Inverse Trig (1 Viewer)

[Lainee]

Given this question,

Find sin(sin^-1 (x))

can I just simply write down

sin(sin^-1 (x))=x

and similarly for cos and tan?

 

[ND]

Yep.

 

[wogboy]

Interesting question. What needs to be defined here is the domain of the sin^-1(x) function, and the domain of the sin(x) function.

At 3U level I think that the domain of sin^-1(x) is -1<=x<=1, and the domain of sin(x) is all real x, right? So therefore according to this definition, sin(sin^-1(x)) = x, where -1<=x<=1, and otherwise it's undefined (simply because sin^-1(x) is undefined if you feed in a number bigger than 1 or smaller than -1).

Same applies for cos, but not for tan. You can say that tan(tan^-1(x)) = x for all real x.

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You can continue reading this post if you do 4U, otherwise stop here.

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If you were to allow the domains of sin^-1(x) and sin(x) to be all complex x, then you have an entirely different story, you can say that sin(sin^-1(x)) = x for all x, and same with cos. This is because if you feed any number >1 or <-1 into the sin^-1(x) function (e.g. sin^-1(5) ), you will get a complex number as a result.

 

[KeypadSDM]

Originally posted by wogboy
If you were to allow the domains of sin^-1(x) and sin(x) to be all complex x, then you have an entirely different story, you can say that sin(sin^-1(x)) = x for all x, and same with cos. This is because if you feed any number >1 or <-1 into the sin^-1(x) function (e.g. sin^-1(5) ), you will get a complex number as a result.

But if Sin^-1[x] has a value (I.e. not undefined) then you should be able to perform the Sin function back on it to get just x.

An intermediary process gave complex answers, the end result is purely real. Doesn't that mean the answer is x without any specific limits?

 

[Lainee]

Hmm... basically, I asked this question because I wanted to distinguish it from sin^-1(sin x) which is only equal to x when -pi/2<=x<=pi/2. In an exam situation, I wanted to know whether for sin(sin^-1(x)) I can automatically assume its equal to x because I can sin anything. So like KeypadSDM said, it sorta means the answer is x without any limits, and unlike sin^-1(sin x).

I understand how it works though, wogboy but thanks anyway for your explanation. I think to prove it I was told to differentiate it and integrate it again - I'm not very clear on it. That's why I wanted to just assume it was equal to x, cause I don't know how to prove it that way. But I can always write an essay on it trying to explain it if I get stuck I suppose.

 

[KeypadSDM]

Originally posted by Lainee
In an exam situation, I wanted to know whether for sin(sin^-1(x)) I can automatically assume its equal to x because I can sin anything.

You can sin anything, but you can't Sin^-1 everything (With real results)

 

[Grey Council]

sin and inverse sin cancel each other out. thats home, self tuition for you. lol

 

[victorling]

Originally posted by Lainee
Given this question,

Find sin(sin^-1 (x))

can I just simply write down

sin(sin^-1 (x))=x

and similarly for cos and tan?

yep
it is the way

 

[Giant Lobster]

sin(sin^-1(x)) = x

but sin^-1(sinx) not necessarily = x
e.g. consider x = pi

but arcsin(sinx) = x