Question

Hi, how would I represent the domain of \(\large \rm \sqrt{sin(x)}\) in a set builder notation
is it {x| sinx>=0} I cannot go any further right?

Answer 1

@SithsAndGiggles I need you help here to clear this ^_^

Answer 2

Do have a brain........??

Answer 3

i was trying to draw the graph of that function, I came up with a concaved down graph
from [0, pi] using differentiation. that's where i got the question lol

Answer 4

if u havea brain then you'll definitely get the answer USE YOUR BRAIN

Answer 5

just kidding don't take it seriously plz

Answer 6

Eh no worries! i didn't really care what you were saying hehe

Answer 7

@amistre64 ^_^

Answer 8

really so sweet of u

Answer 9

^_^!

Answer 10

@surjithayer what do you think!

Answer 11

what is the domain for sqrt(u) ? then compare that to sin(x)

Answer 12

[0, oo)

Answer 13

since its periodic, we may need to play with some n parts

Answer 14

then sin(x) >= 0 when x= ?

Answer 15

x=npi

Answer 16

sin(x) is positive between 0 and pi ...

Answer 17

yes!
and it's positive ....(-2pi,-pi)U(0, pi)U(2pi, 3pi)....

Answer 18

yes so


[0pi , 1pi]
[2pi , 3pi]
[4pi , 5pi]

even pi to odd pi

2n pi to (2n+1) pi

Answer 19

0 is allowed, so the equal brackets are used

Answer 20

so i would right it as {x| x is in [2npi, (2n+1)pi]

Answer 21

write*

Answer 22

set builder ...

{x: (2n)pi <= x <= (2n+1)pi, for n in N}

Answer 23

n in Z really ...

Answer 24

might be some union effect in there, but thats the brunt of it to me

Answer 25

yes n should be integer

Answer 26

i thought of unions but there are troublesome!

Answer 27

the idea is to make a clear and concise definition for x. if its clear from that context then its fine

Answer 28

here when you say sinx>=0 for x=npi
it is not true for all n in N, since we have > so you guys meant sinx=0 here right?

Answer 29

\[x=\left[2 n \pi,\left( 2n+1 \right)\pi \right]\],for n in N
\[x=\left[ \left( 2 n+1 \right)\pi,2n \pi \right],for~ negative~ integers.\]
so union of these two.

Answer 30

since we are only allowed arguments that are 0 or positive

then sin(x) can only output values that are greater or equal to 0

Answer 31

oh?
i think x in [2npi, (2n+1)pi] n in Z works for any integers?

Answer 32

hence

sqrt(u) is defined when u>=0

u = sin(x) so .... sin(x) >=0 , what values of x make this true?

Answer 33

i got that part but when we say x=npi (n in N) it is not true for any n

Answer 34

-6,-5
-4,-3
-2,-1
0,1
2,3
4,5

Answer 35

-2pi <= x <= -1pi is fits all the same

Answer 36

sin x is positive in first and second quadrant only.
( -2 pi,-3 pi) is in 3rd and 4th quadrants.

Answer 37

eh yes! yes! i got things mixed up
i got it, thanks guys for clearing this^_^

Answer 38

this is causing trouble to me sometimes lol

Answer 39

for cosine it would be x in [npi/2, (2n+1)pi/2] right? n in Z

Answer 40

sqrt(cos(x)) ??

Answer 41

yes

Answer 42

same process, solve for cos(x) >= 0

which is the right side of the circle ... -pi/2 to pi/2

Answer 43

if we take our sin(x) domain and subtract pi/2 from it ... then we are set