Example Question - trigonometric function
Here are examples of questions we've helped users solve.
Calculate the Derivative of a Trigonometric Function
<p>\frac{dy}{dx} = \frac{d}{dx}(sin^3(x) + \csc^5(x) + \tan^5(\sqrt{x^2 + 1}))</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{d}{dx}(\sqrt{x^2 + 1})</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{1}{2\sqrt{x^2 + 1}}(2x)</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5x\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))(\frac{1}{\sqrt{x^2 + 1}})</p>
Analysis of a Trigonometric Function
<p>Задача связана с анализом графика тригонометрической функции и определением соответствующих углов и значения функции.</p> <p>Для решения используем свойства тригонометрических функций и их графиков.</p> <p>Сначала нужно определить, для каких x функция \( f(x) = 3\cos(x) - 1 \) положительна.</p> <p>Функция \( \cos(x) \) положительна в первом и четвертом квадрантах, т.е. \( \cos(x) > 0 \) для \( -\pi/2 + 2k\pi < x < \pi/2 + 2k\pi \), где \( k \) — целое число.</p> <p>Теперь рассмотрим уравнение \( 3\cos(x) - 1 = 0 \) и найдем его корни.</p> <p>\( 3\cos(x) = 1 \)</p> <p>\( \cos(x) = \frac{1}{3} \)</p> <p>Поскольку \( \cos(x) \) является убывающей функцией от \( 0 \) до \( \pi \), и \( \frac{1}{3} \) находится между \( \cos(0) = 1 \) и \( \cos(\pi/2) = 0 \), корень уравнения будет находиться в этом интервале. Обозначим его \( x_0 \), и тогда \( x_0 \in (0, \pi/2) \).</p> <p>Функция \( f(x) \) будет положительной слева и справа от точки \( x_0 \) на интервалах, где \( \cos(x) > 0 \). Тогда:</p> <p>\( f(x) > 0 \) для \( 2k\pi < x < x_0 + 2k\pi \) и \( x_0 + 2k\pi < x < \pi + 2k\pi \).</p> <p>Для нахождения точной величины \( x_0 \), нужно решить тригонометрическое уравнение, для чего можно воспользоваться численными методами или приближенными вычислениями, так как точное аналитическое решение в элементарных функциях получить невозможно.</p>
Finding the Derivative of a Function Involving a Product of a Polynomial and a Trigonometric Function
Для нахождения производной данной функции \( f(x) = (4x^4 + 2) \cos(x) \), воспользуемся правилом произведения. <p> \( f'(x) = (4x^4 + 2)' \cos(x) + (4x^4 + 2) \cdot (\cos(x))' \) </p> <p> \( f'(x) = (16x^3) \cos(x) + (4x^4 + 2) \cdot (-\sin(x)) \) </p> <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p> Итак, производная функции \( f(x) \): <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p>
Analysis of a Trigonometric Function Graph
\[ \begin{align*} \text{Given } & f(x) = -4\sin x - \cos 2x, \text{ for } 0 \leq x \leq \pi.\\ \text{Find } f'(x) & = -4\cos x + \sin 2x \cdot 2 \text{ by using the chain rule.}\\ & = -4\cos x + 2\sin 2x \text{ where } \sin 2x = 2\sin x \cos x.\\ \text{Thus, } f'(x) & = -4\cos x + 4\sin x \cos x.\\ \text{For stationary points, set } f'(x) & = 0.\\ & -4\cos x + 4\sin x \cos x = 0.\\ & \cos x (-4 + 4\sin x) = 0.\\ \text{For } \cos x = 0, & \text{ we get } x = \frac{\pi}{2}.\\ \text{For } -4 + 4\sin x = 0, & \text{ we get } \sin x = 1.\\ & \text{No solution for } 0 \leq x \leq \pi \text{ as } \sin x = 1 \text{ only at } x = \frac{\pi}{2}.\\ \text{Stationary point at } & x = \frac{\pi}{2}.\\ \text{To classify this stationary point, find } f''(x).\\ & f''(x) = 4\sin x + 4\cos x \cdot \cos x - 4\sin^2 x.\\ \text{At } x = \frac{\pi}{2}, \text{ } & f''\left(\frac{\pi}{2}\right) = 4\cdot0 + 4\cdot0 - 4\cdot1 = -4.\\ \text{Since } f''\left(\frac{\pi}{2}\right) < 0, & \text{ the point is a maximum.} \end{align*} \]
