The Student Room Group

On sec(x)

How come things like sec(-1/2), would equal 2pi/3 rather than pi/3 since cos(pi/3) = 1/2? Does it have anything to do with -1/2, because I thought that only refers to cast diagrams, or is there an entirely different reason?
Original post by JTDunks
How come things like sec(-1/2), would equal 2pi/3 rather than pi/3 since cos(pi/3) = 1/2? Does it have anything to do with -1/2, because I thought that only refers to cast diagrams, or is there an entirely different reason?


sec(-1/2) is neither 2pi/3 nor pi/3

sec(-1/2) means 1/cos(-1/2) so if you can tell me what cos(-1/2) is the just divide 1 by it. It’s not a neat answer.
(edited 9 months ago)
Reply 2
Original post by RDKGames
sec(-1/2) is neither 2pi/3 nor pi/3

sec(-1/2) means 1/cos(-1/2) so if you can tell me what cos(-1/2) is the just divide 1 by it. It’s not a neat answer.

Sorry, I just checked over my question that I was doing and turns out I was wrong about sec, regardless I'm still lost, but the question I'm at had y = cos^-1(-1/2) which should equal pi/3 in radians right? However it turned out that the textbook answer resulted in x = -2pi/3 and 2pi/3 (because the domain is -pi<x<pi). That also adds to my question as the three trig values, secx, cosecx and cotx, are equal to 1/cosx, 1/ secx, 1/tanx, so would that make secx, cosecx, cotx equal to cos^-1(x), sin^-1(x), tan^-1(x), but if that's so then wouldn't they basically be arccosx, arcsinx, arctanx as well?
(edited 9 months ago)
Original post by JTDunks
Sorry, I just checked over my question that I was doing and turns out I was wrong about sec, regardless I'm still lost, but the question I'm at had y = cos^-1(-1/2) which should equal pi/3 in radians right? However it turned out that the textbook answer resulted in x = -2pi/3 and 2pi/3 (because the domain is -pi<x<pi). That also adds to my question as the three trig values, secx, cosecx and cotx, are equal to 1/cosx, 1/ secx, 1/tanx, so would that make secx, cosecx, cotx equal to cos^-1(x), sin^-1(x), tan^-1(x), but if that's so then wouldn't they basically be arccosx, arcsinx, arctanx as well?


Steer away from using notation like cos^-1(x) because on one hand its either arccos(x) or sec(x) on the other.

arccos(x) and sec(x) are NOT the same function. One is a functional inverse and the other is algebraic inverse. Easy so mix them up when writing cos^-1(x) …. They should be kept separate.

Using cos^-1 at GCSE is OK because there is no concept of sec to cause the confusion. At A-Level this should now be addressed more carefully when writing maths.
(edited 9 months ago)
Reply 4
Original post by RDKGames
Steer away from using notation like cos^-1(x) because on one hand its either arccos(x) or sec(x) on the other.

arccos(x) and sec(x) are NOT the same function. One is a functional inverse and the other is algebraic inverse. Easy so mix them up when writing cos^-1(x) …. They should be kept separate.

Using cos^-1 at GCSE is OK because there is no concept of sec to cause the confusion. At A-Level this should now be addressed more carefully when writing maths.

Oh okay, many thanks, also coming back to it then, cos(x) = -1/2 becomes x = arccos(-1/2) right? So then it'd have to equate to -pi/3 right? But I'm still hung up on why its equal to 2pi/3 and -2pi/3? This twould be my final question, many thanks for clearing up the previous questions.
Reply 5
Original post by JTDunks
Oh okay, many thanks, also coming back to it then, cos(x) = -1/2 becomes x = arccos(-1/2) right? So then it'd have to equate to -pi/3 right? But I'm still hung up on why its equal to 2pi/3 and -2pi/3? This twould be my final question, many thanks for clearing up the previous questions.


From the cast diagram or the sin/cos curves or the usual trig addition identities ... Start with
cos(pi/3) = cos(-pi/3) = 1/2
and the addition identity cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
cosIpi-x) = -cos(x)
so
cos(2pi/3) = cos(pi-pi/3) = -cos(pi/3) = -1/2
which also equals cos(-2pi/3). The negative part in -1/2 means it makes an angle +/-pi/3 with the negative x-axis (cast) so it makes an angle +/-2pi//3 with the positive x-axis.

It probably helps to upload why you think otherwise if youre still unsure.
(edited 9 months ago)
Reply 6
Original post by JTDunks
Oh okay, many thanks, also coming back to it then, cos(x) = -1/2 becomes x = arccos(-1/2) right? So then it'd have to equate to -pi/3 right? But I'm still hung up on why its equal to 2pi/3 and -2pi/3? This twould be my final question, many thanks for clearing up the previous questions.

I suspect you think that "if cos(x) = y then cos(-x) = -y" (where in this example x=pi/3 and y=1/2). This is not the case.
Reply 7
Original post by DFranklin
I suspect you think that "if cos(x) = y then cos(-x) = -y" (where in this example x=pi/3 and y=1/2). This is not the case.

So what would be the case?

Original post by mqb2766
From the cast diagram or the sin/cos curves or the usual trig addition identities ... Start with
cos(pi/3) = cos(-pi/3) = 1/2
and the addition identity cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
cosIpi-x) = -cos(x)
so
cos(2pi/3) = cos(pi-pi/3) = -cos(pi/3) = -1/2
which also equals cos(-2pi/3). The negative part in -1/2 means it makes an angle +/-pi/3 with the negative x-axis (cast) so it makes an angle +/-2pi//3 with the positive x-axis.

It probably helps to upload why you think otherwise if youre still unsure.

Wait I just figured out where I went wrong and now see why I kept thinking it should be pi/3, I drew a graph of cos(x) which then showed me what I wanted, sorry for the Cuffufle, thanks for your help though. 👍👍
Reply 8
Original post by JTDunks
So what would be the case?

cos(-x) = cos(x) (i.e. flipping the sign of x doesn't change the sign of cos(x)).

If you look at the graph you can see that cos(pi/2+x)=-cos(pi/2-x), which is probably the easiest thing to use here.
(edited 9 months ago)
Reply 9
Original post by DFranklin
cos(-x) = cos(x) (i.e. flipping the sign of x doesn't change the sign of cos(x)).

If you look at the graph you can see that cos(pi/2+x)=-cos(pi/2-x), which is probably the easiest thing to use here.

Yeah I tried this and realised where I went wrong now, thank you!

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