Example Question - cosine
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Trigonometric Expression Simplification
<p>Для решения данной задачи, используйте тригонометрические формулы.</p> <p>Упростим выражение \(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)}\), используя тригонометрические тождества:</p> <p>\(1 - \cos(2\alpha) = 1 - (1 - 2\sin^2(\alpha)) = 2\sin^2(\alpha)\)</p> <p>\(\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)\)</p> <p>Теперь подставим упрощённые выражения:</p> <p>\(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)} = \frac{2\sin^2(\alpha)}{2\sin(\alpha)\cos(\alpha)}\)</p> <p>Сократите выражение на \(2\sin(\alpha)\):</p> <p>\(\frac{2\sin^2(\alpha)}{2\sin(\alpha)\cos(\alpha)} = \frac{\sin(\alpha)}{\cos(\alpha)}\)</p> <p>Используя определение тангенса, получаем:</p> <p>\(\frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)\)</p> <p>Итак, \(\frac{1 - \cos(2\alpha)}{\sin(2\alpha)} = \tan(\alpha)\)</p>
Trigonometric Functions in Right Triangles
<p>Para el inciso a, para calcular \(x\) usamos la definición de seno en un triángulo rectángulo: \( \text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}} \).</p> <p>Entonces, tenemos que \( \text{sen}(\theta) = 0.28 = \frac{x}{24} \).</p> <p>\( x = 0.28 \cdot 24 \)</p> <p>\( x = 6.72 \text{ cm} \)</p> <p>Para el inciso b, para calcular la longitud del lado opuesto al ángulo \( \beta \), usaremos la definición de coseno: \( \text{cos}(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}} \).</p> <p>Tenemos \( \text{cos}(\beta) = 0.324 = \frac{\text{adyacente}}{35} \).</p> <p>La longitud del lado adyacente es entonces \( \text{adyacente} = 0.324 \cdot 35 \).</p> <p>\( \text{adyacente} = 11.34 \text{ cm} \)</p> <p>Finalmente, para el inciso c, para calcular la longitud de \( \overline{AB} \), utilizamos la definición de tangente: \( \text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}} \).</p> <p>\( \text{tg}(\alpha) = 1.3 = \frac{\text{opuesto}}{10} \).</p> <p>La longitud del lado opuesto a \( \alpha \) es \( \text{opuesto} = 1.3 \cdot 10 \).</p> <p>\( \text{opuesto} = 13 \text{ cm} \)</p>
Trigonometry Problems Involving Right Triangles
Para el inciso a: <p>\(\text{Usamos la definición de seno en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Hipotenusa}}\)</p> <p>\(sen(\theta) = \frac{x}{24}\)</p> <p>\(0.28 = \frac{x}{24}\)</p> <p>\(x = 0.28 \times 24\)</p> <p>\(x = 6.72 \text{ cm}\)</p> Para el inciso b: <p>\(\text{Usamos la definición de coseno en un triángulo rectángulo:} \frac{\text{Cateto adyacente}}{\text{Hipotenusa}}\)</p> <p>\(cos(\beta) = \frac{35}{x}\)</p> <p>\(0.324 = \frac{35}{x}\)</p> <p>\(x = \frac{35}{0.324}\)</p> <p>\(x \approx 108.025 \text{ cm}\)</p> Para el inciso c: <p>\(\text{Usamos la definición de tangente en un triángulo rectángulo:} \frac{\text{Cateto opuesto}}{\text{Cateto adyacente}}\)</p> <p>\(tg(\alpha) = \frac{x}{10}\)</p> <p>\(1.3 = \frac{x}{10}\)</p> <p>\(x = 1.3 \times 10\)</p> <p>\(x = 13 \text{ cm}\)</p>
Trigonometric Functions in Right Triangles
