Respuesta :
[tex]\bf sin^{-1}\left( \frac{1}{4} \right)=\theta \qquad this~means\qquad sin(\theta )=\cfrac{\stackrel{opposite}{1}}{\stackrel{hypotenuse}{4}}
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\textit{so, let's find the adjacent side for that angle then}
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\textit{using the pythagorean theorem}\\\\[/tex]
[tex]\bf c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm \sqrt{4^2-1^2}=a\implies \pm\sqrt{16-1}=a\implies \pm\sqrt{15}=a\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse} \qquad cos(\theta )=\cfrac{\pm\sqrt{15}}{4}\impliedby cos\left[ sin^{-1}\left( \frac{1}{4} \right) \right][/tex]
[tex]\bf c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm \sqrt{4^2-1^2}=a\implies \pm\sqrt{16-1}=a\implies \pm\sqrt{15}=a\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse} \qquad cos(\theta )=\cfrac{\pm\sqrt{15}}{4}\impliedby cos\left[ sin^{-1}\left( \frac{1}{4} \right) \right][/tex]
