Euler's Eyes
2026-10-23
第六章练习——变换
这是演示练习最多的章节。在打开解决方案之前，决定使用哪一系列公式：加法、乘法、除法或转换为乘积。
加法公式
计算以下值：a)$\cos 15^\circ$； b）$\operatorname{sen} 105^\circ$； c) $\operatorname{tg} 75^\circ$； d) $\sec 285^\circ$。
查看分辨率
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) $\sec 285^\circ = \sec 75^\circ = \dfrac{1}{\cos 75^\circ} = \dfrac{1}{\cos(45^\circ + 30^\circ)} = \dfrac{4}{\sqrt{6} - \sqrt{2}}$ $\left(= \sqrt{6} + \sqrt{2}\right)$
计算$\operatorname{cotg} 165^\circ$、$\sec 255^\circ$和$\operatorname{cossec} 15^\circ$。
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) 根据$\operatorname{tg} A = 2$和$\operatorname{tg} B = 1$，找到$\operatorname{tg}(A - B)$。
Ver resposta
$\dfrac{1}{3}$
(MAPOFEI-75) 计算表达式$\operatorname{sen} 105^\circ - \cos 75^\circ$的值。
Ver resposta
$\dfrac{\sqrt{2}}{2}$
给定 $\operatorname{sen} x = \dfrac{3}{5}$ 和 $\cos y = \dfrac{5}{13}$，在已知 $0 < x < \dfrac{\pi}{2}$ 和 $\dfrac{3\pi}{2} < y < 2\pi$ 的情况下计算 $\cos(x + y)$。
查看分辨率
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) $\cos(x + y) = \cos x \cos y - \operatorname{sen} x \operatorname{sen} y = \dfrac{4}{5} \cdot \dfrac{5}{13} - \dfrac{3}{5} \cdot \left(-\dfrac{12}{13}\right) = \dfrac{56}{65}$.
已知$\operatorname{tg} a = \dfrac{2}{3}$和$\operatorname{sen} b = \dfrac{4}{5}$，用$\dfrac{\pi}{2} < b < \pi$计算$\operatorname{tg}(a + b)$。
查看分辨率
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)
$$\operatorname{tg}(a + b) = \frac{\operatorname{tg} a + \operatorname{tg} b}{1 - \operatorname{tg} a \operatorname{tg} b} = \frac{\frac{2}{3} - \frac{4}{3}}{1 - \frac{2}{3}\left(-\frac{4}{3}\right)} = \frac{-\frac{2}{3}}{\frac{17}{9}} = -\frac{6}{17}.$$已知$\operatorname{sen} x = \dfrac{15}{17}$、$\operatorname{sen} y = -\dfrac{3}{5}$、$0 < x < \dfrac{\pi}{2}$和$\pi < y < \dfrac{3\pi}{2}$，计算$\operatorname{sen}(x + y)$、$\cos(x + y)$和$\operatorname{tg}(x + y)$。
Ver resposta
$\operatorname{sen}(x + y) = -\dfrac{84}{85}$, $\cos(x + y) = \dfrac{13}{85}$, $\operatorname{tg}(x + y) = -\dfrac{84}{13}$
证明身份：
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) $\cos^2(a + b) + \cos^2 b - 2\cos(a + b)\cos a \cos b = \operatorname{sen}^2 a$.
查看分辨率
a) $1^\circ$ 成员 $= (\operatorname{sen} a \cos b + \operatorname{sen} b \cos a)(\operatorname{sen} a \cos b - \operatorname{sen} b \cos a) = \operatorname{sen}^2 a \cos^2 b - \operatorname{sen}^2 b \cos^2 a$
$= (1 - \cos^2 a)\cos^2 b - (1 - \cos^2 b)\cos^2 a = \cos^2 b - \cos^2 a = 2^\circ$ 成员。
b) 通过第一行开发行列式：
$$1^\circ \text{ membro} = \begin{vmatrix} \operatorname{sen} y & \operatorname{sen} z \\ \cos y & \cos z \end{vmatrix} - \begin{vmatrix} \operatorname{sen} x & \operatorname{sen} z \\ \cos x & \cos z \end{vmatrix} + \begin{vmatrix} \operatorname{sen} x & \operatorname{sen} y \\ \cos x & \cos y \end{vmatrix}$$$$= \operatorname{sen}(y - z) - \operatorname{sen}(x - z) + \operatorname{sen}(x - y) = \operatorname{sen}(x - y) + \operatorname{sen}(y - z) + \operatorname{sen}(z - x) = 2^\circ \text{ membro}.$$c) $1^\circ$ 成员 $= (\cos a \cos b - \operatorname{sen} a \operatorname{sen} b)^2 + \cos^2 b - 2(\cos a \cos b - \operatorname{sen} a \operatorname{sen} b)\cos a \cos b$
$= \cos^2 a \cos^2 b - 2\operatorname{sen} a \operatorname{sen} b \cos a \cos b + \operatorname{sen}^2 a \operatorname{sen}^2 b + \cos^2 b - 2\cos^2 a \cos^2 b + 2\operatorname{sen} a \operatorname{sen} b \cos a \cos b$
$= \operatorname{sen}^2 a \operatorname{sen}^2 b - \cos^2 a \cos^2 b + \cos^2 b = \operatorname{sen}^2 a(1 - \cos^2 b) - (1 - \operatorname{sen}^2 a)\cos^2 b + \cos^2 b$
$= \operatorname{sen}^2 a - \operatorname{sen}^2 a \cos^2 b - \cos^2 b + \operatorname{sen}^2 a \cos^2 b + \cos^2 b = \operatorname{sen}^2 a = 2^\circ$ 成员。
