Номер 770, страница 218 - гдз по алгебре 9 класс учебник Солтан, Солтан

Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019

Авторы: Солтан Г. Н., Солтан А. Е., Жумадилова А. Ж.

Тип: Учебник

Издательство: Кокшетау

Год издания: 2019 - 2026

ISBN: 978-601-317-424-2

Рекомендовано Министерством образования и науки Республики Казахстан

IV. Тригонометрия. 29. Преобразование произведения тригонометрических функций в сумму и разность - номер 770, страница 218.

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            "text" => "<p><strong>770.</strong> Докажите, что верно равенство:</p><p><strong>a)</strong> $\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ + \operatorname{tg} 20^\circ \cdot \operatorname{tg} 25^\circ = 1;$</p><p><strong>б)</strong> $\operatorname{ctg} 10^\circ - \operatorname{tg} 35^\circ - \operatorname{ctg} 10^\circ \cdot \operatorname{tg} 35^\circ = 1;$</p><p><strong>в)</strong> $\cos 15^\circ \cdot \cos 7^\circ - \sin 79^\circ \cdot \cos 11^\circ - \sin 86^\circ \cdot \cos 4^\circ = -1;$</p><p><strong>г)</strong> $\sin 17^\circ \cdot \sin 73^\circ - \sin 21^\circ \cdot \cos 13^\circ + \sin 4^\circ \cdot \sin 86^\circ = 0.$</p>"
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            "text" => "<p><strong>770.</strong> Докажите, что верно равенство:</p><p><strong>a)</strong> $\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ + \operatorname{tg} 20^\circ \cdot \operatorname{tg} 25^\circ = 1;$</p><p><strong>б)</strong> $\operatorname{ctg} 10^\circ - \operatorname{tg} 35^\circ - \operatorname{ctg} 10^\circ \cdot \operatorname{tg} 35^\circ = 1;$</p><p><strong>в)</strong> $\cos 15^\circ \cdot \cos 7^\circ - \sin 79^\circ \cdot \cos 11^\circ - \sin 86^\circ \cdot \cos 4^\circ = -1;$</p><p><strong>г)</strong> $\sin 17^\circ \cdot \sin 73^\circ - \sin 21^\circ \cdot \cos 13^\circ + \sin 4^\circ \cdot \sin 86^\circ = 0.$</p>"
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            "text" => "<p><strong>а)</strong> Для доказательства воспользуемся формулой тангенса суммы: $\operatorname{tg}(\alpha + \beta) = \frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.<br>Пусть $\alpha = 20^{\circ}$ и $\beta = 25^{\circ}$. Тогда их сумма $\alpha + \beta = 20^{\circ} + 25^{\circ} = 45^{\circ}$.<br>Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.<br>Подставим значения в формулу:<br>$\operatorname{tg}(20^{\circ} + 25^{\circ}) = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$<br>$1 = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$<br>Умножим обе части на знаменатель $(1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$:<br>$1 \cdot (1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}) = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$<br>$1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ} = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$<br>Перенесем слагаемое $(-\operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$ в правую часть:<br>$1 = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ} + \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}$<br>Равенство доказано.<br>Ответ:</p><p><strong>б)</strong> Преобразуем левую часть равенства. Воспользуемся формулой приведения $\operatorname{ctg}\alpha = \operatorname{tg}(90^{\circ} - \alpha)$.<br>$\operatorname{ctg}10^{\circ} = \operatorname{tg}(90^{\circ} - 10^{\circ}) = \operatorname{tg}80^{\circ}$.<br>Теперь исходное равенство принимает вид:<br>$\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = 1$.<br>Рассмотрим формулу тангенса разности: $\operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.<br>Пусть $\alpha = 80^{\circ}$ и $\beta = 35^{\circ}$. Тогда их разность $\alpha - \beta = 80^{\circ} - 35^{\circ} = 45^{\circ}$.<br>Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.<br>$\operatorname{tg}(80^{\circ} - 35^{\circ}) = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$<br>$1 = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$<br>Умножим обе части на знаменатель $(1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$:<br>$1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}$<br>Перенесем слагаемое $(\operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$ в правую часть:<br>$1 = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}$<br>Подставив обратно $\operatorname{tg}80^{\circ} = \operatorname{ctg}10^{\circ}$, получаем исходное равенство. Равенство доказано.<br>Ответ:</p><p><strong>в)</strong> Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:<br>$\sin 79^{\circ} = \sin(90^{\circ} - 11^{\circ}) = \cos 11^{\circ}$<br>$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$<br>Подставим эти выражения в исходное равенство:<br>$\cos 15^{\circ} \cdot \cos 7^{\circ} - \cos 11^{\circ} \cdot \cos 11^{\circ} - \cos 4^{\circ} \cdot \cos 4^{\circ} = \cos 15^{\circ} \cdot \cos 7^{\circ} - \cos^2 11^{\circ} - \cos^2 4^{\circ}$.