Example Question - step by step simplification
Here are examples of questions we've helped users solve.
Solving Complex Square Root Expression Step by Step
Pour résoudre l'expression donnée dans l'image, commençons par simplifier étape par étape : \[ \frac{{4 \sqrt{32} \times 6 \sqrt{7} \times \sqrt{100}}}{{4 \sqrt{16}}} \] Tout d'abord, simplifions les racines carrées et les nombres en dehors des racines carrées quand c'est possible : \[ \frac{{4 \times \sqrt{32} \times 6 \times \sqrt{7} \times \sqrt{100}}}{{4 \times \sqrt{16}}} \] \[ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \] \[ \sqrt{100} = 10 \] \[ \sqrt{16} = 4 \] Maintenant, remplaçons les racines simplifiées dans l'expression : \[ \frac{{4 \times 4\sqrt{2} \times 6 \times \sqrt{7} \times 10}}{{4 \times 4}} \] Simplifions les nombres hors des racines : \[ \frac{{16 \times 6 \times 10 \times \sqrt{2} \times \sqrt{7}}}{{16}} \] Simplifions la fraction en annulant les 16 (multiplicateur et diviseur) : \[ \frac{{6 \times 10 \times \sqrt{2} \times \sqrt{7}}}{{1}} \] \[ 6 \times 10 = 60 \] Multiplions le reste ensemble : \[ 60 \times \sqrt{2} \times \sqrt{7} = 60\sqrt{14} \] Voilà, l'expression simplifiée est : \[ 60\sqrt{14} \] Cela donne la réponse finale en français.
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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