Given sinx=0.9 , what is cosx ?
\(\bf \textit{Pythagorean Identities} \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ cos(\theta)=\pm\sqrt{1-sin^2(\theta)}\\\\ -----------------------------\\\\ thus \\\\\\ cos(x)=\pm \sqrt{1-(0.9)^2}\)
The answer is cos x = 0.44