<br>Теперь используем формулу произведения косинусов $\cos\alpha\cos\beta = \frac{1}{2}(\cos(\alpha+\beta) + \cos(\alpha-\beta))$ и формулу понижения степени $\cos^2\alpha = \frac{1+\cos(2\alpha)}{2}$.<br>$\cos 15^{\circ} \cdot \cos 7^{\circ} = \frac{1}{2}(\cos(15^{\circ}+7^{\circ}) + \cos(15^{\circ}-7^{\circ})) = \frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ})$<br>$\cos^2 11^{\circ} = \frac{1 + \cos(2 \cdot 11^{\circ})}{2} = \frac{1 + \cos 22^{\circ}}{2}$<br>$\cos^2 4^{\circ} = \frac{1 + \cos(2 \cdot 4^{\circ})}{2} = \frac{1 + \cos 8^{\circ}}{2}$<br>Подставим полученные выражения в левую часть:<br>$\frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ}) - \frac{1 + \cos 22^{\circ}}{2} - \frac{1 + \cos 8^{\circ}}{2}$<br>$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - (1 + \cos 22^{\circ}) - (1 + \cos 8^{\circ})}{2}$<br>$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - 1 - \cos 22^{\circ} - 1 - \cos 8^{\circ}}{2}$<br>$= \frac{(\cos 22^{\circ} - \cos 22^{\circ}) + (\cos 8^{\circ} - \cos 8^{\circ}) - 1 - 1}{2} = \frac{-2}{2} = -1$<br>Левая часть равна -1, что совпадает с правой частью. Равенство доказано.<br>Ответ:</p><p><strong>г)</strong> Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:<br>$\sin 73^{\circ} = \sin(90^{\circ} - 17^{\circ}) = \cos 17^{\circ}$<br>$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$<br>Подставим эти выражения в исходное равенство:<br>$\sin 17^{\circ} \cdot \cos 17^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \sin 4^{\circ} \cdot \cos 4^{\circ}$.<br>Используем формулу синуса двойного угла $\sin(2\alpha) = 2\sin\alpha\cos\alpha$, откуда $\sin\alpha\cos\alpha = \frac{1}{2}\sin(2\alpha)$:<br>$\sin 17^{\circ} \cdot \cos 17^{\circ} = \frac{1}{2}\sin(2 \cdot 17^{\circ}) = \frac{1}{2}\sin 34^{\circ}$<br>$\sin 4^{\circ} \cdot \cos 4^{\circ} = \frac{1}{2}\sin(2 \cdot 4^{\circ}) = \frac{1}{2}\sin 8^{\circ}$<br>Подставим в выражение:<br>$\frac{1}{2}\sin 34^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \frac{1}{2}\sin 8^{\circ} = \frac{1}{2}(\sin 34^{\circ} + \sin 8^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ}$.<br>Теперь применим формулу суммы синусов $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:<br>$\sin 34^{\circ} + \sin 8^{\circ} = 2\sin\frac{34^{\circ}+8^{\circ}}{2}\cos\frac{34^{\circ}-8^{\circ}}{2} = 2\sin\frac{42^{\circ}}{2}\cos\frac{26^{\circ}}{2} = 2\sin 21^{\circ}\cos 13^{\circ}$.<br>Подставим результат обратно в выражение:<br>$\frac{1}{2}(2\sin 21^{\circ}\cos 13^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ} = \sin 21^{\circ}\cos 13^{\circ} - \sin 21^{\circ}\cos 13^{\circ} = 0$.<br>Левая часть равна 0, что совпадает с правой частью. Равенство доказано.<br>Ответ:</p>"
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            "text" => "<p><strong>а)</strong> Для доказательства воспользуемся формулой тангенса суммы: $\operatorname{tg}(\alpha + \beta) = \frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.<br>Пусть $\alpha = 20^{\circ}$ и $\beta = 25^{\circ}$. Тогда их сумма $\alpha + \beta = 20^{\circ} + 25^{\circ} = 45^{\circ}$.<br>Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.<br>Подставим значения в формулу:<br>$\operatorname{tg}(20^{\circ} + 25^{\circ}) = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$<br>$1 = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$<br>Умножим обе части на знаменатель $(1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$:<br>$1 \cdot (1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}) = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$<br>$1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ} = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$<br>Перенесем слагаемое $(-\operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$ в правую часть:<br>$1 = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ} + \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}$<br>Равенство доказано.<br>Ответ:</p><p><strong>б)</strong> Преобразуем левую часть равенства. Воспользуемся формулой приведения $\operatorname{ctg}\alpha = \operatorname{tg}(90^{\circ} - \alpha)$.<br>$\operatorname{ctg}10^{\circ} = \operatorname{tg}(90^{\circ} - 10^{\circ}) = \operatorname{tg}80^{\circ}$.<br>Теперь исходное равенство принимает вид:<br>$\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = 1$.<br>Рассмотрим формулу тангенса разности: $\operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.<br>Пусть $\alpha = 80^{\circ}$ и $\beta = 35^{\circ}$. Тогда их разность $\alpha - \beta = 80^{\circ} - 35^{\circ} = 45^{\circ}$.<br>Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.