(MAPOFEI-75) 证明身份$\operatorname{tg}(45^\circ + x) \cdot \operatorname{cotg}(45^\circ - x) = \dfrac{1 + \operatorname{sen} 2x}{1 - \operatorname{sen} 2x}$。
查看演示
对于 $t = \operatorname{tg} x$：$\operatorname{tg}(45^\circ + x) = \dfrac{1 + t}{1 - t}$ 和 $\operatorname{cotg}(45^\circ - x) = \dfrac{1}{\operatorname{tg}(45^\circ - x)} = \dfrac{1 + t}{1 - t}$。该产品是
$$\left(\frac{1 + t}{1 - t}\right)^2 = \left(\frac{\cos x + \operatorname{sen} x}{\cos x - \operatorname{sen} x}\right)^2 = \frac{1 + 2\operatorname{sen} x \cos x}{1 - 2\operatorname{sen} x \cos x} = \frac{1 + \operatorname{sen} 2x}{1 - \operatorname{sen} 2x},$$使用 $\operatorname{sen} 2x = 2\operatorname{sen} x \cos x$，以下主题中的双圆弧公式。
如果$a$和$b$是锐角且为正角，则证明$\operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b$。
查看分辨率
让$X = \operatorname{sen}(a + b) - \operatorname{sen} a - \operatorname{sen} b = \operatorname{sen} a \cos b + \operatorname{sen} b \cos a - \operatorname{sen} a - \operatorname{sen} b = \operatorname{sen} a(\cos b - 1) + \operatorname{sen} b(\cos a - 1)$。
作为$0 < a < \frac{\pi}{2}$，我们有$\operatorname{sen} a > 0$和$\cos a < 1$；同样，$\operatorname{sen} b > 0$和$\cos b < 1$。然后
$$X = \underbrace{\operatorname{sen} a}_{> 0} \cdot \underbrace{(\cos b - 1)}_{< 0} + \underbrace{\operatorname{sen} b}_{> 0} \cdot \underbrace{(\cos a - 1)}_{< 0} < 0 \;\Longrightarrow\; \operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b.$$(EPUSP-63) 证明，如果 $\dfrac{\pi}{4} < a < \dfrac{\pi}{2}$ 和 $\dfrac{\pi}{4} < b < \dfrac{\pi}{2}$，则 $\operatorname{sen}(a + b) < \operatorname{sen} a + \dfrac{4}{5}\operatorname{sen} b$。
查看演示
$\operatorname{sen}(a + b) = \operatorname{sen} a \cos b + \operatorname{sen} b \cos a$。在给定区间内，余弦递减且为正，因此为 $0 < \cos a < \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} \approx 0{,}707 < \frac{4}{5}$，$\cos b$ 也是如此。如 $\operatorname{sen} a > 0$ 和 $\operatorname{sen} b > 0$：
$$\operatorname{sen} a \cos b < \operatorname{sen} a \qquad\text{e}\qquad \operatorname{sen} b \cos a < \frac{4}{5}\operatorname{sen} b.$$添加，$\operatorname{sen}(a + b) < \operatorname{sen} a + \frac{4}{5}\operatorname{sen} b$。
证明非直角三角形的内角 $A$、$B$ 和 $C$ 验证关系 $\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = \operatorname{tg} A \cdot \operatorname{tg} B \cdot \operatorname{tg} C$。
查看分辨率
（如果为 $A + B = 90^\circ$，则三角形将在 $C$ 处成直角；因此为 $1 - \operatorname{tg} A \operatorname{tg} B \ne 0$。）
证明身份$4 \cdot \operatorname{sen}(x + 60^\circ) \cdot \cos(x + 30^\circ) = 3\cos^2 x - \operatorname{sen}^2 x$。
查看演示
产品的形式为 $(u + v)(u - v)$、$u = \frac{\sqrt{3}}{2}\cos x$ 和 $v = \frac{1}{2}\operatorname{sen} x$：
$$4\left(\frac{3}{4}\cos^2 x - \frac{1}{4}\operatorname{sen}^2 x\right) = 3\cos^2 x - \operatorname{sen}^2 x.$$研究实函数的变分：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) $h(x) = \dfrac{1 + \operatorname{tg} x}{1 - \operatorname{tg} x}$。
查看分辨率
a) $f(x) = \operatorname{sen}(2x + x) = \operatorname{sen} 3x$，然后是 $D(f) = \mathbb{R}$、$p = \dfrac{2\pi}{3}$ 和 $\operatorname{Im}(f) = [-1, 1]$。
b) $g(x) = \cos x \cos\frac{\pi}{4} + \operatorname{sen} x \operatorname{sen}\frac{\pi}{4} = \cos\left(x - \frac{\pi}{4}\right)$，然后是 $D(g) = \mathbb{R}$、$p = 2\pi$ 和 $\operatorname{Im}(g) = [-1, 1]$。
c) $h(x) = \dfrac{\operatorname{tg}\frac{\pi}{4} + \operatorname{tg} x}{1 - \operatorname{tg}\frac{\pi}{4}\operatorname{tg} x} = \operatorname{tg}\left(x + \frac{\pi}{4}\right)$，然后是 $D(h) = \left\{x \in \mathbb{R} \;\middle|\; x \ne \frac{\pi}{4} + k\pi\right\}$、$p = \pi$ 和 $\operatorname{Im}(h) = \mathbb{R}$。对于 $x = -\frac{\pi}{4}$、$h = \operatorname{tg} 0 = 0$。
