2B) = -2\operatorname{sen} 2A \operatorname{sen} 2B$. Therefore, the sum is worth $-4\operatorname{sen} 2A \operatorname{sen} 2B \operatorname{sen} 2C$. If it is null, some factor is null, for example $\operatorname{sen} 2A = 0$ with $0 < 2A < 2\pi$, that is, $2A = \pi$ and $A = \frac{\pi}{2}$.
Show that every triangle whose angles verify the relationship $\operatorname{sen} 3A + \operatorname{sen} 3B + \operatorname{sen} 3C = 0$ has an angle of $60^\circ$.
View a demo
With $\frac{3(A + B)}{2} = \frac{3\pi}{2} - \frac{3C}{2}$, we have $\operatorname{sen}\frac{3(A + B)}{2} = -\cos\frac{3C}{2}$ and $\operatorname{sen}\frac{3C}{2} = -\cos\frac{3(A + B)}{2}$. Then
$$\operatorname{sen} 3A + \operatorname{sen} 3B + \operatorname{sen} 3C = 2\operatorname{sen}\frac{3(A + B)}{2}\cos\frac{3(A - B)}{2} + 2\operatorname{sen}\frac{3C}{2}\cos\frac{3C}{2}$$$$= -2\cos\frac{3C}{2}\left[\cos\frac{3(A - B)}{2} + \cos\frac{3(A + B)}{2}\right] = -4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}.$$If the sum is null, some factor is null, say $\cos\frac{3A}{2} = 0$; as $0 < \frac{3A}{2} < \frac{3\pi}{2}$, it can only be $\frac{3A}{2} = \frac{\pi}{2}$, that is, $A = \frac{\pi}{3} = 60^\circ$.