Mathematics, 20.07.2019 10:30 ptrlvn01
Find the exact value of cos(arcsin(1/ explain your reasoning.
Answers: 1
![\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]](/tpl/images/0111/3474/f9d1c.png)
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Find the exact value of cos(arcsin(1/ explain your reasoning....
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