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Domains! (1 Viewer)

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How do you work out the domains of inverse trig functions? like if you have f(x)= sin^-1 (pi/x) how do you work out its domain?
and also, when u have y= sin^-1 (cos x), I know you have two answers, but how do you know which domain it belongs to?
Any help would be appreciated!!!
 

Affinity

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remember anyting inside a inverse sine or a inverse cos must be between -1 and 1
and inverse tan is defined for all values

so for f it would be x is from -pi to pi

for the second one there are not just 2, but infinitely many valid domains.
 
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:: ck ::

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f(x)= sin^-1 (pi/x)

Originally posted by Affinity
so for f it would be x is from -pi to pi

correct me if im wrong

-1 =< pi/x =< 1
taking : -1 =< pi/x

-x^2 =< pix
0 =< x^2 + pix
0 =< x(x+pi)

.'. x > 0 [or we have unknown in denom], x <= -pi

taking : pi/x =< 1
pix =< x^2
x^2 - pix >= 0
x(x-pi) >= 0

.'. x >= pi, x < 0 [for the same reasons as above]

So the two cases for the domain is
1. x > 0, x <= -pi
2. x < 0, x >= pi

edit : bleh i think the x >0 or x <0 part is wrong now.. arghh !!!!

Originally posted by Affinity
for the second one there are not just 2, but infinitely many valid domains.
are u sure?

i thought it would be 0 =< x =< pi

according to the property : sin-1 (sinx) = x for -pi/2 =< x <= pi/2
 
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:: ck ::

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sum1 plz clear up the first one for me........ im lost now
 

CM_Tutor

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Domain of f(x) = sin<sup>-1</sup>(pi / x) is {x: x => pi or x <= -pi}
 

:: ck ::

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can u tell me why x < 0 and x > 0 arent included [i know if u shove x values here in the expression doesnt hold true]

but i thought whenever u have unknown in denominator u have to square it
 

CM_Tutor

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pi / x is undefined for x = 0, so when you split -1 <= pi / x <= 1 into the two cases, you actually have:

0 < pi / x <= 1, and hence pi / x <= 1 for x > 0

and

-1 <= pi / x < 0, and hence pi / x => -1 for x < 0.

Thus, your solution of (for the first one) of x => pi or x < 0, when subjected to the constraint x > 0, becomes just
x => pi. Similarly the second case is just x <= - pi
 

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