Finding the Derivative of a Trigonometric Function Raised to a Power
\[ \begin{align*} f(x) &= \sin^2(x) \\ \frac{d}{dx}f(x) &= \frac{d}{dx}[\sin^2(x)] \\ &= 2\sin(x) \cdot \cos(x) \\ &= \sin(2x) \end{align*} \]
Solving a Trigonometric Function Using Identities
The expression given in the image is a mathematical function involving trigonometric identities: \[ \frac{\tan 315^\circ - \cos 1020^\circ}{\sin 150^\circ - \tan (-135^\circ)} \] Let's solve it step by step using trigonometric identities and properties. Firstly, we can simplify each trigonometric function by using standard angles and periodic properties. The tangent and sine functions have a period of 360°, meaning \(\tan(\theta) = \tan(\theta + k \cdot 360^\circ)\) and \(\sin(\theta) = \sin(\theta + k \cdot 360^\circ)\) where \(k\) is any integer. The cosine function also has the same period. So: - \(\tan 315^\circ\) is equivalent to \(\tan (360^\circ - 45^\circ)\), which equals \(\tan (-45^\circ)\). Since \(\tan\) is an odd function, \(\tan (-\theta) = -\tan (\theta)\), so \(\tan 315^\circ = -\tan 45^\circ = -1\). - \(\cos 1020^\circ\) is equivalent to \(\cos (3 \cdot 360^\circ - 60^\circ)\), which equals \(\cos (-60^\circ)\). Since \(\cos\) is an even function, \(\cos (-\theta) = \cos (\theta)\), so \(\cos 1020^\circ = \cos 60^\circ = \frac{1}{2}\). - \(\sin 150^\circ\) is equivalent to \(\sin (180^\circ - 30^\circ)\), which equals \(\sin 30^\circ\). Thus, \(\sin 150^\circ = \frac{1}{2}\). - \(\tan (-135^\circ)\) is equivalent to \(-\tan (135^\circ)\), which is \(-\tan (180^\circ - 45^\circ)\) and equals \(-(-1)\) because \(\tan (180^\circ - \theta) = -\tan (\theta)\). So, \(\tan (-135^\circ) = 1\). Now we can plug these values into the original expression: \[ \frac{-1 - \frac{1}{2}}{\frac{1}{2} - 1} = \frac{-\frac{3}{2}}{-\frac{1}{2}} \] Upon simplifying the fraction: \[ \frac{-\frac{3}{2}}{-\frac{1}{2}} = \frac{-3}{2} \div \frac{-1}{2} = \frac{-3}{2} \cdot \frac{-2}{1} = \frac{3 \cdot 2}{2 \cdot 1} = \frac{6}{2} = 3 \] Hence, the simplified value of the given expression is 3.
Solving for Angle x in a Right Triangle
The problem in the image shows a right triangle (as indicated by the small square which denotes a 90-degree angle). You are asked to find the value of x, which is the measure of the angle opposed to the side of length 14. To solve for the angle x, you can use the trigonometric function tangent (tan), because it relates the opposite side to the adjacent side in a right triangle. \[ \tan(x) = \frac{opposite}{adjacent} \] \[ \tan(x) = \frac{14}{18} \] Now you'll use the arctangent function to find the angle whose tangent is 14/18. \[ x = \arctan\left(\frac{14}{18}\right) \] Using a calculator: \[ x \approx \arctan(0.7778) \] \[ x \approx 37.87 \text{ degrees} \] Rounded to the nearest degree, x is approximately 38 degrees.