<p>Para la parte a), usando la definición de seno en un triángulo rectángulo:</p> <p>\[\text{sen}(\theta) = \frac{\text{opuesto}}{\text{hipotenusa}}\]</p> <p>Si \(\text{sen}(\theta) = 0,28\), entonces la hipotenusa es \(\frac{24}{0,28}\):</p> <p>\[x = \frac{24}{0,28}\]</p> <p>\[x = 85,71\ (2 d.p.)\]</p> <p>Para la parte b), usando la definición de coseno:</p> <p>\[\cos(\theta) = \frac{\text{adyacente}}{\text{hipotenusa}}\]</p> <p>Si \(\cos(\beta) = 0,324\), entonces la longitud adyacente es \(\frac{35}{0,324}\):</p> <p>\[x = \frac{35}{0,324}\]</p> <p>\[x = 108,02\ (2 d.p.)\]</p> <p>Para la parte c), usando la definición de tangente:</p> <p>\[\text{tg}(\theta) = \frac{\text{opuesto}}{\text{adyacente}}\]</p> <p>Si \(\text{tg}(\alpha) = 1,3\), entonces la longitud adyacente es \(\frac{10}{1,3}\):</p> <p>\[x = \frac{10}{1,3}\]</p> <p>\[x = 7,69\ (2 d.p.)\]</p>
Differential Equation Problem Solving
Dado que la imagen proporciona un problema de ecuación diferencial, primero identificamos la ecuación dada y luego seguimos los pasos para resolverla. La ecuación proporcionada es: \[ \left(\frac{1}{1 + y^2} + \cos x - 2xy\right)\frac{dy}{dx} = y(\sqrt{y} + \sen x), \quad y(0) = 1 \] Para resolver esta ecuación diferencial, vamos a realizar los siguientes pasos: <p>Paso 1: Separar las variables $\frac{dy}{dx}$ y $x$, a un lado de la igualdad y las variables $dy$ y $y$, al otro lado.</p> <p>Paso 2: Integramos ambos lados para encontrar una solución implícita.</p> La ecuación es no lineal y no se separa fácilmente en términos de $x$ y $y$, lo cual significa que requerimos aplicar técnicas avanzadas de resolución que pueden involucrar métodos numéricos o cambios de variables adecuados. En este caso, la solución explícita puede no ser directa o posible de expresar en términos de funciones elementales.</p> <p>Nota: La resolución detallada de dicha ecuación diferencial va más allá del alcance de esta respuesta simplificada y requerirá herramientas matemáticas más avanzadas como son las ecuaciones diferenciales implícitas, uso de series de potencias, o aproximaciones numéricas.</p>
Solving an Equation Involving Absolute Values and Trigonometric Functions
<p>Для решения уравнения, содержащего модули, необходимо рассмотреть различные случаи, в зависимости от знака выражений внутри модулей.</p> <p>\textbf{Случай 1:} $x \geq -4$ и $x \geq -1$</p> <p>Уравнение примет вид:</p> <p>$x + 4 + x + 1 + 3 \cos(\pi x) = 0$</p> <p>$2x + 5 + 3 \cos(\pi x) = 0$</p> <p>Дальнейшее аналитическое решение уравнения затруднительно из-за комбинации косинуса и линейных членов. Необходимо использовать численные методы или графический подход для нахождения корней.</p> <p>\textbf{Случай 2:} $x < -4$ и $x \geq -1$</p> <p>Уравнение примет вид:</p> <p>$-(x + 4) + x + 1 + 3 \cos(\pi x) = 0$</p> <p>$-3 + 3 \cos(\pi x) = 0$</p> <p>$3 \cos(\pi x) = 3$</p> <p>$\cos(\pi x) = 1$</p> <p>Это возможно, если $\pi x = 2k\pi$, где $k$ - целое число.</p> <p>$x = 2k$, где $x < -4$, что несовместимо с этим условием, следовательно, решений нет.</p> <p>\textbf{Случай 3:} $x \geq -4$ и $x < -1$</p> <p>Уравнение примет вид:</p> <p>$x + 4 - (x + 1) + 3 \cos(\pi x) = 0$</p> <p>$3 + 3 \cos(\pi x) = 0$</p> <p>Аналогично случаю 2, получаем $x = 2k$, что не удовлетворяет условию $x < -1$. Следовательно, решений нет.</p> <p>\textbf{Случай 4:} $x < -4$ и $x < -1$</p> <p>Уравнение примет вид:</p> <p>$-(x + 4) - (x + 1) + 3 \cos(\pi x) = 0$</p> <p>$-2x - 5 + 3 \cos(\pi x) = 0$</p> <p>$-2x + 3 \cos(\pi x) = 5$</p> <p>И опять, аналитическое решение затруднительно, и потребуются численные методы или графический подход.</p> <p>Итак, общее решение уравнения будет включать численный поиск корней для случаев 1 и 4, так как аналитическое решение невозможно из-за комбинации алгебраических и тригонометрических функций.</p>
