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How do you simplify the expression #cos(arcsin(x/5)) #?

Answer 1

#cos(arcsin(\frac{x}{5})) = \frac{sqrt{25-x^2}}{5}#.

Firstly, note that for the range of #arcsin(\frac{x}{5})#, the cosine function is purely positive.

Thus, I can say,

#cos(arcsin(\frac{x}{5})) = sqrt{1-sin^2(arcsin(\frac{x}{5}))}#, and also for the domain of x, #sin(arcsin(\frac{x}{5})) = \frac{x}{5}# since the arcsine function outputs values ranging from #-\frac{pi}{2}# to #\frac{pi}{2}#.

This can be verified graphically, the red one being #cos(arcsin(\frac{x}{5}))#, green #cos(\frac{x}{5})#, and blue #arcsin(\frac{x}{5})#:

Therefore,

#cos(arcsin(\frac{x}{5})) = sqrt{1-\frac{x^2}{25})#

#= \frac{sqrt{25-x^2}}{5}#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.