Finding Absolute Maxima and Minima of a Trigonometric Function
To find the absolute maximum and minimum of the function \( f(x) = 3\sin(x) + \sin(2x) \) over the interval \( \left[ \frac{\pi}{2}, \frac{7\pi}{2} \right] \), we need to: 1. Find the critical points of \( f(x) \) by taking the derivative and setting it to zero. 2. Evaluate \( f(x) \) at the critical points and the endpoints of the interval. 3. Compare the values to determine the absolute maximum and minimum. Let's start by finding the derivative of \( f(x) \): \( f'(x) = 3\cos(x) + 2\cos(2x) \cdot 2 \) \( f'(x) = 3\cos(x) + 4\cos(2x) \) Now, we set the derivative equal to zero to find the critical points: \( 3\cos(x) + 4\cos(2x) = 0 \) \( 3\cos(x) + 4(2\cos^2(x) - 1) = 0 \) (since \( \cos(2x) = 2\cos^2(x) - 1 \)) \( 3\cos(x) + 8\cos^2(x) - 4 = 0 \) \( 8\cos^2(x) + 3\cos(x) - 4 = 0 \) This is a quadratic in terms of \( \cos(x) \). Let's solve for \( \cos(x) \): Let \( u = \cos(x) \), so we have: \( 8u^2 + 3u - 4 = 0 \) Solve for \( u \) using the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) \( u = \frac{-3 \pm \sqrt{9 + 4 \cdot 8 \cdot 4}}{2 \cdot 8} \) \( u = \frac{-3 \pm \sqrt{137}}{16} \) Only the solutions that satisfy \( -1 \le u \le 1 \) (since \( u \) is \( \cos(x) \)) are valid for cosine. Thus, we may disregard solutions that do not fall in this range. Now, we need to find the corresponding \( x \) values. Since the derivative does not explicitly solve for \( x \), we need to use the inverse cosine function \( \arccos(u) \) to find the values of \( x \), and consider the interval \( \left[ \frac{\pi}{2}, \frac{7\pi}{2} \right] \). Evaluate \( f(x) \) at the critical points found from the derivative and the endpoints of the interval: 1. \( x = \frac{\pi}{2} \) 2. \( x = \frac{7\pi}{2} \) 3. \( x = \arccos \left( \frac{-3 + \sqrt{137}}{16} \right) \) 4. \( x = \arccos \left( \frac{-3 - \sqrt{137}}{16} \right) \) The \( \arccos \) function returns values in the range \( [0, \pi] \). Therefore, if the solutions are within the interval \( \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right] \), we must also consider the symmetric points along the interval \( \left[ \frac{3\pi}{2}, \frac{7\pi}{2} \right] \) due to the periodic nature of the cosine function. For each valid \( x \) value (including the endpoints and critical points that fall within the interval), calculate \( f(x) \): 5. \( f\left(\frac{\pi}{2}\right) \) 6. \( f\left(\frac{7\pi}{2}\right) \) 7. \( f\left(\arccos \left( \frac{-3 + \sqrt{137}}{16} \right)\right) \) 8. \( f\left(\arccos \left( \frac{-3 - \sqrt{137}}{16} \right)\right) \) Check for additional valid solutions using symmetry if necessary. Finally, compare all of these values to determine which is the largest (absolute maximum) and which is the smallest (absolute minimum). Since I am not able to perform actual calculations, you will need to complete these calculations manually to find the absolute maximum and minimum values of \( f(x) \). Remember to provide the answers in terms of \( x \), and ensure they fall within the specified interval.
