Example Question - cosine function
Here are examples of questions we've helped users solve.
Trigonometric Values of Special Angles
\begin{align*} \sin(45^\circ) &= \frac{\sqrt{2}}{2},\\ \cos(45^\circ) &= \frac{\sqrt{2}}{2},\\ \tan(45^\circ) &= \frac{\sin(45^\circ)}{\cos(45^\circ)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1. \end{align*}
Solving Trigonometric Equation
La ecuación proporcionada en la imagen es \( 2 \cos(\theta) = \sqrt{3} \). Para resolver esta ecuación, primero debemos aislar \( \cos(\theta) \) dividiendo ambos lados de la ecuación entre 2: \[ \cos(\theta) = \frac{\sqrt{3}}{2} \] Ahora buscamos un ángulo agudo cuyo coseno sea \( \frac{\sqrt{3}}{2} \). Sabemos en trigonometría que el coseno de 30 grados, o \( \frac{\pi}{6} \) radianes, es igual a \( \frac{\sqrt{3}}{2} \). Por lo tanto, el ángulo agudo y positivo \(\theta\) que satisface la ecuación original es: \[ \theta = 30^\circ \] \[ \text{o} \] \[ \theta = \frac{\pi}{6} \text{ radianes} \] Dado que hemos encontrado un ángulo agudo (menor a 90 grados o \( \frac{\pi}{2} \) radianes), y que es positivo, hemos respondido a la pregunta de manera correcta.
Comparison Test for Convergence of Series with Cosine Function
在使用比较检验法来确定级数 \(\sum_{k=1}^{\infty} \frac{2 + \cos k}{k^3}\) 的收敛性或发散性时，我们应该选一个与给定级数的每一项都相关的级数进行比较，且该级数的收敛性或发散性是已知的。 因为余弦函数的值域是 \([-1, 1]\)，所以对于所有的 \(k\)，有 \(-1 \leq \cos k \leq 1\)。这意味着 \(\cos k\) 对 \(2 + \cos k\) 的值的影响是有限的，并且每一项 \(\frac{2 + \cos k}{k^3}\) 将会被包含在 \(\frac{1}{k^3}\) 和 \(\frac{3}{k^3}\) 之间。 我们可以忽略 \(2 + \cos k\) 中的 \(\cos k\) 因素，因为 \(2 + \cos k\) 的最小值是 \(2 - 1 = 1\)，最大值是 \(2 + 1 = 3\)。所以这个级数的每一项至少和 \(\frac{1}{k^3}\) 一样大，至多和 \(\frac{3}{k^3}\) 一样大。其实直接用 \(\frac{2}{k^3}\) 比较就足够了，因为它能良好地代表原级数的行为，而且我们知道 \(p\)-级数 \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) 当 \(p>1\) 时收敛。 所以，我们可以将给定的级数 \(\sum_{k=1}^{\infty} \frac{2 + \cos k}{k^3}\) 与 \(p\)-级数 \(\sum_{k=1}^{\infty} \frac{2}{k^3}\) 比较，因为 \(p=3 > 1\)，后者是收敛的。根据比较检验法，如果我们比较的级数收敛，且所有的比较项都大于等于原级数的对应项（正项级数），那么原级数也收敛。 综上所述，根据比较检验法来确定原级数的收敛性，我们应选 (B) \(\sum_{k=1}^{\infty} \frac{2}{k^3}\) 作为比较级数。
Calculus - Derivative of y = 3 cos(2x)
The image shows a calculus problem asking to find the derivative dy/dx of the function y with respect to x. The function given is: y = 3 cos(2x) To find the derivative of this function, we use the chain rule, which in this case involves differentiating the cosine function and then accounting for the inner function, 2x. The derivative of cos(u) with respect to u is -sin(u), and then we multiply this by the derivative of the inner function, which is 2 in this case. The derivative of y with respect to x is thus: dy/dx = -3 sin(2x) * 2 Simplify the expression: dy/dx = -6 sin(2x)
Understanding Inverse Trigonometric Functions
The question in the image asks to evaluate the expression: \[ \cos(\cos^{-1}(-0.9)) \] By definition, the inverse cosine function, \(\cos^{-1}(x)\), returns the angle whose cosine is \(x\). Therefore, when we apply the cosine function to the result of an inverse cosine function, we get back the original value inside the inverse cosine. For the given problem: \[ \cos(\cos^{-1}(-0.9)) = -0.9 \] This is because the inverse cosine of \(-0.9\) gives us the angle whose cosine is \(-0.9\), and taking the cosine of that angle returns us back to the original value of \(-0.9\). So the correct answer is: -0.9
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