Question

demonstre que:

(senx + sen2x)² + (cosx + cos2x)² = 2(1+cosx)

Answer 1

$\large\begin{array}{l} \mathsf{(sen\,x+sen\,2x)^2+(cos\,x+cos\,2x)^2}\\\\\\ \textsf{Expandindo os quadrados,}\\\\ =\mathsf{sen^2\,x+2\,sen\,x\,sen\,2x+sen^2\,2x+cos^2\,x+2\,cos\,x\,cos\,2x+cos^2\,2x}\\\\ =\mathsf{(sen^2\,x+cos^2\,x)+2\,sen\,x\,sen\,2x+2\,cos\,x\,cos\,2x+(cos^2\,2x+sen^2\,2x)} \end{array}$

$\large\begin{array}{l} =\mathsf{1+2\,sen\,x\,sen\,2x+2\,cos\,x\,cos\,2x+1}\\\\ =\mathsf{2+2\,sen\,x\,sen\,2x+2\,cos\,x\,cos\,2x}\\\\ =\mathsf{2+2\cdot (sen\,x\,sen\,2x+cos\,x\,cos\,2x)}\\\\ =\mathsf{2+2\cdot (cos\,x\,cos\,2x+sen\,x\,sen\,2x)}\\\\ =\mathsf{2+2\cdot (cos\,2x\,cos\,x+sen\,2x\,sen\,x)\qquad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Note que a express\~ao em par\^enteses \'e a expans\~ao do}\\\textsf{cosseno da diferen\c{c}a entre dois arcos:}\\\\ \mathsf{cos(\alpha-\beta)=cos\,\alpha\,cos\,\beta+sen\,\alpha\,sen\,\beta}\\\\ \textsf{para }\mathsf{\alpha=2x~~e~~\beta=x.}\\\\\\ \textsf{Ent\~ao a express\~ao (i) equivale a}\\\\ \mathsf{=2+2\,cos(2x-x)}\\\\ \mathsf{=2+2\,cos\,x}\\\\ \mathsf{=2\cdot (1+cos\,x)\qquad\checkmark} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$