Finding Absolute Maximum and Minimum of a Trigonometric Function
The question asks for the absolute maximum and minimum of the function \( f(x) = 3\cos^2\left(\frac{x}{2}\right) \) over the interval \(\left[\frac{\pi}{4}, \pi\right]\). To solve this: 1. Compute the derivative of the function to find the critical points where the derivative is zero or undefined. Critical points can also occur at the endpoints of the interval. 2. Evaluate the function at the critical points and at the endpoints of the interval to determine the absolute maximum and minimum. Let's find the derivative \(f'(x)\): \( f(x) = 3\cos^2\left(\frac{x}{2}\right) \) Using the chain rule, \( f'(x) = 3 \cdot 2\cos\left(\frac{x}{2}\right)(-\sin\left(\frac{x}{2}\right)) \left(\frac{1}{2}\right) = -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \) Setting the derivative equal to zero, \( -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = 0 \) This equation is satisfied when either \( \sin\left(\frac{x}{2}\right) = 0 \) or \( \cos\left(\frac{x}{2}\right) = 0 \). \( \sin\left(\frac{x}{2}\right) = 0 \) at \( x = 0, 2\pi, 4\pi, \ldots \) but within \(\left[\frac{\pi}{4}, \pi\right]\), this doesn't give any solution. \( \cos\left(\frac{x}{2}\right) = 0 \) at \( x = \pi, 3\pi, \ldots \), and only \( x = \pi \) falls within our interval. The critical points are just at the endpoints of the interval since there are no other points in \(\left[\frac{\pi}{4}, \pi\right]\) where the derivative is zero or undefined. Now, evaluate \(f(x)\) at \( x = \frac{\pi}{4} \) and \( x = \pi \): \( f\left(\frac{\pi}{4}\right) = 3\cos^2\left(\frac{\pi}{8}\right) \) and \(f(\pi) = 3\cos^2\left(\frac{\pi}{2}\right)\). Note that \( \cos\left(\frac{\pi}{2}\right) = 0 \), so \( f(\pi) = 0 \). To find \(f\left(\frac{\pi}{4}\right)\), we compute the cosine of \( \pi/8 \). This value is not one of the standard angles, so we can't find an exact value easily without using a calculator. However, since \(0 < \frac{\pi}{8} < \frac{\pi}{2}\), \( \cos\left(\frac{\pi}{8}\right) \) is positive and less than 1, so \( 3\cos^2\left(\frac{\pi}{8}\right) \) is also positive and less than 3. Therefore, the absolute maximum of \( f(x) \) is at \( x = \frac{\pi}{4} \) and is greater than 0, and the absolute minimum of \( f(x) \) is 0 at \( x = \pi \). Based only on the information available, you can't give an exact value for the absolute maximum without a calculator, but you can state where it occurs and that the absolute minimum is 0 at \( x = \pi \).
Solving Integral of Exponential and Trigonometric Function
The image shows the integral: \[ \int e^{2x} \cos(4x) \,dx \] To solve this integral, we'll have to use integration by parts multiple times or apply the tabular integration by parts method. However, there's an alternative approach that's often quicker and involves recognizing that the integral is the real part of a complex exponential. Here are the steps for the alternative method: 1. Recognize that \(e^{ix} = \cos(x) + i\sin(x)\), where \(i\) is the imaginary unit. 2. Write \(\cos(4x) = \Re(e^{4ix})\), where \(\Re\) denotes the real part: \[ \int e^{2x} \cos(4x) \,dx = \Re\left(\int e^{2x} \cdot e^{4ix} \,dx\right) \] 3. Combine the exponentials: \[ \Re\left(\int e^{2x} \cdot e^{4ix} \,dx\right) = \Re\left(\int e^{(2+4i)x} \,dx\right) \] 4. Integrate the complex exponential: \[ \Re\left(\int e^{(2+4i)x} \,dx\right) = \Re\left(\frac{1}{2+4i} e^{(2+4i)x} + C\right) \] where \(C\) is the constant of integration. 5. Simplify the expression by finding the real part after multiplying by the conjugate to get rid of the \(i\): \[ \frac{1}{2+4i} = \frac{2-4i}{(2+4i)(2-4i)} = \frac{2-4i}{4+16} = \frac{2-4i}{20} = \frac{1}{10} - \frac{1}{5}i \] 6. Insert this into the integrated function: \[ \Re\left(\frac{1}{10} - \frac{1}{5}i\right)e^{(2+4i)x} + C = \left(\frac{1}{10} e^{2x} \cos(4x) - \frac{1}{5} e^{2x} \sin(4x)\right) + C \] 7. The final result only requires the real part: \[ \int e^{2x} \cos(4x) \,dx = \frac{1}{10} e^{2x} \cos(4x) - \frac{1}{5} e^{2x} \sin(4x) + C \] Hence, this is the evaluated integral of the original function.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com