*注意：*严格来说，如果$\operatorname{tg} x$不存在($x = \frac{\pi}{2} + k\pi$)，$h$也没有被定义；在这些点上，简化表达式将给出 $\operatorname{tg}\left(\frac{3\pi}{4} + k\pi\right) = -1$。
研究实函数的变分：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) $D(h) = \left\{x \in \mathbb{R} \mid x \ne \dfrac{\pi}{4} + k\pi\right\}$, $p(h) = \pi$, $\operatorname{Im}(h) = \mathbb{R}$
函数 $f\colon \mathbb{R} \to \mathbb{R}$ 的周期是多少
$$f(x) = \operatorname{sen} x \cos 2x \cos 3x - \operatorname{sen} x \operatorname{sen} 2x \operatorname{sen} 3x + \cos x \operatorname{sen} 2x \cos 3x + \cos x \operatorname{sen} 3x \cos 2x\ ?$$Ver resposta
$p(f) = \dfrac{\pi}{3}$（表达式减少为 $f(x) = \operatorname{sen} 6x$）
乘法公式
根据$\operatorname{tg} x = \dfrac{3}{4}$和$\pi < x < \dfrac{3\pi}{2}$，计算$\operatorname{sen} 2x$。
查看分辨率
在第三象限，正弦和余弦为负：
$$\operatorname{sen} x = -\sqrt{\frac{\operatorname{tg}^2 x}{1 + \operatorname{tg}^2 x}} = -\sqrt{\frac{\frac{9}{16}}{1 + \frac{9}{16}}} = -\frac{3}{5} \qquad \cos x = -\sqrt{\frac{1}{1 + \operatorname{tg}^2 x}} = -\sqrt{\frac{1}{1 + \frac{9}{16}}} = -\frac{4}{5}$$$$\operatorname{sen} 2x = 2\operatorname{sen} x \cos x = 2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right) = \frac{24}{25}.$$(FEI-77) 已知 $\operatorname{tg} x + \operatorname{cotg} x = 3$ 计算 $\operatorname{sen} 2x$。
Ver resposta
$\operatorname{sen} 2x = \dfrac{2}{3}$
根据$\operatorname{cotg} x = \dfrac{12}{5}$和$0 < x < \dfrac{\pi}{2}$，计算$\cos 2x$。
查看分辨率
(MAUÁ-77) 为$\operatorname{sen}\alpha = \dfrac{2}{3}$，其中$0 < \alpha < \dfrac{\pi}{2}$： a) 计算$\operatorname{sen}\left(\dfrac{\pi}{2} + 2\alpha\right)$； b) 计算$\cos\left(\dfrac{\pi}{4} + \alpha\right)$。
Ver resposta
a) $\dfrac{1}{9}$ · b) $\dfrac{\sqrt{10} - 2\sqrt{2}}{6}$
根据$\sec x = \dfrac{25}{24}$和$\dfrac{3\pi}{2} < x < 2\pi$，计算$\operatorname{tg} 2x$。
查看分辨率
如果$\cos x = \dfrac{3}{5}$和$\dfrac{3\pi}{2} < x < 2\pi$，则计算$\operatorname{sen} 3x$。
Ver resposta
$\operatorname{sen} 3x = -\dfrac{44}{125}$
如果$\operatorname{sen} x = \dfrac{12}{13}$和$\dfrac{\pi}{2} < x < \pi$，则计算$\cos 3x$。
Ver resposta
$\cos 3x = \dfrac{2035}{2197}$
如果$\sec x = \dfrac{4}{3}$和$0 < x < \dfrac{\pi}{2}$，则计算$\operatorname{tg} 3x$。
Ver resposta
$\operatorname{tg} 3x = -\dfrac{5\sqrt{7}}{9}$
(MAPOFEI-75) 计算$\operatorname{sen}^2\dfrac{\pi}{12} - \cos^2\dfrac{\pi}{12} + \operatorname{tg}\dfrac{\pi}{3} + \operatorname{tg}\dfrac{14\pi}{3}$。
Ver resposta
$-\dfrac{\sqrt{3}}{2}$
证明：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) $\operatorname{tg} 4a = \dfrac{4\operatorname{tg} a - 4\operatorname{tg}^3 a}{\operatorname{tg}^4 a - 6\operatorname{tg}^2 a + 1}$。
查看演示
我们做$4a = 2 \cdot (2a)$。
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) 与 $t = \operatorname{tg} a$、$\operatorname{tg} 2a = \dfrac{2t}{1 - t^2}$ 和
$$\operatorname{tg} 4a = \frac{2\operatorname{tg} 2a}{1 - \operatorname{tg}^2 2a} = \frac{\frac{4t}{1 - t^2}}{1 - \frac{4t^2}{(1 - t^2)^2}} = \frac{4t(1 - t^2)}{(1 - t^2)^2 - 4t^2} = \frac{4t - 4t^3}{t^4 - 6t^2 + 1}.$$由有限归纳原理可知
$$\cos a \cdot \cos 2a \cdot \cos 4a \cdot \ldots \cdot \cos 2^{n-1}a = \frac{\operatorname{sen} 2^n a}{2^n \cdot \operatorname{sen} a} \qquad (\operatorname{sen} a \ne 0).$$查看演示
Base ($n = 1$)： 第一个成员是 $\cos a$，第二个成员是 $\dfrac{\operatorname{sen} 2a}{2\operatorname{sen} a} = \dfrac{2\operatorname{sen} a \cos a}{2\operatorname{sen} a} = \cos a$。
步骤： 假设等式对于 $n$ 有效。将两个成员乘以 $\cos 2^n a$：
$$\cos a \cdots \cos 2^{n-1}a \cdot \cos 2^n a = \frac{\operatorname{sen} 2^n a \cos 2^n a}{2^n \operatorname{sen} a} = \frac{\frac{1}{2}\operatorname{sen} 2^{n+1}a}{2^n \operatorname{sen} a} = \frac{\operatorname{sen} 2^{n+1}a}{2^{n+1}\operatorname{sen} a},$$这是 $n + 1$ 的等式（我们使用 $\operatorname{sen} 2\theta = 2\operatorname{sen}\theta\cos\theta$ 和 $\theta = 2^n a$）。根据归纳原理，它对所有$n \ge 1$都有效。
