Example Question - mathematical expression simplification
Here are examples of questions we've helped users solve.
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.
Complex Number Simplification
The question is to show that \( \left( \frac{1 + i \sqrt{3}}{1 - i \sqrt{3}} \right)^2 = 2 \). We will begin by simplifying the expression inside the parentheses before squaring it. \( \frac{1 + i \sqrt{3}}{1 - i \sqrt{3}} \) To simplify, we can multiply both the numerator and the denominator by the conjugate of the denominator: \( \frac{(1 + i \sqrt{3})(1 + i \sqrt{3})}{(1 - i \sqrt{3})(1 + i \sqrt{3})} \) Expanding the numerator and the denominator: Numerator: \( (1 + i \sqrt{3})(1 + i \sqrt{3}) = 1 + i \sqrt{3} + i \sqrt{3} + (i \sqrt{3})^2 \) \( = 1 + 2i \sqrt{3} - 3 \) [since \( i^2 = -1 \)] \( = -2 + 2i \sqrt{3} \) Denominator: \( (1 - i \sqrt{3})(1 + i \sqrt{3}) = 1 - (i \sqrt{3})^2 \) \( = 1 - (-3) \) \( = 1 + 3 \) \( = 4 \) Now we have: \( \frac{-2 + 2i \sqrt{3}}{4} \) We can simplify this further by dividing both terms in the numerator by the denominator: \( \frac{-2}{4} + \frac{2i \sqrt{3}}{4} \) \( = -\frac{1}{2} + \frac{i \sqrt{3}}{2} \) Now we square this result to find the original expression's square: \( \left( -\frac{1}{2} + \frac{i \sqrt{3}}{2} \right)^2 \) Multiplying this out gives us: \( = \left( -\frac{1}{2} \right)^2 + 2 \cdot \left( -\frac{1}{2} \right) \cdot \left( \frac{i \sqrt{3}}{2} \right) + \left( \frac{i \sqrt{3}}{2} \right)^2 \) \( = \frac{1}{4} - \frac{i \sqrt{3}}{2} + \frac{-3i^2}{4} \) \( = \frac{1}{4} - \frac{i \sqrt{3}}{2} + \frac{3}{4} \) [since \( i^2 = -1 \)] \( = 1 - \frac{i \sqrt{3}}{2} \) However, the imaginary term does not cancel out, which makes the final expression complex. It appears there may be a mistake either in the original question or in the simplification process, as the expected result is supposed to be \( 2 \), a real number, but the working out has resulted in a complex number. It is possible that there is a typo in the question or additional constraints that were not included in the image provided.
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