Pré-Vestibular ⇒ (Poliedro) Transformação de arcos Tópico resolvido

[felipef]

Simplificar a expressão:
[tex3]\cos ^{2}(a+b)+\cos ^{2}b-2.\cos (a+b).\cos a.\cos b[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\cos ^{2}(a+b)+\cos ^{2}b-2.\cos (a+b).\cos a.\cos b[/tex3]

[LucasPinafi]

[tex3]\omega = \cos^2 ( a + b) + \cos^2 b- 2 \cos (a+b) \cdot \cos a \cdot \cos b \\ \omega = [ \cos a \cos b - \sen a \sen b ]^2 + \cos^2 b -2[\cos a \cos b - \sen a \sen b ] \cos a \cos b \\ \omega = \cos^2 a \cos^2 b -\cancel{2 \sen a \sen b \cos a \cos b} + \sin^2 a \sin^2 b+\cos^2 b - 2 \cos^2 a \cos^2 b +\cancel{ 2 \sen a \sen b \cos a \cos b} \\ \omega = \sin^2 a \sin^2 b -\cos^2 a \cos^2 b + \cos^2 b \\ \omega = \sin^2 a \sin^2 b +\cos^2 b (1- \cos^2 a) \\ \omega = \sin^2 a \sin^2 b + \cos^2 b \sin^2 a = \sin^2 a (\sin^2 b +\cos^2 b) = \sin^2 a \\ \omega = \sin^2 a [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\omega = \cos^2 ( a + b) + \cos^2 b- 2 \cos (a+b) \cdot \cos a \cdot \cos b \\ \omega = [ \cos a \cos b - \sen a \sen b ]^2 + \cos^2 b -2[\cos a \cos b - \sen a \sen b ] \cos a \cos b \\ \omega = \cos^2 a \cos^2 b -\cancel{2 \sen a \sen b \cos a \cos b} + \sin^2 a \sin^2 b+\cos^2 b - 2 \cos^2 a \cos^2 b +\cancel{ 2 \sen a \sen b \cos a \cos b} \\ \omega = \sin^2 a \sin^2 b -\cos^2 a \cos^2 b + \cos^2 b \\ \omega = \sin^2 a \sin^2 b +\cos^2 b (1- \cos^2 a) \\ \omega = \sin^2 a \sin^2 b + \cos^2 b \sin^2 a = \sin^2 a (\sin^2 b +\cos^2 b) = \sin^2 a \\ \omega = \sin^2 a [/tex3]