Question

Find the exact value for arcsin(sin 630)

Answer 1

find a point on the unit circle between \(-90\) and \(90\) that is coterminal with \(630\)

Answer 2

arcsin and sin are inverse function, thus they cancel out (since the existence of inverse implies bijectivity)

Answer 3

cancel my foot!

Answer 4

But the answer is supposed to be 201pi-603

Answer 5

???

Answer 6

are you working in degrees or radians?

Answer 7

wait i mean 201pi-630

Answer 8

in radians

Answer 9

like hell

Answer 10

"cancel my foot!" LOL

Answer 11

when you write
\[\arcsin(\sin(630))\] it is pretty much assumed you are working in degrees

Answer 12

cancel
\[\frac{8}{2}=4\] because the two's cancel
\[x^2-2x+2x-4=x^2-4\] because the \(2x\) cancel
\[\sqrt{x^2}=x\] because the square root and the square cancel
\[e^{\ln(x)}=x\] because the \(e\) and the \(\ln(x)\) cancel
what the heck

Answer 13

Is that the word of the day :)

Answer 14

\[\huge\cancel{\text{cancel}}\]

Answer 15

so is the answer just 630?

Answer 16

no it is \(-90\)

Answer 17

\[630-360=270\\
270-360=-90\]so \(-90\) is coterminal with \(630\) and that is the number you are looking for, the number between \(-90\) and \(90\) coterminal with \(630\)

Answer 18

another way of looking at it is that
\[\sin(630)=-1\] and
\[\arcsin(-1)=-90\)

Answer 19

if it is in radians (which is a little weird) then the answer is \(201\pi-630\)

Answer 20

\[\arcsin(-1)=-90\]
not sure where the \(\pi\) comes from

Answer 21

how could it be in radians?

Answer 22

I'm just going by what they said...one would think that it would be in degrees though

Answer 23

if it is in radians then the answer would be \(-\frac{\pi}{2}\) or am i missing something?

Answer 24

Zarkon, how did you get that answer?

Answer 25

yeah enquiring minds want to know

Answer 26

the 630 radians (which is not a multiple of pi)

Answer 27

ooooh

Answer 28

just take 630 mod PI

Answer 29

you need to keep subtracting \(2\pi\) to get back between \(-\pi/2\) and \(\pi/2\)

Answer 30

more trickonometry

Answer 31

yes

Answer 32

ohhh okayyy thanks all of you!!

Answer 33

\[\sin(630)=\sin(630-100*2\pi)\]
This would mean we are now looking at 630-200pi which is in the second quadrant
if we have are angle is between pi/2 and pi then we can say sin(pi-that angle)=sin(of some angle between 0 and pi/2)
so \[\sin(630)=\sin(630-100*2\pi)=\sin(pi-[630-100 \cdot 2 \pi])\]

Answer 34

Why did you choose to multiply 100 by \[2\Pi \]?

Answer 35

sin(x)=sin(x-k*2pi)

Answer 36

for any integer k

Answer 37

where x is in radians of course

Answer 38

if we had x in degrees then sin(x)=sin(x-k*360 deg)

Answer 39

k represents the number of rotations we make a round the circle

Answer 40

Where did you get that equation from?

Answer 41

It is just something from going around and around again

Answer 42

but yeah it is a formula somewhere

Answer 43

it is a formula here

Answer 44

ohh okay thanks!!!! :D

Answer 45

say that is angle we are looking at right there