How do you find the range and domain of #sin(arctan x)#?

Answer 1

Range for #sin(arctanx)# is #[-1,1]#, and domain is #(-oo,oo)#.

#arctanx# is always from #-pi/2# to #pi/2#, but here #x# can take any value from #(-oo.oo)#, which is domain for #sin(arctanx)#.
But #sin(arctanx)# can take values only from #[-1,1]#
Hence range for #sin(arctanx)# is #[-1,1]#.
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Answer 2

To find the range and domain of ( \sin(\arctan(x)) ), first consider the domain of ( \arctan(x) ), which is all real numbers. The range of ( \arctan(x) ) is ( (-\frac{\pi}{2}, \frac{\pi}{2}) ), meaning it outputs angles between ( -\frac{\pi}{2} ) and ( \frac{\pi}{2} ).

Then, applying the sine function to these angles, the range of ( \sin(\arctan(x)) ) is between ( -1 ) and ( 1 ).

So, the domain of ( \sin(\arctan(x)) ) is all real numbers, and the range is between ( -1 ) and ( 1 ).

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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.

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.

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.

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.

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