Example Question - trigonometric expression simplification
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Simplifying Trigonometric Expression
La imagen muestra una expresión matemática que parece ser una identidad trigonométrica, pero está incompleta o incorrecta tal como está. La expresión es: \[ \frac{\sin(\alpha + \beta)}{\cos(\alpha) \cos(\beta)} \] Para resolver la expresión habría que aplicar la fórmula de adición para el seno, la cual es: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \] Si sustituimos esto en la expresión original, obtenemos: \[ \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha) \cos(\beta)} \] Ahora, dividimos cada término en el numerador por el producto \(\cos(\alpha) \cos(\beta)\): \[ = \frac{\sin(\alpha)\cos(\beta)}{\cos(\alpha) \cos(\beta)} + \frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha) \cos(\beta)} \] Simplificamos los términos donde sea posible: \[ = \frac{\sin(\alpha)}{\cos(\alpha)} + \frac{\sin(\beta)}{\cos(\beta)} \] Finalmente, sabiendo que \(\frac{\sin(\alpha)}{\cos(\alpha)}\) es \(\tan(\alpha)\) y \(\frac{\sin(\beta)}{\cos(\beta)}\) es \(\tan(\beta)\), la expresión queda como: \[ \tan(\alpha) + \tan(\beta) \] Por lo tanto, la expresión original se puede expresar como la suma de las tangentes de los dos ángulos \(\alpha\) y \(\beta\). Sin embargo, cabe destacar que la imagen no incluyó la parte derecha de la igualdad, así que estoy suponiendo que deseabas conocer cómo simplificar el lado izquierdo de la expresión presentada.
Trigonometric Expression Simplification
Certainly! The expression given in the image is: \( \frac{\sin(\frac{3\pi}{2} + \theta) + \cot(-\theta)}{1 - \sin(2\pi - \theta)} \) Let's simplify the numerator and denominator of this fraction step by step using trigonometric identities: 1. \(\sin(\frac{3\pi}{2} + \theta)\) can be simplified using the identity that \(\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\): Since \(\sin(\frac{3\pi}{2}) = -1\) and \(\cos(\theta) = \cos(\theta)\), we get: \(\sin(\frac{3\pi}{2} + \theta) = \sin(\frac{3\pi}{2})\cos(\theta) + \cos(\frac{3\pi}{2})\sin(\theta) = -1 \cdot \cos(\theta) + 0 \cdot \sin(\theta) = -\cos(\theta)\) 2. \(\cot(-\theta) = \frac{\cos(-\theta)}{\sin(-\theta)}\) can be further simplified using the facts that \(\cos\) is an even function and \(\sin\) is an odd function, which results in \(\cos(-\theta) = \cos(\theta)\) and \(\sin(-\theta) = -\sin(\theta)\): \(\cot(-\theta) = \frac{\cos(\theta)}{-\sin(\theta)} = -\cot(\theta)\) Putting these together, the numerator becomes: \(-\cos(\theta) - \cot(\theta)\) For the denominator: 1. \(\sin(2\pi - \theta)\) can be simplified using the identity for \(\sin(\pi - x) = \sin(x)\), as \(2\pi - \theta\) is the same as \(\pi - (\pi - \theta)\): \(\sin(2\pi - \theta) = \sin(\pi - (\pi - \theta)) = \sin(\pi - \theta)\) Since \(\sin(\pi - x) = \sin(x)\), we get: \(\sin(2\pi - \theta) = \sin(\theta)\) So the original expression simplifies to: \( \frac{-\cos(\theta) - \cot(\theta)}{1 - \sin(\theta)} \) Now let's simplify further: Since \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\), we can rewrite the numerator in terms of sines and cosines, getting: \( \frac{-\cos(\theta) - \frac{\cos(\theta)}{\sin(\theta)}}{1 - \sin(\theta)} \) To combine the terms in the numerator, we get a common denominator of \(\sin(\theta)\): \( \frac{-\cos(\theta)\sin(\theta) - \cos(\theta)}{\sin(\theta)(1 - \sin(\theta))} \) Now, we factor out \(-\cos(\theta)\) from the numerator: \( \frac{-\cos(\theta)(\sin(\theta) + 1)}{\sin(\theta)(1 - \sin(\theta))} \) Observing that \(\sin(\theta) + 1\) is the additive inverse of \(1 - \sin(\theta)\), we can cancel the corresponding terms in the numerator and denominator, yielding: \( \frac{-\cos(\theta)}{\sin(\theta)} \) Finally, this expression can be simplified to \( -\cot(\theta) \) So the simplified form of the original expression is \( -\cot(\theta) \).
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