(POLY-61) 使用 $\cos 2x$ 的图形绘制函数 $y = 2 \cdot \operatorname{sen}^2 x$ 的图形。
查看分辨率
从身份$\cos 2x = 1 - 2\operatorname{sen}^2 x$，我们有$y = 2\operatorname{sen}^2 x \Longrightarrow y = 1 - \cos 2x$。
| $x$ | $0$ | $\frac{\pi}{4}$ | $\frac{\pi}{2}$ | $\frac{3\pi}{4}$ | $\pi$ |
|---|---|---|---|---|---|
| $2x$ | $0$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ | $2\pi$ |
| $\cos 2x$ | $1$ | $0$ | $-1$ | $0$ | $1$ |
| $y$ | $0$ | $1$ | $2$ | $1$ | $0$ |
研究实函数的变分：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) $p(h) = \dfrac{\pi}{2}$, $D(h) = \mathbb{R}$, $\operatorname{Im}(h) = \left[\dfrac{1}{2}, 1\right]$
实函数的周期是多少：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
$p(f) = \pi$, $p(g) = \dfrac{\pi}{2}$, $p(h) = \dfrac{\pi}{2}$
除法公式
计算$\dfrac{\pi}{8}$的循环函数。
查看分辨率
由于 $\frac{\pi}{8}$ 在第一象限，所有符号均为正：
$$\operatorname{sen}\frac{\pi}{8} = \sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2} \qquad \cos\frac{\pi}{8} = \sqrt{\frac{1 + \cos\frac{\pi}{4}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$$$\operatorname{tg}\frac{\pi}{8} = \sqrt{\frac{1 - \cos\frac{\pi}{4}}{1 + \cos\frac{\pi}{4}}} = \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} = \sqrt{2} - 1.$$（另外三个是相反的：$\operatorname{cotg}\frac{\pi}{8} = \sqrt{2} + 1$、$\sec\frac{\pi}{8} = \frac{2}{\sqrt{2 + \sqrt{2}}}$、$\operatorname{cossec}\frac{\pi}{8} = \frac{2}{\sqrt{2 - \sqrt{2}}}$。）
如果$\operatorname{sen} x = \dfrac{24}{25}$和$\dfrac{\pi}{2} < x < \pi$，计算$\dfrac{x}{2}$的循环函数。
查看分辨率
$\cos x = -\sqrt{1 - \frac{576}{625}} = -\frac{7}{25}$。如 $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$，所有迹象均为正：
$$\operatorname{sen}\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \qquad \cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \qquad \operatorname{tg}\frac{x}{2} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{16}{9}} = \frac{4}{3}.$$如果$\operatorname{tg} x = \dfrac{5}{12}$，则计算$\operatorname{sen}\dfrac{x}{2}$。
查看分辨率
那么 $\operatorname{sen}\frac{x}{2}$ 有 4 种可能性：$\;+\dfrac{1}{\sqrt{26}}$、$\;-\dfrac{1}{\sqrt{26}}$、$\;+\dfrac{5}{\sqrt{26}}$ 或 $\;-\dfrac{5}{\sqrt{26}}$。
(FAUSP-69) 已知$x$是第一象限的圆弧和$\cos x = \dfrac{1}{3}$，确定$\operatorname{sen}\dfrac{x}{2}$和$\operatorname{tg}\dfrac{x}{2}$。
Ver resposta
$\operatorname{sen}\dfrac{x}{2} = \dfrac{\sqrt{3}}{3}$, $\operatorname{tg}\dfrac{x}{2} = \dfrac{\sqrt{2}}{2}$
已知$\cos x = \dfrac{24}{25}$和$0 < x < \dfrac{\pi}{2}$，计算$\operatorname{sen}\dfrac{x}{4}$、$\cos\dfrac{x}{4}$和$\operatorname{tg}\dfrac{x}{4}$。
Ver resposta
$\operatorname{sen}\dfrac{x}{4} = \sqrt{\dfrac{10 - 7\sqrt{2}}{20}}$, $\cos\dfrac{x}{4} = \sqrt{\dfrac{10 + 7\sqrt{2}}{20}}$, $\operatorname{tg}\dfrac{x}{4} = \sqrt{\dfrac{10 - 7\sqrt{2}}{10 + 7\sqrt{2}}}$
根据$\sec x = 4$和$\dfrac{3\pi}{2} < x < 2\pi$，计算$\operatorname{tg}\left(\dfrac{\pi + x}{2}\right)$。
Ver resposta
$\operatorname{tg}\left(\dfrac{\pi + x}{2}\right) = +\sqrt{\dfrac{5}{3}}$
研究$f(x) = \sqrt{\dfrac{1 - \cos 2x}{2}}$给出的函数$f\colon \mathbb{R} \to \mathbb{R}$的变化。
查看分辨率
从 $\cos 2x = 1 - 2\operatorname{sen}^2 x$ 到 $\dfrac{1 - \cos 2x}{2} = \operatorname{sen}^2 x$；因此$f(x) = \sqrt{\operatorname{sen}^2 x} = |\operatorname{sen} x|$。我们已经看到了$D(f) = \mathbb{R}$、$p = \pi$和$\operatorname{Im}(f) = [0, 1]$。
研究$f(x) = (1 - \cos 2x)^{\frac{1}{2}} \cdot (1 + \cos 2x)^{-\frac{1}{2}}$给出的函数$f\colon \mathbb{R} - \left\{x \;\middle|\; x = \dfrac{\pi}{2} + k\pi\right\} \to \mathbb{R}$的变化。
Ver resposta
$f(x) = |\operatorname{tg} x|$, $\operatorname{Im}(f) = \mathbb{R}_+$, $D(f) = \left\{x \in \mathbb{R} \mid x \ne \dfrac{\pi}{2} + k\pi\right\}$, $p(f) = \pi$
$f(x) = (1 + \cos 4x)^{\frac{1}{2}}$ 给出的函数 $f\colon \mathbb{R} \to \mathbb{R}$ 的周期是多少？