Trigonometric Equation Solving
\[ \begin{align*} &\text{给定方程式为} \quad 2\tan(\theta) = 3\cos(\theta) \\ &\text{解这个方程式，我们首先将其转换成同一个三角函数，} \\ &\text{使用基本恒等式} \quad \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}: \\ &2\frac{\sin(\theta)}{\cos(\theta)} = 3\cos(\theta) \\ &2\sin(\theta) = 3\cos^2(\theta) \\ &\text{因为} \quad \cos^2(\theta) = 1 - \sin^2(\theta), \text{代入上面方程得到} \\ &2\sin(\theta) = 3(1 - \sin^2(\theta)) \\ &3\sin^2(\theta) + 2\sin(\theta) - 3 = 0 \\ &\text{将}\sin(\theta)\text{代换为} x \\ &3x^2 + 2x - 3 = 0 \\ &\text{使用求根公式得到:} \\ &x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &x = \frac{-2 \pm \sqrt{(2)^2 - 4(3)(-3)}}{2(3)} \\ &x = \frac{-2 \pm \sqrt{4 + 36}}{6} \\ &x = \frac{-2 \pm \sqrt{40}}{6} \\ &x = \frac{-2 \pm 2\sqrt{10}}{6} \\ &\text{因为} -1 \leq \sin(\theta) \leq 1, \text{所以我们舍弃} x > 1 \text{和} x < -1 \text{的根。} \\ &x = -1 \text{（不合理）或 } x = \frac{\sqrt{10}-1}{3} \\ &\text{所以}\sin(\theta) = \frac{\sqrt{10}-1}{3} \\ &\text{利用计算器得到} \theta \text{的两个可能的值(确保在 0° 至 360°之间)} \\ &\theta \approx 50.2^\circ \text{或} 360^\circ - 50.2^\circ \approx 309.8^\circ \\ &\text{最终答案为：} \theta \approx 50.2^\circ \text{或} 309.8^\circ \end{align*} \]
Trigonometric Equation Solution
<p>给定方程 \(2 \tan \theta = 3 \cos \theta\)，求解 \(\theta\)，其中 \(0^\circ \leq \theta < 360^\circ\)。</p> <p>将 \(\tan \theta\) 转换为 \(\sin \theta\) 和 \( \cos \theta\) 的比率：</p> <p>\(2 \cdot \frac{\sin \theta}{\cos \theta} = 3 \cos \theta\)</p> <p>交叉相乘得到：</p> <p>\(2 \sin \theta = 3 \cos^2 \theta\)</p> <p>我们知道 \(\cos^2 \theta = 1 - \sin^2 \theta\), 所以代入得：</p> <p>\(2 \sin \theta = 3 (1 - \sin^2 \theta)\)</p> <p>现在我们得到一个关于 \(\sin \theta\) 的二次方程：</p> <p>\(3 \sin^2 \theta + 2 \sin \theta - 3 = 0\)</p> <p>使用求根公式得到 \(\sin \theta\) 的值：</p> <p>\(\sin \theta = \frac{-2 \pm \sqrt{2^2 - 4\cdot3\cdot(-3)}}{2\cdot3}\)</p> <p>\(\sin \theta = \frac{-2 \pm \sqrt{4+36}}{6}\)</p> <p>\(\sin \theta = \frac{-2 \pm \sqrt{40}}{6}\)</p> <p>\(\sin \theta = \frac{-2 \pm 2\sqrt{10}}{6}\)</p> <p>由于 \(0^\circ \leq \theta < 360^\circ\)，我们需要考虑正弦函数的所有可能值：</p> <p>负数解不在给定的区间内，我们采用：</p> <p>\(\sin \theta = \frac{-2 + 2\sqrt{10}}{6}\)</p> <p>然后利用反正弦函数找到对应的角度值，在 \(0^\circ \leq \theta < 180^\circ\) 时为主值，\(180^\circ \leq \theta < 360^\circ\) 时为同周期值：</p> <p>\(\theta = \arcsin\left(\frac{-2 + 2\sqrt{10}}{6}\right)\) or \(\theta = 180^\circ - \arcsin\left(\frac{-2 + 2\sqrt{10}}{6}\right)\)</p> <p>计算得到角度值并保留到最近的 \(0.1^\circ\)：</p> <p>\(\theta \approx 43.9^\circ\) or \(\theta \approx 136.1^\circ\)</p> <p>所以答案是 C)。</p>
Solving an Integral Involving a Trigonometric Function
<p>The integral of \( f(x) = -3 \cos(4x) \) can be found using the standard integration techniques for trigonometric functions.</p> <p>Let \( u = 4x \). Therefore, \( du = 4dx \) or \( \frac{du}{4} = dx \).</p> <p>The integral becomes:</p> <p>\( \int -3 \cos(4x) dx = \int -3 \cos(u) \frac{du}{4} \)</p> <p>\( = -\frac{3}{4} \int \cos(u) du \)</p> <p>\( = -\frac{3}{4} \sin(u) + C \)</p> <p>Substituting back \( u = 4x \):</p> <p>\( = -\frac{3}{4} \sin(4x) + C \)</p> <p>Where \( C \) is the constant of integration.</p>
