Step-by-step explanation:
Given: [tex]y = cos(sin^{-1}x))[/tex]
As the domain for the inverse sine function is 1 ≤ x ≤ 1 as this is the range for the sine function.
The range for the function is the same as the range for the cosine function, 1 ≤ y ≤ 1
As
[tex]sin^{2}x +cos^{2}x = 1[/tex]
So, using the identity [tex]cos(x) = \pm \sqrt{1-sin^{2}(x) }[/tex]
[tex]y = \pm \sqrt{1-sin^{2}(sin^{-1}(x)) }[/tex]
As the sin and inverse sin function do cancel each other, so only x² is left.
Hence,
[tex]y = \pm \sqrt{1-x^{2}[/tex] ; [tex]-1\leq x\leq 1[/tex]
Keywords: trigonometric function
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