# c3 inverse trig functionsWatch

Thread starter 4 years ago
#1
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 , as it's inverse cos, I know the domain has to be so because the value isn't within this range, I used the symmetry properties of the cos graph, so I used and I suspect this is where I went wrong because I've got the wrong answer.....could anyone help me a little?

Thanks
0
4 years ago
#2
(Original post by marcsaccount)
For inverse trig functions, say if I want the , as it's inverse cos, I know the domain has to be so because the value isn't within this range...
Perhaps you are confusing range and domain?

If

then

the domain is

and

the range is
0
Thread starter 4 years ago
#3
(Original post by BabyMaths)
Perhaps you are confusing range and domain?

If

then

the domain is

and

the range is
Thanks, yes I was. So as it's within the correct range, can I just put it through the calculator?
0
4 years ago
#4
(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.
1
Thread starter 4 years ago
#5
(Original post by BabyMaths)
Yes.

OK, thanks for your help
1
Thread starter 4 years ago
#6
(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?
0
4 years ago
#7
(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.
0
Thread starter 4 years ago
#8
(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 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.
0
4 years ago
#9
remember that

Cos( 90 - x )Â° or

Cos( Ï€/2 - x )

are both equivalent to Sin x.

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

and also Cos( -Î² ) is the same as Cos( Î² )
1
Thread starter 4 years ago
#10
(Original post by the bear)
remember that

Cos( 90 - x )Â° or

Cos( Ï€/2 - x )

are both equivalent to Sin x.

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

and also Cos( -Î² ) is the same as Cos( Î² )

thanks bear
0
4 years ago
#11
(Original post by marcsaccount)
thanks bear
you're welcome

0
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