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$D(f) = \mathbb{R}$、$\operatorname{Im}(f) = [0, \sqrt{2}]$和$p(f) = \dfrac{\pi}{2}$，如$f(x) = \sqrt{2\cos^2 2x} = \sqrt{2}\,|\cos 2x|$。 （印刷版模板显示 $\dfrac{\pi}{4}$，与 $f\left(x + \frac{\pi}{4}\right) = \sqrt{2}\,|\operatorname{sen} 2x|$ 不匹配。）
a) 对于每个实数 $\alpha \ne \dfrac{k\pi}{2}$ ($k \in \mathbb{Z}$)，证明 $\dfrac{1}{\operatorname{sen}\alpha} = \operatorname{cotg}\dfrac{\alpha}{2} - \operatorname{cotg}\alpha$。
b) 使用之前的结果表明
$$\frac{1}{\operatorname{sen} a} + \frac{1}{\operatorname{sen} 2a} + \frac{1}{\operatorname{sen} 4a} + \cdots + \frac{1}{\operatorname{sen} 2^n a} = \operatorname{cotg}\frac{a}{2} - \operatorname{cotg} 2^n a.$$查看演示
a) 对于 $\operatorname{sen}\alpha = 2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}$ 和 $\cos\alpha = 2\cos^2\frac{\alpha}{2} - 1$：
$$\operatorname{cotg}\frac{\alpha}{2} - \operatorname{cotg}\alpha = \frac{\cos\frac{\alpha}{2}}{\operatorname{sen}\frac{\alpha}{2}} - \frac{2\cos^2\frac{\alpha}{2} - 1}{2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{2\cos^2\frac{\alpha}{2} - 2\cos^2\frac{\alpha}{2} + 1}{2\operatorname{sen}\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{1}{\operatorname{sen}\alpha}.$$b) 将 (a) 与 $\alpha = a, 2a, 4a, \ldots, 2^n a$ 一起应用，每次分期付款都会有所不同：
$$\left(\operatorname{cotg}\frac{a}{2} - \operatorname{cotg} a\right) + \left(\operatorname{cotg} a - \operatorname{cotg} 2a\right) + \cdots + \left(\operatorname{cotg} 2^{n-1}a - \operatorname{cotg} 2^n a\right).$$中间项抵消（“伸缩和”），我们剩下 $\operatorname{cotg}\frac{a}{2} - \operatorname{cotg} 2^n a$。
转化为产品
转化为乘积：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) $y = \operatorname{sen} a + \operatorname{sen} b + \operatorname{sen} c - \operatorname{sen}(a + b + c)$。
查看分辨率
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) $y = (\operatorname{sen} a + \operatorname{sen} b) - [\operatorname{sen}(a + b + c) - \operatorname{sen} c] = 2\operatorname{sen}\dfrac{a + b}{2}\cos\dfrac{a - b}{2} - 2\operatorname{sen}\dfrac{a + b}{2}\cos\dfrac{a + b + 2c}{2}$
$= -2\operatorname{sen}\dfrac{a + b}{2}\left[\cos\dfrac{a + b + 2c}{2} - \cos\dfrac{a - b}{2}\right] = -2\operatorname{sen}\dfrac{a + b}{2}\left(-2\operatorname{sen}\dfrac{a + c}{2}\operatorname{sen}\dfrac{b + c}{2}\right)$
$= 4\operatorname{sen}\dfrac{a + b}{2}\operatorname{sen}\dfrac{a + c}{2}\operatorname{sen}\dfrac{b + c}{2}$
转化为乘积：a)$y = 1 + \operatorname{sen} 2x$； b）$y = 1 + \cos x$； c) $y = 1 + \cos a + \cos 2a$； d) $y = \operatorname{sen} a + 2\operatorname{sen} 3a + \operatorname{sen} 5a$。
查看分辨率
a) $y = \operatorname{sen}\dfrac{\pi}{2} + \operatorname{sen} 2x = 2\operatorname{sen}\left(\dfrac{\pi}{4} + x\right)\cos\left(\dfrac{\pi}{4} - x\right) = 2\operatorname{sen}^2\left(\dfrac{\pi}{4} + x\right)$（因为 $\cos\left(\frac{\pi}{4} - x\right) = \operatorname{sen}\left(\frac{\pi}{4} + x\right)$）
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) $y = (\operatorname{sen} 5a + \operatorname{sen} a) + 2\operatorname{sen} 3a = 2\operatorname{sen} 3a \cos 2a + 2\operatorname{sen} 3a = 2\operatorname{sen} 3a[\cos 2a + \cos 0]$
$= 2\operatorname{sen} 3a[2\cos a \cos a] = 4\operatorname{sen} 3a \cos^2 a$
转化为乘积：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) $y = \dfrac{\operatorname{sen} a + \operatorname{sen} b}{\cos a + \cos b}$。
查看分辨率
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) $y = \dfrac{2\operatorname{sen}\frac{a + b}{2}\cos\frac{a - b}{2}}{2\cos\frac{a + b}{2}\cos\frac{a - b}{2}} = \operatorname{tg}\dfrac{a + b}{2}$
转化为产品：
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) $y = \dfrac{1 + \operatorname{sen} a}{1 - \operatorname{sen} a}$.
