Fórum de Matemática | DÚVIDAS? Nós respondemos! https://forumdematematica.org/ |
|
Simplificar a equação trigonométrica ABAIXO https://forumdematematica.org/viewtopic.php?f=23&t=12737 |
Página 1 de 1 |
Autor: | danjr5 [ 20 mai 2017, 19:37 ] |
Título da Pergunta: | Re: Simplificar a equação trigonométrica ABAIXO |
Olá! Seja bem-vindo diegorodrigo2022!! \(\mathbf{U = \frac{(1 - \sin \alpha + \cos \alpha)^2}{(\sin \alpha + \tan \alpha)(\cos \alpha - \cot \alpha)}}\) \(\mathbf{U = \frac{(1 - \sin \alpha)^2 + 2 \cdot (1 - \sin \alpha) \cdot \cos \alpha + \cos^2 \alpha}{\left ( \sin \alpha + \frac{\sin \alpha}{\cos \alpha} \right )\left ( \cos \alpha - \frac{\cos \alpha}{\sin \alpha} \right )}}\) \(\displaystyle \mathbf{U = \frac{1 - 2 \cdot \sin \alpha + \sin^2 \alpha + 2 \cdot \cos \alpha - 2 \cdot \sin \alpha \cdot \cos \alpha + \cos^2 \alpha}{\frac{\sin \alpha \cdot \cos \alpha + \sin \alpha}{\cos \alpha} \cdot \frac{\sin \alpha \cdot \cos \alpha - \cos \alpha}{sin \alpha}}}\) \(\displaystyle \mathbf{U = \frac{2 - 2 \cdot \sin \alpha + 2 \cdot \cos \alpha - 2 \cdot \sin \alpha \cdot \cos \alpha}{\frac{\sin \alpha \cdot \cos \alpha + \sin \alpha}{\cos \alpha} \cdot \frac{\sin \alpha \cdot \cos \alpha - \cos \alpha}{sin \alpha}}}\) \(\displaystyle \mathbf{U = \frac{2 \cdot (1 - \sin \alpha) + 2 \cdot \cos \alpha \cdot (1 - \sin \alpha)}{\frac{\sin \alpha \cdot (\cos \alpha + 1)}{\cos \alpha} \cdot \frac{\cos \alpha \cdot (\sin \alpha - 1)}{sin \alpha}}}\) \(\mathbf{U = \frac{(1 - \sin \alpha) \cdot 2 \cdot (1 + \cos \alpha)}{(\cos \alpha + 1) \cdot (\sin \alpha - 1)}}\) \(\mathbf{U = \frac{(1 - \sin \alpha) \cdot 2}{(\sin \alpha - 1)}}\) \(\mathbf{U = \frac{- 1 \cdot 2}{1}}\) \(\fbox{\fbox{\mathbf{U = - 2}}}\) |
Página 1 de 1 | Os Horários são TMG [ DST ] |
Powered by phpBB® Forum Software © phpBB Group https://www.phpbb.com/ |