Question

sabendo que senx + cosx = 1/3, determine o valor de sen2x.

Answer 1

Olá, Mattheus.

$(\text{sen}\,x + \cos x)^2=\text{sen}\²x + \underbrace{2\,\text{sen}\,x \cos x}_{=\text{sen}\,2x}+\cos^2 x=1+\text{sen}\,2x\Rightarrow\\\\\\ (\frac13)^2=1+\text{sen}\,2x\Rightarrow\text{sen}\,2x=\frac19-1=\frac{1-9}9\Rightarrow\boxed{\text{sen}\,2x=-\frac89}$