Trigonometric Functions Calculation
写真に写っている問題を解くために、次のように進めます。 与えられた方程式は次の通りです。 \[ \sin \alpha = \frac{1}{3}, \quad \cos \beta = -\frac{2}{5} \] 求めるべきは次の式の値です。 \[ \sin(\alpha - \beta), \quad \cos(\alpha + \beta) \] まず初めに、与えられた条件から、余弦定理や正弦定理を使わずに三角関数の値を求めるために、以下のピタゴラスの定理に基づく基本的な関係を用います。 \[ \sin^2 \theta + \cos^2 \theta = 1 \] この関係を使うことで、与えられた \(\sin \alpha\) と \(\cos \beta\) の値から、残りの三角関数の値を導き出すことが可能です。 \[ \sin \alpha = \frac{1}{3} \] \[ \sin^2 \alpha= \left(\frac{1}{3}\right)^2 = \frac{1}{9} \] \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{1}{9} = \frac{8}{9} \] \[ \cos \alpha = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3} \] 同様に、 \[ \cos \beta = -\frac{2}{5} \] \[ \cos^2 \beta = \left(-\frac{2}{5}\right)^2 = \frac{4}{25} \] \[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \frac{4}{25} = \frac{21}{25} \] \[ \sin \beta = \pm \sqrt{\frac{21}{25}} = \pm \frac{\sqrt{21}}{5} \] 三角関数の正負については、角 \(\alpha\) と \(\beta\) がどの象限にあるかによって決まりますが、この問題にはそのような情報が与えられていません。ですので、正負の両方のケースを検討する必要があります。しかし、通常、この手の問題では、最も一般的な第1象限（すなわち、全ての三角関数の値が正）を仮定します。 三角関数の加法定理に基づき、次の式を使用します。 \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] 得られた三角関数の値を以上の式に適用します（ここでは正の値を使用します）。 まずは \(\sin(\alpha - \beta)\) から求めます。 \[ \sin(\alpha - \beta) = \left(\frac{1}{3}\right)\left(-\frac{2}{5}\right) - \left(\frac{2\sqrt{2}}{3}\right)\left(\frac{\sqrt{21}}{5}\right) \] \[ \sin(\alpha - \beta) = -\frac{2}{15} - \frac{2\sqrt{42}}{15} = -\frac{2(1 + \sqrt{42})}{15} \] 次に \(\cos(\alpha + \beta)\) を求めます。 \[ \cos(\alpha + \beta) = \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{2}{5}\right) - \left(\frac{1}{3}\right)\left(\frac{\sqrt{21}}{5}\right) \] \[ \cos(\alpha + \beta) = -\frac{4\sqrt{2}}{15} - \frac{\sqrt{21}}{15} = -\frac{4\sqrt{2} + \sqrt{21}}{15} \] 以上の計算を踏まえて、最終的な結果は以下の通りになります。 \[ \sin(\alpha - \beta) = -\frac{2(1 + \sqrt{42})}{15} \] \[ \cos(\alpha + \beta) = -\frac{4\sqrt{2} + \sqrt{21}}{15} \]
Trigonometric Ratios in the Second Quadrant
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), and asks us to find the values of the remaining trigonometric ratios. Given: \[ \cos \theta = -\frac{\sqrt{3}}{2} \] Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant). Let's use the Pythagorean identity to find sine: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute the given value of cos θ: \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{3}{4} = 1 \] Solving for \(\sin^2 \theta\): \[ \sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \] Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive: \[ \sin \theta = \frac{1}{2} \] The tangent of θ is the ratio of sine to cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \] \[ \tan \theta = -\frac{1}{\sqrt{3}} \] \[ \tan \theta = -\frac{\sqrt{3}}{3} \] For the reciprocals: \[ \csc \theta = \frac{1}{\sin \theta} = 2 \] \[ \sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \] \[ \cot \theta = \frac{1}{\tan \theta} = -\sqrt{3} \] So, the six trigonometric ratios for the given angle θ in the second quadrant are: \[ \sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3} \] \[ \csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3} \]
