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Inverse trig functions

How to do something like arcsin[sin(pi/3)]. Wtf does this even mean

safe
Original post by QuantumSuicide
How to do something like arcsin[sin(pi/3)]. Wtf does this even mean

safe


You're taking the inverse of a function, so wouldn't you just be left with the initial value?

It's just f1{f(x)}f^{-1} \{ f(x) \} isn't it?

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Reply 2
Original post by majmuh24
You're taking the inverse of a function, so wouldn't you just be left with the initial value?

It's just f1{f(x)}f^{-1} \{ f(x) \} isn't it?

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Seems like it, but this doesn't work for something like arcsin[sin(2pi/3)]
Original post by QuantumSuicide
Seems like it, but this doesn't work for something like arcsin[sin(2pi/3)]

Arcsin only returns the principle angle.
Reply 4
How do I even go about solving these?

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Reply 5
Original post by QuantumSuicide
How to do something like arcsin[sin(pi/3)]. Wtf does this even mean

safe


what did you get for c1 c2 s1.
What other subs do you do?
Reply 6
Original post by QuantumSuicide
How to do something like arcsin[sin(pi/3)]. Wtf does this even mean

safe


Well, you know what sin(pi/3) is because it's a standard angle. And you know what arcsin does to a value.

All you have to do is decide how the answer relates to pi/3 i.e. is it within the range of values for a principal angle.
Reply 7
Original post by davros
Well, you know what sin(pi/3) is because it's a standard angle. And you know what arcsin does to a value.

All you have to do is decide how the answer relates to pi/3 i.e. is it within the range of values for a principal angle.


I'm just unsure about how to draw the quadrants, where the angles go etc for these inverse trig expressions

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Reply 8
Original post by QuantumSuicide
I'm just unsure about how to draw the quadrants, where the angles go etc for these inverse trig expressions

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Start by writing down what range of angles can be returned by arcsin.
Reply 9
Original post by davros
Start by writing down what range of angles can be returned by arcsin.


So should i just bang arcsin(sin(pi/3)) in my calculator? I get pi/3

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(edited 10 years ago)
Reply 10
Original post by QuantumSuicide
So should i just bang arcsin(sin(pi/3)) in my calculator? I get pi/3

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Well that's certainly one way of answering the question :smile:
Original post by davros
Well that's certainly one way of answering the question :smile:


:frown:
I'm really not understanding this lol. If anyone could take me through an example with the quadrant method, that would be great.

How would I do arctan(root3)

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Reply 12
Original post by QuantumSuicide
:frown:
I'm really not understanding this lol. If anyone could take me through an example with the quadrant method, that would be great.

How would I do arctan(root3)

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I don't get why you're trying to confuse yourself?
Youre aware that arcsin(sin(X)) = X? Where does the quadrant method come into this at all?
Original post by lmorgan95
I don't get why you're trying to confuse yourself?
Youre aware that arcsin(sin(X)) = X? Where does the quadrant method come into this at all?


Arcsin(x) only returns the principal value.

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Reply 14
Original post by majmuh24
Arcsin(x) only returns the principal value.

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What value is he trying to get to though? You can get to the other values from the principle.
Oh right so is he asking for an explanation of how to get to the secondary+ values?
Original post by lmorgan95
What value is he trying to get to though? You can get to the other values from the principle.
Oh right so is he asking for an explanation of how to get to the secondary+ values?


Yeah, but if you take values outside the range of arcsin(x), arcsin(sin(x)) doesn't just give you x.

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Reply 16
Original post by QuantumSuicide
:frown:
I'm really not understanding this lol. If anyone could take me through an example with the quadrant method, that would be great.

How would I do arctan(root3)

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Let arctan3=αarctan \sqrt{3} =\alpha

then

tan(arctan3)=tanα \tan \left (arctan \sqrt{3}\right )=\tan \alpha

So
3=tanα\sqrt{3}=\tan \alpha

THe angle for which the tan gent is root(3) is degree 60

so
α=π3+kπ\alpha=\frac{\pi}{3}+k\cdot \pi

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