Edexcel Year 2 Chapter 6 exercise 6E question 6C

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doozy123
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#1
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#1
From what I can't understand why solutionbanks says that tan k stays the same when k changes to -pi/2

If you can prove why tan stays positive please can you explain

thanks in advanced
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mqb2766
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#2
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
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doozy123
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#3
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#3
(Original post by mqb2766)
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression
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mqb2766
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#4
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#4
(Original post by doozy123)
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression
The key thing is that the expressions for cos and tan, defined in terms of sin in the first quadrant, also apply in the fourth quadrant. You can write
cos(k) = sqrt(1-sin^2(k))
tan(k) = sin(k)/sqrt(1-sin^2(k))
and its valid in both the first and fourth quadrant. Here, they have simply replaced sin() with x.

I think thats what youre saying in the attachment by considering -x? If the original x is positive (first quadrant), then -x would refer to the value of sin in the fourth quadrant and its all good.
Last edited by mqb2766; 3 months ago
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doozy123
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#5
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#5
(Original post by mqb2766)
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression

(Original post by mqb2766)
The key thing is that the expressions for cos and tan, defined in terms of sin in the first quadrant, also apply in the fourth quadrant. You can write
cos(k) = sqrt(1-sin^2(k))
tan(k) = sin(k)/sqrt(1-sin^2(k))
and its valid in both the first and fourth quadrant. Here, they have simply replaced sin() with x.

I think thats what youre saying in the attachment by considering -x? If the original x is positive (first quadrant), then -x would refer to the value of sin in the fourth quadrant and its all good.
in essence that is the correct working for cos(k) and tan(k) when 0<k<pi/2. however tan(k) is, like you said, sin/cos but sin in the 4th quadrant is negative (only cos is positive in the 4th quadrant), cos is the same

so doesnt that mean that tan(k) is negative when -pi/2<k<0

thanks for your reply
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mqb2766
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#6
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#6
(Original post by doozy123)
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression


in essence that is the correct working for cos(k) and tan(k) when 0<k<pi/2. however tan(k) is, like you said, sin/cos but sin in the 4th quadrant is negative (only cos is positive in the 4th quadrant), cos is the same

so doesnt that mean that tan(k) is negative when -pi/2<k<0

thanks for your reply
the
sin(k) = x
is valid here in both the 1st and 4th quadrants, so it can be -1 to 1. You seem to be insisting you write -x when its in the fourth quadrant and this is the only problem I can see with your attachment. The expressions
cos(k) = sqrt(1-sin^2(k)) = sqrt(1-x^2)
tan(k) = sin(k)/sqrt(1-sin^2(k)) = x/sqrt(1-x^2)
are valid in both the 1st and 4th quadrant and x = sin(k) so varies -1 .. 1.

The expressions are not valid in the other two quadrants as cos() becomes negative and you'd have to consider the negated root for it so
cos(k) = -sqrt(1-x^2)
and similar for tan().
Last edited by mqb2766; 3 months ago
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doozy123
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#7
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#7
(Original post by mqb2766)
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression

(Original post by mqb2766)
The key thing is that the expressions for cos and tan, defined in terms of sin in the first quadrant, also apply in the fourth quadrant. You can write
cos(k) = sqrt(1-sin^2(k))
tan(k) = sin(k)/sqrt(1-sin^2(k))
and its valid in both the first and fourth quadrant. Here, they have simply replaced sin() with x.

I think thats what youre saying in the attachment by considering -x? If the original x is positive (first quadrant), then -x would refer to the value of sin in the fourth quadrant and its all good.
in essence that is the correct working for cos(k) and tan(k) when 0<k<pi/2. however tan(k) is, like you said, sin/cos but sin in the 4th quadrant is negative (only cos is positive in the 4th quadrant), cos is the same

so doesnt that mean that tan(k) is negative when -pi/2<k<0

thanks for your reply

(Original post by mqb2766)
the
sin(k) = x
is valid here in both the 1st and 4th quadrants, so it can be -1 to 1. You seem to be insisting you write -x when its in the fourth quadrant and this is the only problem I can see with your attachment. The expressions
cos(k) = sqrt(1-sin^2(k)) = sqrt(1-x^2)
tan(k) = sin(k)/sqrt(1-sin^2(k)) = x/sqrt(1-x^2)
are valid in both the 1st and 4th quadrant and x = sin(k) so varies -1 .. 1.

