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Calculus1 22 Online
OpenStudy (anonymous):

Is there anyone on here that is overly good with inverse sin, cos, tan that could tell me the rules for things like: Sin^-1(sin5) or cos(cos^-1(4)) I'm attempting to study for a final and can't seem to find it in my notes

OpenStudy (anonymous):

can you write in more good

OpenStudy (anonymous):

If we have sin-1(sin(x)) = x & cos^-1(cos(x))= x or cos(cos^-1(x)) = x

OpenStudy (anonymous):

so sin^-1(sin5) = ----------?

OpenStudy (anonymous):

5? so sin^-1(sinx)=x but what does sin(sin^-1x)=?

OpenStudy (anonymous):

correct !! ,both are same ..

OpenStudy (anonymous):

okay but aren't there restrictions on it? like with the domain?

OpenStudy (anonymous):

oops sin^-1(sin(x)) = x for [-pi/2 , pi/2] & sin(sin^-1(x))=x for [-1 , 1] & for cos^-1(cos(x)) = x for [0 , pi] & cos(cos^-1(x)) = x for [-1 , 1]

OpenStudy (anonymous):

and how would you do that without a calculator?

OpenStudy (anonymous):

5 is in degree so sin-1(sin(5)) = 5 ,

OpenStudy (anonymous):

& similarly for cos^-1(cos(4)) = 4..

OpenStudy (anonymous):

You can check it by converting 5 degree in radian that is pi/36 = 180/5 = 5

OpenStudy (anonymous):

and if i got sin(sin^-1(6)) how would I find that?

OpenStudy (anonymous):

6 degree

OpenStudy (anonymous):

but there are restrictions sin(sin^-1(x)) is different from sin^-1(sinx). One of them is way more complicated than the other, i just can't remember which one it is

OpenStudy (anonymous):

im pretty good at inverse trig ?????

OpenStudy (anonymous):

when you do a problem such as sin^-1(sinx) or sin(sin^-1x) what are the restrictions? I know one of them the x can't be greater then 1 and the other normally just equals x because the sins cancel out. and I know when you do either cos^-1(cosx) or cos(cos^-1x) theres a weird long process to find the answer i just cant remember what it is

OpenStudy (anonymous):

i'm trying to review for a final exam and my notes are all gibberish

OpenStudy (anonymous):

am I making sense? Lol i feel like I'm not explaining it very well

OpenStudy (anonymous):

we have always f^-1(f(x)) = x here f(x) = sin x and inverse of sine function is sin^-1. so when we apply the inverse on the sine function that is sine^-1(sin(x)) = x just reply x by 5 you will get the final answer is 5. okay. I hope this hint will help you. If you need help please repply.

OpenStudy (anonymous):

I got the answer to those problems, i just need to know the general rules/restrictions so i will know what to look for on the exam

OpenStudy (anonymous):

can you please ask your problem more clearly?

OpenStudy (anonymous):

idk he gave us a list of problems on the review: sin^-1(-1/2) cos^-1(-1/2) tan^-1(-sqrt3) sec^-1(-1/3) sin^-1(sin(3pi/4)) cos^-1(cos(4pi/3)) sin^-1(sin5)c cos^-1(cos5) sin^-1(sin(sqrt2/2) sin^-1(cos4) sin(cos^-1(sqrt52) I need to know how to solve these kinds of problems, when they are undefined, when does x just equal x as opposed to when do I have to solve the inside then the outside. I just need to know how to do them.

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