Example Question - secant function
Here are examples of questions we've helped users solve.
Definite Integral of the Square of Secant Function
<p>La función secante al cuadrado, \( \sec^2(x) \), es la derivada de la función tangente, \( \tan(x) \). Por lo tanto, para resolver esta integral definida, podemos aplicar la regla fundamental del cálculo que dice que la integral de la derivada de una función en un intervalo es igual a la diferencia de los valores de la función en los extremos del intervalo:</p> <p>\[ \int_{0}^{4} \sec^2 x \, dx = \tan(x) \Big|_{0}^{4} \]</p> <p>Calculamos los valores de \( \tan(x) \) en \( x = 4 \) y \( x = 0 \) y encontramos la diferencia:</p> <p>\[ = \tan(4) - \tan(0) \]</p> <p>Sin embargo, notarás que hay un problema al querer evaluar \( \tan(4) \), ya que 4 radianes está muy cerca de \( \frac{\pi}{2} \approx 1.5708 \) y la función tangente no está definida en \( \frac{\pi}{2} + n\pi \), donde \( n \) es cualquier número entero. Al sustituir \( x = 4 \) en la función tangente, obtendremos un valor no definido, lo que significa que la integral no converge en ese punto.</p> <p>Por ende, debido a la discontinuidad en \( x = 4 \), no podemos obtener un valor numérico real para esta integral definida con los límites dados de \( 0 \) a \( 4 \). Si se cometió un error al escribir los límites de integración y se trataba de un intervalo donde la función tangente es continua, los pasos a seguir para resolverla serían correctos.</p>
Proving Trigonometric Identity Involving Secant and Tangent Functions
<p>Given: \(\sec(\theta) = x + \frac{1}{4x}\)</p> <p>To prove: \(\sec^2(\theta) + \tan^2(\theta) = 2x \text{ or } \frac{1}{2x}\)</p> <p>We know that: \(\sec^2(\theta) = 1 + \tan^2(\theta)\)</p> <p>So, \(\sec^2(\theta) + \tan^2(\theta) = 2 \sec^2(\theta) - 1\)</p> <p>Using given \(\sec(\theta) = x + \frac{1}{4x}\), we get:</p> <p>\(\sec^2(\theta) = \left(x + \frac{1}{4x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{4x} + \frac{1}{16x^2}\)</p> <p>\(\sec^2(\theta) = x^2 + \frac{1}{2} + \frac{1}{16x^2}\)</p> <p>Hence \(\sec^2(\theta) + \tan^2(\theta) = 2x^2 + 1 + \frac{1}{8x^2} - 1\)</p> <p>\(\sec^2(\theta) + \tan^2(\theta) = 2x^2 + \frac{1}{8x^2}\)</p> <p>We realize there is a mistake because we cannot get \(2x\) or \(\frac{1}{2x}\) directly from \(2x^2 + \frac{1}{8x^2}\), thus the original statement seems incorrect. We need to reassess the problem or check the given identity.</p>
Trigonometric Identities and Functions
To solve the problem, we'll start with the given information: \[ \sec(\theta) = 4 \] \[ \cot(\theta) > 0 \] We want to find the exact values of \(\tan(\theta)\) and \(\sin(\theta)\). First, recall the trigonometric identities involving \(\sec(\theta)\) and \(\tan(\theta)\): \[ \sec(\theta) = \frac{1}{\cos(\theta)} \] \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] From \(\sec(\theta) = 4\), we know that \(\cos(\theta) = \frac{1}{4}\). Since we are given that \(\cot(\theta) > 0\), which means \(\cos(\theta)\) and \(\sin(\theta)\) must have the same sign, and because \(\cos(\theta) = \frac{1}{4}\) is positive, \(\sin(\theta)\) must also be positive. Now we need to find \(\sin(\theta)\) knowing \(\cos(\theta)\). Using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Substituting \(\cos(\theta) = \frac{1}{4}\) into this equation: \[ \sin^2(\theta) + \left(\frac{1}{4}\right)^2 = 1 \] \[ \sin^2(\theta) + \frac{1}{16} = 1 \] \[ \sin^2(\theta) = 1 - \frac{1}{16} \] \[ \sin^2(\theta) = \frac{15}{16} \] Since \(\sin(\theta)\) is positive (as discussed earlier): \[ \sin(\theta) = \sqrt{\frac{15}{16}} \] \[ \sin(\theta) = \frac{\sqrt{15}}{4} \] Now we can find \(\tan(\theta)\) using \(\sin(\theta)\) and \(\cos(\theta)\): \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] \[ \tan(\theta) = \frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}} \] \[ \tan(\theta) = \sqrt{15} \] So the exact values are: \[ \tan(\theta) = \sqrt{15} \] \[ \sin(\theta) = \frac{\sqrt{15}}{4} \]
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com