Question

Find the exact value of cos(arcsin(1/4)). explain your reasoning.

Answer 1

it's 12/13

Explanation:

Answer 2

`\bf sin^(-1)\left( (1)/(4) \right)=\theta \qquad this~means\qquad sin(\theta )=\cfrac{\stackrel{opposite}{1}}{\stackrel{hypotenuse}{4}} \\\\\\ \textit{so, let's find the adjacent side for that angle then} \\\\\\ \textit{using the pythagorean theorem}\\\\`


`\bf c^2=a^2+b^2\implies \pm √(c^2-b^2)=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm √(4^2-1^2)=a\implies \pm√(16-1)=a\implies \pm√(15)=a\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse} \qquad cos(\theta )=\cfrac{\pm√(15)}{4}\impliedby cos\left[ sin^(-1)\left( (1)/(4) \right) \right]`