Respuesta :

[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -------------------------------\\\\ \sqrt{16-x^2}\implies \sqrt{16-[4sin(x)]^2}\implies \sqrt{16-[4^2sin^2(x)]} \\\\\\ \sqrt{16-16sin^2(x)}\implies \sqrt{16[1-sin^2(x)]}\implies \sqrt{16cos^2(x)} \\\\\\ \sqrt{4^2cos^2(x)}\implies \sqrt{[4cos(x)]^2}\implies 4cos(x)[/tex]