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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) 将乘积 $\cos 2x \cdot \cos 4x$ 转换为等价和。
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$\dfrac{1}{2}(\cos 6x + \cos 2x)$
（FEFAAP-77）证明$(\operatorname{sen} A + \cos A)^4 = 4\cos^4\left(A - \dfrac{\pi}{4}\right)$。
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通过练习 173a，$\operatorname{sen} A + \cos A = \sqrt{2}\cos\left(A - \frac{\pi}{4}\right)$。求四次方：$(\sqrt{2})^4\cos^4\left(A - \frac{\pi}{4}\right) = 4\cos^4\left(A - \frac{\pi}{4}\right)$。
计算表达式$y = \operatorname{sen}\dfrac{13\pi}{12} \cdot \cos\dfrac{11\pi}{12}$的数值。
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执行$\dfrac{p + q}{2} = \dfrac{13\pi}{12}$和$\dfrac{p - q}{2} = \dfrac{11\pi}{12}$，我们得到$p = \dfrac{24\pi}{12} = 2\pi$和$q = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$。所以
$$y = \frac{1}{2}\left(2\operatorname{sen}\frac{13\pi}{12}\cos\frac{11\pi}{12}\right) = \frac{1}{2}\left(\operatorname{sen} 2\pi + \operatorname{sen}\frac{\pi}{6}\right) = \frac{1}{2}\left(0 + \frac{1}{2}\right) = \frac{1}{4}.$$计算表达式的数值：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) $y = \operatorname{sen}\dfrac{5\pi}{24} \cdot \cos\dfrac{\pi}{24}$。
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a) $-\dfrac{1}{4}(2 + \sqrt{2})$ · b) $-\dfrac{1}{4}$ · c) $\dfrac{\sqrt{2} + 1}{4}$
证明$\cos 40^\circ \cdot \cos 80^\circ \cdot \cos 160^\circ = -\dfrac{1}{8}$。
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因为$\cos 200^\circ = \cos 160^\circ$（补充弧：$200^\circ = 360^\circ - 160^\circ$）。
证明$\operatorname{tg} 81^\circ - \operatorname{tg} 63^\circ - \operatorname{tg} 27^\circ + \operatorname{tg} 9^\circ = 4$。
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将 $(\operatorname{tg} 81^\circ + \operatorname{tg} 9^\circ) - (\operatorname{tg} 63^\circ + \operatorname{tg} 27^\circ)$ 分组并将 $\operatorname{tg} p + \operatorname{tg} q = \dfrac{\operatorname{sen}(p + q)}{\cos p \cos q}$ 与 $p + q = 90^\circ$ 一起使用：
$$\operatorname{tg} 81^\circ + \operatorname{tg} 9^\circ = \frac{1}{\cos 81^\circ \cos 9^\circ} = \frac{2}{\cos 90^\circ + \cos 72^\circ} = \frac{2}{\cos 72^\circ} \qquad \operatorname{tg} 63^\circ + \operatorname{tg} 27^\circ = \frac{2}{\cos 36^\circ}.$$该表达式的值为 $\dfrac{2}{\cos 72^\circ} - \dfrac{2}{\cos 36^\circ} = \dfrac{2(\cos 36^\circ - \cos 72^\circ)}{\cos 36^\circ \cos 72^\circ}$。现在：
- $\cos 36^\circ \cos 72^\circ = \dfrac{1}{4}$：乘以$\operatorname{sen} 36^\circ$、$\operatorname{sen} 36^\circ \cos 36^\circ \cos 72^\circ = \frac{1}{2}\operatorname{sen} 72^\circ \cos 72^\circ = \frac{1}{4}\operatorname{sen} 144^\circ = \frac{1}{4}\operatorname{sen} 36^\circ$；
- $\cos 36^\circ - \cos 72^\circ = 2\operatorname{sen} 54^\circ \operatorname{sen} 18^\circ = 2\cos 36^\circ \cos 72^\circ = \dfrac{1}{2}$.
因此，表达式的值为$\dfrac{2 \cdot \frac{1}{2}}{\frac{1}{4}} = 4$。
证明，如果 $A$、$B$ 和 $C$ 是三角形的内角，则关系成立：
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) $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A \cos B \cos C$.