The expressions are not valid in the other two quadrants as cos() becomes negative and you'd have to consider the negated root for it so
cos(k) = -sqrt(1-x^2)
and similar for tan().
perhaps i am phrasing it wrong with -x. when the range of k is restricted to -pi/2<k<0 that means x has to be negative, therefore -1<x<0.
that is why i am saying sin(k)=-(x) as as k is within this interval it is negative whether you look at the graphs or cast diagrams, and like we've agreed cos(k) stays the same.

if that isnt prove enough, we are explicitly told the range of k -pi/2<k<0
when you look at the graphs of sin cos and tan
sin is negative between that domain
cos is positive for that domain
tan is negative for that domain
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doozy123
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#8
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#8
(Original post by mqb2766)
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression

(Original post by mqb2766)
The key thing is that the expressions for cos and tan, defined in terms of sin in the first quadrant, also apply in the fourth quadrant. You can write
cos(k) = sqrt(1-sin^2(k))
tan(k) = sin(k)/sqrt(1-sin^2(k))
and its valid in both the first and fourth quadrant. Here, they have simply replaced sin() with x.

I think thats what youre saying in the attachment by considering -x? If the original x is positive (first quadrant), then -x would refer to the value of sin in the fourth quadrant and its all good.
in essence that is the correct working for cos(k) and tan(k) when 0<k<pi/2. however tan(k) is, like you said, sin/cos but sin in the 4th quadrant is negative (only cos is positive in the 4th quadrant), cos is the same

so doesnt that mean that tan(k) is negative when -pi/2<k<0

thanks for your reply

(Original post by mqb2766)
the
sin(k) = x
is valid here in both the 1st and 4th quadrants, so it can be -1 to 1. You seem to be insisting you write -x when its in the fourth quadrant and this is the only problem I can see with your attachment. The expressions
cos(k) = sqrt(1-sin^2(k)) = sqrt(1-x^2)
tan(k) = sin(k)/sqrt(1-sin^2(k)) = x/sqrt(1-x^2)
are valid in both the 1st and 4th quadrant and x = sin(k) so varies -1 .. 1.

The expressions are not valid in the other two quadrants as cos() becomes negative and you'd have to consider the negated root for it so
cos(k) = -sqrt(1-x^2)
and similar for tan().
perhaps i am phrasing it wrong with -x. when the range of k is restricted to -pi/2<k<0 that means x has to be negative, therefore -1<x<0.
that is why i am saying sin(k)=-(x) as as k is within this interval it is negative whether you look at the graphs or cast diagrams, and like we've agreed cos(k) stays the same.

if that isnt proof enough, we are explicitly told the range of k -pi/2<k<0
when you look at the graphs of sin cos and tan
sin is negative between that domain
cos is positive for that domain
tan is negative for that domain
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mqb2766
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#9
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#9
(Original post by doozy123)
perhaps i am phrasing it wrong with -x. when the range of k is restricted to -pi/2<k<0 that means x has to be negative, therefore -1<x<0.
that is why i am saying sin(k)=-(x) as as k is within this interval it is negative whether you look at the graphs or cast diagrams, and like we've agreed cos(k) stays the same.

if that isnt prove enough, we are explicitly told the range of k -pi/2<k<0
when you look at the graphs of sin cos and tan
sin is negative between that domain
cos is positive for that domain
tan is negative for that domain
The only thing I disagree with is the bold and I think you should write
sin(k) = x
not -x. Obviously the x in the previous line is negative in the fourth quadrant and positive in the first quadrant. I don't understand why you keep writing the negative sign in front of it. Obviously sin() is odd, so
sin(-k) = -sin(k)
but here we're taking a different k in quadrant, not expressing it terms of the original k which lay in quadrant 1.