<br>$\operatorname{tg}(80^{\circ} - 35^{\circ}) = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$<br>$1 = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$<br>Умножим обе части на знаменатель $(1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$:<br>$1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}$<br>Перенесем слагаемое $(\operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$ в правую часть:<br>$1 = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}$<br>Подставив обратно $\operatorname{tg}80^{\circ} = \operatorname{ctg}10^{\circ}$, получаем исходное равенство. Равенство доказано.<br>Ответ:</p><p><strong>в)</strong> Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:<br>$\sin 79^{\circ} = \sin(90^{\circ} - 11^{\circ}) = \cos 11^{\circ}$<br>$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$<br>Подставим эти выражения в исходное равенство:<br>$\cos 15^{\circ} \cdot \cos 7^{\circ} - \cos 11^{\circ} \cdot \cos 11^{\circ} - \cos 4^{\circ} \cdot \cos 4^{\circ} = \cos 15^{\circ} \cdot \cos 7^{\circ} - \cos^2 11^{\circ} - \cos^2 4^{\circ}$.<br>Теперь используем формулу произведения косинусов $\cos\alpha\cos\beta = \frac{1}{2}(\cos(\alpha+\beta) + \cos(\alpha-\beta))$ и формулу понижения степени $\cos^2\alpha = \frac{1+\cos(2\alpha)}{2}$.<br>$\cos 15^{\circ} \cdot \cos 7^{\circ} = \frac{1}{2}(\cos(15^{\circ}+7^{\circ}) + \cos(15^{\circ}-7^{\circ})) = \frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ})$<br>$\cos^2 11^{\circ} = \frac{1 + \cos(2 \cdot 11^{\circ})}{2} = \frac{1 + \cos 22^{\circ}}{2}$<br>$\cos^2 4^{\circ} = \frac{1 + \cos(2 \cdot 4^{\circ})}{2} = \frac{1 + \cos 8^{\circ}}{2}$<br>Подставим полученные выражения в левую часть:<br>$\frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ}) - \frac{1 + \cos 22^{\circ}}{2} - \frac{1 + \cos 8^{\circ}}{2}$<br>$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - (1 + \cos 22^{\circ}) - (1 + \cos 8^{\circ})}{2}$<br>$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - 1 - \cos 22^{\circ} - 1 - \cos 8^{\circ}}{2}$<br>$= \frac{(\cos 22^{\circ} - \cos 22^{\circ}) + (\cos 8^{\circ} - \cos 8^{\circ}) - 1 - 1}{2} = \frac{-2}{2} = -1$<br>Левая часть равна -1, что совпадает с правой частью. Равенство доказано.<br>Ответ:</p><p><strong>г)</strong> Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:<br>$\sin 73^{\circ} = \sin(90^{\circ} - 17^{\circ}) = \cos 17^{\circ}$<br>$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$<br>Подставим эти выражения в исходное равенство:<br>$\sin 17^{\circ} \cdot \cos 17^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \sin 4^{\circ} \cdot \cos 4^{\circ}$.<br>Используем формулу синуса двойного угла $\sin(2\alpha) = 2\sin\alpha\cos\alpha$, откуда $\sin\alpha\cos\alpha = \frac{1}{2}\sin(2\alpha)$:<br>$\sin 17^{\circ} \cdot \cos 17^{\circ} = \frac{1}{2}\sin(2 \cdot 17^{\circ}) = \frac{1}{2}\sin 34^{\circ}$<br>$\sin 4^{\circ} \cdot \cos 4^{\circ} = \frac{1}{2}\sin(2 \cdot 4^{\circ}) = \frac{1}{2}\sin 8^{\circ}$<br>Подставим в выражение:<br>$\frac{1}{2}\sin 34^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \frac{1}{2}\sin 8^{\circ} = \frac{1}{2}(\sin 34^{\circ} + \sin 8^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ}$.<br>Теперь применим формулу суммы синусов $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:<br>$\sin 34^{\circ} + \sin 8^{\circ} = 2\sin\frac{34^{\circ}+8^{\circ}}{2}\cos\frac{34^{\circ}-8^{\circ}}{2} = 2\sin\frac{42^{\circ}}{2}\cos\frac{26^{\circ}}{2} = 2\sin 21^{\circ}\cos 13^{\circ}$.<br>Подставим результат обратно в выражение:<br>$\frac{1}{2}(2\sin 21^{\circ}\cos 13^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ} = \sin 21^{\circ}\cos 13^{\circ} - \sin 21^{\circ}\cos 13^{\circ} = 0$.<br>Левая часть равна 0, что совпадает с правой частью. Равенство доказано.<br>Ответ:</p>"
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                1 => App\Models\Task {#1280
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                2 => App\Models\Task {#1408
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                3 => App\Models\Task {#1536
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                4 => App\Models\Task {#1664
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                    "field_task_type" => Illuminate\Database\Eloquent\Collection {#1665 …2}
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                5 => App\Models\Task {#1736
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                6 => App\Models\Task {#1896
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                    "field_task_type" => Illuminate\Database\Eloquent\Collection {#1897 …2}
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№770 (с. 218)
Условие. №770 (с. 218)
скриншот условия
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 218, номер 770, Условие