查看分辨率
预备知识。 I) $A + B + C = \pi \Rightarrow B + C = \pi - A$，然后 $\operatorname{sen}(B + C) = \operatorname{sen} A$ 和 $\cos(B + C) = -\cos A$。 II) $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$，然后是 $\operatorname{sen}\dfrac{B + C}{2} = \cos\dfrac{A}{2}$ 和 $\cos\dfrac{B + C}{2} = \operatorname{sen}\dfrac{A}{2}$。
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) 作为 $\cos^2\alpha = \dfrac{\cos 2\alpha + 1}{2}$：
$\cos^2 A + \dfrac{\cos 2B + 1}{2} + \dfrac{\cos 2C + 1}{2} = 1 + \cos^2 A + \cos(B + C)\cos(B - C) = 1 + \cos^2 A - \cos A \cos(B - C)$
$= 1 - \cos A[\cos(B - C) - \cos A] = 1 - \cos A[\cos(B - C) + \cos(B + C)] = 1 - 2\cos A \cos B \cos C$.
证明，如果 $A$、$B$、$C$ 是三角形的内角，则关系成立：
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) $\dfrac{1}{\operatorname{tg} A \operatorname{tg} B} + \dfrac{1}{\operatorname{tg} B \operatorname{tg} C} + \dfrac{1}{\operatorname{tg} C \operatorname{tg} A} = 1$ $\left(A, B, C \ne \dfrac{\pi}{2}\right)$.
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我们使用练习的预备部分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) $\operatorname{sen}^2 A + \operatorname{sen}^2 B + \operatorname{sen}^2 C = 3 - (\cos^2 A + \cos^2 B + \cos^2 C) = 3 - (1 - 2\cos A \cos B \cos C) = 2(1 + \cos A \cos B \cos C)$，用于练习 181d。
e) 年份 142，$\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = \operatorname{tg} A \operatorname{tg} B \operatorname{tg} C$。将两个成员除以 $\operatorname{tg} A \operatorname{tg} B \operatorname{tg} C$（不为空），我们获得了所请求的关系。
证明，如果$a + b + c = \dfrac{\pi}{2}$，则： a) $\operatorname{tg} a \operatorname{tg} b + \operatorname{tg} b \operatorname{tg} c + \operatorname{tg} c \operatorname{tg} a = 1$; b) $\cos^2 a + \cos^2 b + \cos^2 c - 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 2$。
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a) $a + b = \frac{\pi}{2} - c \Rightarrow \operatorname{tg}(a + b) = \operatorname{cotg} c \Rightarrow \dfrac{\operatorname{tg} a + \operatorname{tg} b}{1 - \operatorname{tg} a \operatorname{tg} b} = \dfrac{1}{\operatorname{tg} c} \Rightarrow \operatorname{tg} c(\operatorname{tg} a + \operatorname{tg} b) = 1 - \operatorname{tg} a \operatorname{tg} b$，这是请求的关系（假设已定义切线）。
b) 年份 186，$\operatorname{sen}^2 a + \operatorname{sen}^2 b + \operatorname{sen}^2 c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$。将每个$\operatorname{sen}^2$替换为$1 - \cos^2$：$3 - (\cos^2 a + \cos^2 b + \cos^2 c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$，这相当于所请求的关系。
证明，如果 $(\operatorname{sen} 2A, \operatorname{sen} 2B, \operatorname{sen} 2C)$ 是等差数列，那么 $(\operatorname{tg}(A + B), \operatorname{tg}(C + A), \operatorname{tg}(B + C))$ 也是如此。
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根据假设，$\operatorname{sen} 2B - \operatorname{sen} 2A = \operatorname{sen} 2C - \operatorname{sen} 2B$，即
$$2\operatorname{sen}(B - A)\cos(B + A) = 2\operatorname{sen}(C - B)\cos(C + B).$$写出$B - A = (B + C) - (A + C)$和$C - B = (C + A) - (A + B)$并按差值的正弦展开：
$$[\operatorname{sen}(B + C)\cos(A + C) - \operatorname{sen}(A + C)\cos(B + C)]\cos(B + A) = [\operatorname{sen}(C + A)\cos(A + B) - \operatorname{sen}(A + B)\cos(C + A)]\cos(C + B).$$除以 $\cos(A + B)\cos(B + C)\cos(C + A)$：
$$\frac{\operatorname{sen}(B + C)}{\cos(B + C)} - \frac{\operatorname{sen}(C + A)}{\cos(C + A)} = \frac{\operatorname{sen}(C + A)}{\cos(C + A)} - \frac{\operatorname{sen}(A + B)}{\cos(A + B)},$$即$\operatorname{tg}(B + C) - \operatorname{tg}(C + A) = \operatorname{tg}(C + A) - \operatorname{tg}(A + B)$：三个数按等差数列排列。
证明，如果 $(\operatorname{sen}(a + b - c), \operatorname{sen}(a - b + c), \operatorname{sen}(-a + b + c))$ 是等差数列，那么 $(\operatorname{tg} a, \operatorname{tg} b, \operatorname{tg} c)$ 也是如此。
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令它们为$\alpha = a + b - c$、$\beta = a - b + c$和$\gamma = -a + b + c$。假设是$(\operatorname{sen}\beta - \operatorname{sen}\alpha) + (\operatorname{sen}\beta - \operatorname{sen}\gamma) = 0$。转化为产品：
$$\operatorname{sen}\beta - \operatorname{sen}\alpha = 2\operatorname{sen}(c - b)\cos a \qquad \operatorname{sen}\beta - \operatorname{sen}\gamma = 2\operatorname{sen}(a - b)\cos c.$$标志 $\operatorname{sen}(c - b)\cos a = \operatorname{sen}(b - a)\cos c$。除以 $\cos a \cos b \cos c$ 并使用 $\operatorname{tg} p - \operatorname{tg} q = \dfrac{\operatorname{sen}(p - q)}{\cos p \cos q}$：
$$\frac{\operatorname{sen}(c - b)}{\cos b \cos c} = \frac{\operatorname{sen}(b - a)}{\cos a \cos b} \;\Longrightarrow\; \operatorname{tg} c - \operatorname{tg} b = \operatorname{tg} b - \operatorname{tg} a.$$证明如果$a + b + c = \dfrac{\pi}{2}$，则$\operatorname{sen}^2 a + \operatorname{sen}^2 b + \operatorname{sen}^2 c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$。
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作为 $b + c = \frac{\pi}{2} - a$，我们有 $\cos(b + c) = \operatorname{sen} a$。然后
$1^\circ$ 成员 $= \operatorname{sen}^2 a + \dfrac{1 - \cos 2b}{2} + \dfrac{1 - \cos 2c}{2} + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 + \operatorname{sen}^2 a - \dfrac{\cos 2b + \cos 2c}{2} + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c$
$= 1 + \operatorname{sen}^2 a - \cos(b + c)\cos(b - c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 + \operatorname{sen}^2 a - \operatorname{sen} a \cos(b - c) + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c$
$= 1 + \operatorname{sen} a[\cos(b + c) - \cos(b - c)] + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1 - 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c + 2\operatorname{sen} a \operatorname{sen} b \operatorname{sen} c = 1$.
