Ensino Médio ⇒ Identidade trigonométrica (sen e cos) Tópico resolvido

[triplebig]

Demonstre a igualdade:

[tex3]\cos^3\alpha \cdot \frac{\text{sen}3\alpha}{3} + \text{sen}^3\alpha\cdot\frac{\cos3\alpha}{3}=\frac{\text{sen}4\alpha}{4}[/tex3]

[petrasMOD]

[tex3]

\mathsf{cos(2x) = 1-2sen^2x\implies
cos^2x = cos(2x)+sen^2x(I)\\
cos^2x = 1-sen^2x(II)\\
(I)+(II): 2cos^2x=cos(2x)+1\therefore \boxed{cos^2x = \frac{1}{2}(cos(2x))+1}(III)\\
cosx.cos(2x) =\frac{1}{2}(cos(2x+x)+cos(2x-x)) \therefore \boxed{cosxcos(2x)= \frac{1}{2}(cos(3x)+cos(x))}(V)\\


cos^3x=\underbrace{cos^2x}_{(III)}cosx=\frac{1}{2}(\underbrace{cosxcos(2x)}_{(V)}+cosx) = \frac{1}{2}(\frac{1}{2}(cosx+cos(3x))+cosx)=\\
\underline{\frac{1}{4}(cosx+cos(3x))+\frac{1}{2}cosx=\boxed{\frac{cos3x+3cosx}{4}}}(VI)\\

sen^2x=1-sen^2x-cos(2x)\implies 2sen^2x = 1-cos(2x) \therefore \boxed{sen^2x=\frac{1-cos(2x)}{2}}(VII)\\

sen^3x = \underbrace{sen^2x}_{(VII)}.senx = \frac{1-cos(2x)}{2}.senx=\frac{1}{2}(senx-\underbrace{senxcos(2x))}_{sen(a).cosb)=\frac{1}{2}(sen(a+b)+sen(a-b))}=\\
\frac{1}{2}(senx- (\frac{sin(2x+x)+sen(2x-x)}{2})=\frac{1}{2}(senx-\frac{sen(3x)+sen(x)}{2})=\boxed{\frac{3senx-sen(3x)}{4}}(VIII)\\
\therefore \frac{cos3x+3cosx}{4}.\frac{sen(3x)}{3}+\frac{-sen(3x)+3senx}{4}.\frac{cos(3x)}{3}=\\\frac{1}{12}[3(\underbrace{cosxsen(3x)+senxcos(3x)}_{sen(3x+x)}]=\boxed{\frac{sen(4x)}{4}}\color{green}\\



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