770. Докажите, что верно равенство:

a) $\operatorname{tg} 20^\circ + \operatorname{tg} 25^\circ + \operatorname{tg} 20^\circ \cdot \operatorname{tg} 25^\circ = 1;$

б) $\operatorname{ctg} 10^\circ - \operatorname{tg} 35^\circ - \operatorname{ctg} 10^\circ \cdot \operatorname{tg} 35^\circ = 1;$

в) $\cos 15^\circ \cdot \cos 7^\circ - \sin 79^\circ \cdot \cos 11^\circ - \sin 86^\circ \cdot \cos 4^\circ = -1;$

г) $\sin 17^\circ \cdot \sin 73^\circ - \sin 21^\circ \cdot \cos 13^\circ + \sin 4^\circ \cdot \sin 86^\circ = 0.$

Решение. №770 (с. 218)
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 218, номер 770, Решение
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 218, номер 770, Решение (продолжение 2)
Решение 2 (rus). №770 (с. 218)

а) Для доказательства воспользуемся формулой тангенса суммы: $\operatorname{tg}(\alpha + \beta) = \frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.
Пусть $\alpha = 20^{\circ}$ и $\beta = 25^{\circ}$. Тогда их сумма $\alpha + \beta = 20^{\circ} + 25^{\circ} = 45^{\circ}$.
Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.
Подставим значения в формулу:
$\operatorname{tg}(20^{\circ} + 25^{\circ}) = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$
$1 = \frac{\operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}}{1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}}$
Умножим обе части на знаменатель $(1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$:
$1 \cdot (1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}) = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$
$1 - \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ} = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ}$
Перенесем слагаемое $(-\operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ})$ в правую часть:
$1 = \operatorname{tg}20^{\circ} + \operatorname{tg}25^{\circ} + \operatorname{tg}20^{\circ} \cdot \operatorname{tg}25^{\circ}$
Равенство доказано.
Ответ:

б) Преобразуем левую часть равенства. Воспользуемся формулой приведения $\operatorname{ctg}\alpha = \operatorname{tg}(90^{\circ} - \alpha)$.
$\operatorname{ctg}10^{\circ} = \operatorname{tg}(90^{\circ} - 10^{\circ}) = \operatorname{tg}80^{\circ}$.
Теперь исходное равенство принимает вид:
$\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = 1$.
Рассмотрим формулу тангенса разности: $\operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha \cdot \operatorname{tg}\beta}$.
Пусть $\alpha = 80^{\circ}$ и $\beta = 35^{\circ}$. Тогда их разность $\alpha - \beta = 80^{\circ} - 35^{\circ} = 45^{\circ}$.
Мы знаем, что $\operatorname{tg}45^{\circ} = 1$.
$\operatorname{tg}(80^{\circ} - 35^{\circ}) = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$
$1 = \frac{\operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}}{1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}}$
Умножим обе части на знаменатель $(1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$:
$1 + \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ} = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ}$
Перенесем слагаемое $(\operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ})$ в правую часть:
$1 = \operatorname{tg}80^{\circ} - \operatorname{tg}35^{\circ} - \operatorname{tg}80^{\circ} \cdot \operatorname{tg}35^{\circ}$
Подставив обратно $\operatorname{tg}80^{\circ} = \operatorname{ctg}10^{\circ}$, получаем исходное равенство. Равенство доказано.
Ответ:

в) Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:
$\sin 79^{\circ} = \sin(90^{\circ} - 11^{\circ}) = \cos 11^{\circ}$
$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$
Подставим эти выражения в исходное равенство:
$\cos 15^{\circ} \cdot \cos 7^{\circ} - \cos 11^{\circ} \cdot \cos 11^{\circ} - \cos 4^{\circ} \cdot \cos 4^{\circ} = \cos 15^{\circ} \cdot \cos 7^{\circ} - \cos^2 11^{\circ} - \cos^2 4^{\circ}$.
Теперь используем формулу произведения косинусов $\cos\alpha\cos\beta = \frac{1}{2}(\cos(\alpha+\beta) + \cos(\alpha-\beta))$ и формулу понижения степени $\cos^2\alpha = \frac{1+\cos(2\alpha)}{2}$.
$\cos 15^{\circ} \cdot \cos 7^{\circ} = \frac{1}{2}(\cos(15^{\circ}+7^{\circ}) + \cos(15^{\circ}-7^{\circ})) = \frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ})$
$\cos^2 11^{\circ} = \frac{1 + \cos(2 \cdot 11^{\circ})}{2} = \frac{1 + \cos 22^{\circ}}{2}$
$\cos^2 4^{\circ} = \frac{1 + \cos(2 \cdot 4^{\circ})}{2} = \frac{1 + \cos 8^{\circ}}{2}$
Подставим полученные выражения в левую часть:
$\frac{1}{2}(\cos 22^{\circ} + \cos 8^{\circ}) - \frac{1 + \cos 22^{\circ}}{2} - \frac{1 + \cos 8^{\circ}}{2}$
$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - (1 + \cos 22^{\circ}) - (1 + \cos 8^{\circ})}{2}$
$= \frac{\cos 22^{\circ} + \cos 8^{\circ} - 1 - \cos 22^{\circ} - 1 - \cos 8^{\circ}}{2}$
$= \frac{(\cos 22^{\circ} - \cos 22^{\circ}) + (\cos 8^{\circ} - \cos 8^{\circ}) - 1 - 1}{2} = \frac{-2}{2} = -1$
Левая часть равна -1, что совпадает с правой частью. Равенство доказано.
Ответ:

г) Преобразуем левую часть равенства, используя формулы приведения $\sin(90^{\circ} - \alpha) = \cos\alpha$:
$\sin 73^{\circ} = \sin(90^{\circ} - 17^{\circ}) = \cos 17^{\circ}$
$\sin 86^{\circ} = \sin(90^{\circ} - 4^{\circ}) = \cos 4^{\circ}$
Подставим эти выражения в исходное равенство:
$\sin 17^{\circ} \cdot \cos 17^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \sin 4^{\circ} \cdot \cos 4^{\circ}$.
Используем формулу синуса двойного угла $\sin(2\alpha) = 2\sin\alpha\cos\alpha$, откуда $\sin\alpha\cos\alpha = \frac{1}{2}\sin(2\alpha)$:
$\sin 17^{\circ} \cdot \cos 17^{\circ} = \frac{1}{2}\sin(2 \cdot 17^{\circ}) = \frac{1}{2}\sin 34^{\circ}$
$\sin 4^{\circ} \cdot \cos 4^{\circ} = \frac{1}{2}\sin(2 \cdot 4^{\circ}) = \frac{1}{2}\sin 8^{\circ}$
Подставим в выражение:
$\frac{1}{2}\sin 34^{\circ} - \sin 21^{\circ} \cdot \cos 13^{\circ} + \frac{1}{2}\sin 8^{\circ} = \frac{1}{2}(\sin 34^{\circ} + \sin 8^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ}$.
Теперь применим формулу суммы синусов $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$:
$\sin 34^{\circ} + \sin 8^{\circ} = 2\sin\frac{34^{\circ}+8^{\circ}}{2}\cos\frac{34^{\circ}-8^{\circ}}{2} = 2\sin\frac{42^{\circ}}{2}\cos\frac{26^{\circ}}{2} = 2\sin 21^{\circ}\cos 13^{\circ}$.
Подставим результат обратно в выражение:
$\frac{1}{2}(2\sin 21^{\circ}\cos 13^{\circ}) - \sin 21^{\circ} \cdot \cos 13^{\circ} = \sin 21^{\circ}\cos 13^{\circ} - \sin 21^{\circ}\cos 13^{\circ} = 0$.
Левая часть равна 0, что совпадает с правой частью. Равенство доказано.
Ответ:

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