c3 inverse trig functions Watch

marcsaccount
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Hi,

Just revising c3 and I'm stuck on something that I was fine with the first time around!

For inverse trig functions, say if I want the cos^{-1}(\frac{-\sqrt{3}}{2}), as it's inverse cos, I know the domain has to be  0<x<\pi so because the value isn't within this range, I used the symmetry properties of the cos graph, so I used \frac{-\sqrt{3}}{2} =\frac{\sqrt{3}}{2} and I suspect this is where I went wrong because I've got the wrong answer.....could anyone help me a little?

Thanks
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BabyMaths
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(Original post by marcsaccount)
For inverse trig functions, say if I want the cos^{-1}(\frac{-\sqrt{3}}{2}), as it's inverse cos, I know the domain has to be  0<x<\pi so because the value isn't within this range...
Perhaps you are confusing range and domain?

If f(x) = \arccos(x)

then

the domain is -1\le x \le 1

and

the range is 0\le f(x)\le \pi
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marcsaccount
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(Original post by BabyMaths)
Perhaps you are confusing range and domain?

If f(x) = \arccos(x)

then

the domain is -1\le x \le 1

and

the range is 0\le f(x)\le \pi
Thanks, yes I was. So as it's within the correct range, can I just put it through the calculator?
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BabyMaths
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(Original post by marcsaccount)
Thanks, yes I was. So as it's within the correct range, can I just put it through the calculator?
Yes.
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marcsaccount
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(Original post by BabyMaths)
Yes.

OK, thanks for your help
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marcsaccount
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(Original post by BabyMaths)
Yes.
So, in general, I should use the relevant graph to find the correct value in the appropriate domain. Once I've obtained that, I can just put it through my calculator (unless it's a value that I know whereby the calculator isn't needed)?

I basically hate the unit circle so the above will always be ok?
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BabyMaths
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(Original post by marcsaccount)
So, in general, I should use the relevant graph to find the correct value in the appropriate domain. Once I've obtained that, I can just put it through my calculator (unless it's a value that I know whereby the calculator isn't needed)?

I basically hate the unit circle so the above will always be ok?
Could you give an example of the type of problem that you're trying to solve?

Considering a relevant graph is often a good idea.
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marcsaccount
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(Original post by BabyMaths)
Could you give an example of the type of problem that you're trying to solve?

Considering a relevant graph is often a good idea.
Just say, something simple like  cos^{-1}(sin \frac{4\pi}{3}) I know lots of people quote the unit circle, but I'd much rather find the appropriate value from a graph and then take it from there.
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the bear
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remember that

Cos( 90 - x )° or

Cos( π/2 - x )

are both equivalent to Sin x.

thus Sin (4π/3) is the same as Cos( π/2 - 4π/3 )

and also Cos( -β ) is the same as Cos( β )
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marcsaccount
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(Original post by the bear)
remember that

Cos( 90 - x )° or

Cos( π/2 - x )

are both equivalent to Sin x.

thus Sin (4π/3) is the same as Cos( π/2 - 4π/3 )

and also Cos( -β ) is the same as Cos( β )

thanks bear
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the bear
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(Original post by marcsaccount)
thanks bear
you're welcome

:borat:
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