Solving Complex Number Division
Certainly! To find \( \frac{z_1}{z_2} \) given that \( z_1 = \cos \theta + j \sin \theta \) and \( z_2 = \cos \theta - j \sin \theta \), we can directly divide the complex numbers as follows: \[ \frac{z_1}{z_2} = \frac{\cos \theta + j \sin \theta}{\cos \theta - j \sin \theta} \] To divide these two complex numbers, we can multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary parts from the denominator. The conjugate of \( z_2 \) is \( \cos \theta + j \sin \theta \), which is actually the same as \( z_1 \). So, we multiply both top and bottom by this conjugate: \[ \frac{z_1}{z_2} = \frac{(\cos \theta + j \sin \theta)(\cos \theta + j \sin \theta)}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)} \] Multiplying out the numerator, we get: Numerator: \( (\cos^2 \theta + 2j \cos \theta \sin \theta - \sin^2 \theta) \) (using \( j^2 = -1 \)) Multiplying out the denominator, we get: Denominator: \( (\cos^2 \theta - (\sin^2 \theta)) \) Simplify the denominator using \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \): Denominator: \( \cos(2\theta) \) However, the denominator simplifies even further because \( \cos^2 \theta - \sin^2 \theta = 1 \) (from the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \)). Therefore, the denominator actually is: Denominator: \( 1 \) Returning to the numerator, we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) again: Numerator: \( 1 + 2j \cos \theta \sin \theta \) The numerator simplifies using the double angle identity for sine: \( 2\sin \theta \cos \theta = \sin(2\theta) \): Numerator: \( 1 + j \sin(2\theta) \) Putting it all together, we have: \[ \frac{z_1}{z_2} = 1 + j \sin(2\theta) \] This is the final simplified form of \( \frac{z_1}{z_2} \).
Solving Limit Using Sine Addition Formula
The limit provided in the image is: \[ \lim_{x \to a} \frac{\sin x - \sin a}{x - a} \] We can solve this limit by using the Sine Addition Formula in the form of the difference: \[ \sin x - \sin a = 2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right) \] Now let's apply this transformation and compute the limit: \[ \lim_{x \to a} \frac{2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right)}{x - a} = \lim_{x \to a} \left(2 \cos\left(\frac{x + a}{2}\right)\right) \cdot \left(\frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}}\right) \] As \( x \) approaches \( a \), the term \( \frac{x - a}{2} \) approaches 0. Thus, the second factor: \[ \lim_{x \to a} \frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}} \] approaches 1, because it's the definition of the derivative of sine at 0: \[ \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \] Also, since \( \cos\left(\frac{x + a}{2}\right) \) is continuous, we can simply replace \( x \) with \( a \): \[ \lim_{x \to a} \cos\left(\frac{x + a}{2}\right) = \cos\left(\frac{2a}{2}\right) = \cos(a) \] Multiplying these two results yields: \[ \cos(a) \cdot 1 = \cos(a) \] Therefore, the value of the limit is \( \cos(a) \).
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