研究$f(x) = \cos x - \operatorname{sen} x$给出的函数$f\colon \mathbb{R} \to \mathbb{R}$的变化。
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因此 $D(f) = \mathbb{R}$、$p = 2\pi$ 和 $\operatorname{Im}(f) = \left[-\sqrt{2}, \sqrt{2}\right]$ — 这证实了练习 62 中的图表。
研究$f(x) = \operatorname{sen} 2x + \cos 2x$给出的函数$f\colon \mathbb{R} \to \mathbb{R}$的变化。
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$f(x) = \sqrt{2}\operatorname{sen}\left(2x + \dfrac{\pi}{4}\right)$: $D(f) = \mathbb{R}$, $p(f) = \pi$, $\operatorname{Im}(f) = \left[-\sqrt{2}, \sqrt{2}\right]$
函数$f(x) = \dfrac{1 + \operatorname{tg} x}{1 - \operatorname{tg} x}$的周期是多少？
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$p(f) = \pi$
证明如果三角形 $ABC$ 的内角与 $\cos A + \cos B = \operatorname{sen} C$ 的关系成立，则该三角形是直角三角形。
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（因为$\cos\frac{A + B}{2} = \operatorname{sen}\frac{C}{2}$）。作为 $\operatorname{sen}\frac{C}{2} \ne 0$，出现 $\cos\dfrac{A - B}{2} = \cos\dfrac{C}{2}$，其中
$$A - B = C \;\Rightarrow\; A = B + C = \frac{\pi}{2} \qquad\text{ou}\qquad A - B = -C \;\Rightarrow\; B = A + C = \frac{\pi}{2}.$$证明如果三角形 $ABC$ 的内角与 $\operatorname{sen} 4A + \operatorname{sen} 4B + \operatorname{sen} 4C = 0$ 的关系成立，则该三角形是直角三角形。
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作为 $2A + 2B = 2\pi - 2C$，我们有 $\operatorname{sen}(2A + 2B) = -\operatorname{sen} 2C$。然后
$$\operatorname{sen} 4A + \operatorname{sen} 4B + \operatorname{sen} 4C = 2\operatorname{sen}(2A + 2B)\cos(2A - 2B) + 2\operatorname{sen} 2C \cos 2C = 2\operatorname{sen} 2C[\cos 2C - \cos(2A - 2B)].$$例如$2C = 2\pi - 2A - 2B$、$\cos 2C = \cos(2A + 2B)$和$\cos(2A + 2B) - \cos(2A - 2B) = -2\operatorname{sen} 2A \operatorname{sen} 2B$。因此，总和值是$-4\operatorname{sen} 2A \operatorname{sen} 2B \operatorname{sen} 2C$。如果为空，则说明某个因子为空，例如$\operatorname{sen} 2A = 0$和$0 < 2A < 2\pi$，即$2A = \pi$和$A = \frac{\pi}{2}$。
证明每个角度验证关系 $\operatorname{sen} 3A + \operatorname{sen} 3B + \operatorname{sen} 3C = 0$ 的三角形都有一个角度 $60^\circ$。
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对于 $\frac{3(A + B)}{2} = \frac{3\pi}{2} - \frac{3C}{2}$，我们有 $\operatorname{sen}\frac{3(A + B)}{2} = -\cos\frac{3C}{2}$ 和 $\operatorname{sen}\frac{3C}{2} = -\cos\frac{3(A + B)}{2}$。然后
$$\operatorname{sen} 3A + \operatorname{sen} 3B + \operatorname{sen} 3C = 2\operatorname{sen}\frac{3(A + B)}{2}\cos\frac{3(A - B)}{2} + 2\operatorname{sen}\frac{3C}{2}\cos\frac{3C}{2}$$$$= -2\cos\frac{3C}{2}\left[\cos\frac{3(A - B)}{2} + \cos\frac{3(A + B)}{2}\right] = -4\cos\frac{3A}{2}\cos\frac{3B}{2}\cos\frac{3C}{2}.$$如果总和为空，则某个因子为空，例如 $\cos\frac{3A}{2} = 0$；如$0 < \frac{3A}{2} < \frac{3\pi}{2}$，则只能是$\frac{3A}{2} = \frac{\pi}{2}$，即$A = \frac{\pi}{3} = 60^\circ$。