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)

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.

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)

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.

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.

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)

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.

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.

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. 👍👍

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)

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.

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!

- Stationary point for curve y=3sec(2x)
- A level differentiation chain rule
- AQA A-level Further Maths Required Trig Functions?
- A level integrating by parts maths question
- Simplifying an expression again
- Further maths integration question
- Help maths urgent ‼️
- x = sec^2 3y differentiate
- gcses help
- Reverse chain problem doesn’t make sense
- electricity mcq plz help
- trig identities help
- STEP maths I, II, III 1990 solutions
- ah maths help
- STEP 1 maths
- Polar equations
- Help plz urgent maths
- A level maths - trigonometry help
- STEP Maths I, II, III 1993 Solutions
- Integrating by substitution

Last reply 1 week ago

Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths

72

Last reply 1 week ago

Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths

72