Example Question - tangent function
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Values That Make the Tangent Function Undefined
<p>Функция тангенса \( \tan(x) \) не имеет смысла (то есть, становится неопределенной), когда ее аргумент приближается к \((2k+1)\frac{\pi}{2}\), где \( k \) - целое число.</p> <p>Исходя из задачи, нам нужно установить, когда выражение \( \tan\left(\frac{\alpha}{2}\right) \) не имеет смысла. Используя аналогию с общей формулой, получаем, что:</p> <p>\(\frac{\alpha}{2} = (2k+1)\frac{\pi}{2}\)</p> <p>Умножая обе части уравнения на 2, получаем:</p> <p>\(\alpha = (2k+1)\pi\)</p> <p>Таким образом, функция тангенса \( \tan\left(\frac{\alpha}{2}\right) \) не имеет смысла, когда \( \alpha \) равно \((2k+1)\pi\), где \( k \) - целое число.</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 Identity Proof Involving Tangent and Sine Functions
<p>Given that \( \tan^2 \alpha = \cos^2 \phi - \sin^2 \phi \), we need to prove that \( \cos^2 \alpha - \sin^2 \alpha = \tan^2 \phi \).</p> <p>From the given, we can write:</p> <p>\( \tan^2 \alpha = \cos^2 \phi - \sin^2 \phi \)</p> <p>\( \tan^2 \alpha = \cos(2\phi) \) (using the double-angle formula \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\))</p> <p>Also, from the Pythagorean identity \(1 + \tan^2 \theta = \sec^2 \theta\), we have</p> <p>\( \sec^2 \alpha = 1 + \tan^2 \alpha \)</p> <p>\( \sec^2 \alpha = 1 + \cos(2\phi) \)</p> <p>Now, using the identity \( \cos^2 \theta = \frac{1}{\sec^2 \theta} \), we get:</p> <p>\( \cos^2 \alpha = \frac{1}{\sec^2 \alpha} \)</p> <p>\( \cos^2 \alpha = \frac{1}{1 + \cos(2\phi)} \)</p> <p>From the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have</p> <p>\( \sin^2 \alpha = 1 - \cos^2 \alpha \)</p> <p>\( \sin^2 \alpha = 1 - \frac{1}{1 + \cos(2\phi)} \)</p> <p>So, the left-hand side of what we need to prove becomes:</p> <p>\( \cos^2 \alpha - \sin^2 \alpha \)</p> <p>\( = \frac{1}{1 + \cos(2\phi)} - \left(1 - \frac{1}{1 + \cos(2\phi)}\right) \)</p> <p>\( = \frac{1}{1 + \cos(2\phi)} - \frac{1 + \cos(2\phi) - 1}{1 + \cos(2\phi)} \)</p> <p>\( = \frac{1}{1 + \cos(2\phi)} - \frac{\cos(2\phi)}{1 + \cos(2\phi)} \)</p> <p>\( = \frac{1 - \cos(2\phi)}{1 + \cos(2\phi)} \)</p> <p>Now we can use the identity \( \tan^2 \theta = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \) to write:</p> <p>\( \cos^2 \alpha - \sin^2 \alpha = \tan^2 \phi \)</p> <p>Which is what we wanted to prove.</p>
Calculating Length in a Right Triangle
To find the length of \(x\) in the right triangle as depicted in the image, we can use trigonometric ratios. The triangle has an angle of 35 degrees, and we're given the length of the side opposite to this angle, which is 16 units. Since we have the angle and the opposite side, we can use the tangent function (tan) to find the length of the adjacent side (\(x\)). The tangent of an angle in a right triangle is equal to the opposite side divided by the adjacent side. So, we have: \[ \tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} \] Substituting the known values, we get: \[ \tan(35^\circ) = \frac{16}{x} \] To isolate \(x\), we'll multiply both sides by \(x\) and then divide both sides by \(\tan(35^\circ)\): \[ x \cdot \tan(35^\circ) = 16 \] \[ x = \frac{16}{\tan(35^\circ)} \] Using a calculator and making sure it's set to degree mode, we can calculate the value of \(x\): \[ x \approx \frac{16}{0.70020753820971} \] \[ x \approx 22.849 \] So, the length of \(x\) rounded to three decimal places is approximately 22.849 units.
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: \[ \tan(\theta) = \sqrt{15} \] \[ \sin(\theta) = \frac{\sqrt{15}}{4} \]
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