Question

Y=Cosx.arcsinx What is the solution

Answer 1

Step-by-step explanation:

Given: $y = cos(sin^{-1}x))$

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

$sin^{2}x +cos^{2}x = 1$

So, using the identity $cos(x) = \pm \sqrt{1-sin^{2}(x) }$

$y = \pm \sqrt{1-sin^{2}(sin^{-1}(x)) }$

As the sin and inverse sin function do cancel each other, so only x² is left.

Hence,

$y = \pm \sqrt{1-x^{2}$ ; $-1\leq x\leq 1$

Keywords: trigonometric function

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