To put numbers in, Id have something like k=-30 (fourth quadrant).
sin(k) = x = -1/2
cos(k) = sqrt(3)/2
tan(k) = -1/2 / sqrt(3)/2 = -1/sqrt(3)
Because x has a negative value, its all consistent in the fourth quadrant, just as if k=30 and therefore x=1/2.
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doozy123
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#10
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#10
(Original post by mqb2766)
In the 4th quadrant, tan would be negative. But as x is negative, the ans (expression for tan() in b) is the same.
taking that logic, you are saying that as k is between -pi/2 and 0 that x is negative - that is right. but for example cos 30 =cos-30 but tan 30!=tan-30
also if you sub in -x into the expression in part b you get negative of that expression

(Original post by mqb2766)
The key thing is that the expressions for cos and tan, defined in terms of sin in the first quadrant, also apply in the fourth quadrant. You can write
cos(k) = sqrt(1-sin^2(k))
tan(k) = sin(k)/sqrt(1-sin^2(k))
and its valid in both the first and fourth quadrant. Here, they have simply replaced sin() with x.

I think thats what youre saying in the attachment by considering -x? If the original x is positive (first quadrant), then -x would refer to the value of sin in the fourth quadrant and its all good.
in essence that is the correct working for cos(k) and tan(k) when 0<k<pi/2. however tan(k) is, like you said, sin/cos but sin in the 4th quadrant is negative (only cos is positive in the 4th quadrant), cos is the same

so doesnt that mean that tan(k) is negative when -pi/2<k<0

thanks for your reply

(Original post by mqb2766)
the
sin(k) = x
is valid here in both the 1st and 4th quadrants, so it can be -1 to 1. You seem to be insisting you write -x when its in the fourth quadrant and this is the only problem I can see with your attachment. The expressions
cos(k) = sqrt(1-sin^2(k)) = sqrt(1-x^2)
tan(k) = sin(k)/sqrt(1-sin^2(k)) = x/sqrt(1-x^2)
are valid in both the 1st and 4th quadrant and x = sin(k) so varies -1 .. 1.

The expressions are not valid in the other two quadrants as cos() becomes negative and you'd have to consider the negated root for it so
cos(k) = -sqrt(1-x^2)
and similar for tan().
perhaps i am phrasing it wrong with -x. when the range of k is restricted to -pi/2<k<0 that means x has to be negative, therefore -1<x<0.
that is why i am saying sin(k)=-(x) as as k is within this interval it is negative whether you look at the graphs or cast diagrams, and like we've agreed cos(k) stays the same.

if that isnt prove enough, we are explicitly told the range of k -pi/2<k<0
when you look at the graphs of sin cos and tan
sin is negative between that domain
cos is positive for that domain
tan is negative for that domain

(Original post by mqb2766)
The only thing I disagree with is the bold and I think you should write
sin(k) = x
not -x. Obviously the x in the previous line is negative in the fourth quadrant and positive in the first quadrant. I don't understand why you keep writing the negative sign in front of it. Obviously sin() is odd, so
sin(-k) = -sin(k)
but here we're taking a different k in quadrant, not expressing it terms of the original k which lay in quadrant 1.

To put numbers in, Id have something like k=-30 (fourth quadrant).
sin(k) = x = -1/2
cos(k) = sqrt(3)/2
tan(k) = -1/2 / sqrt(3)/2 = -1/sqrt(3)
Because x has a negative value, its all consistent in the fourth quadrant, just as if k=30 and therefore x=1/2.
ok I see where you are coming from with the -x. to be clear x is the same value but i am putting -x to indicate that it is as you are saying, x has a negative value since you know k=arcsinx. i guess you could say i am putting the negative infront of the x to treat x the value itself as positive with a negative infront to help with visualisation in terms of triangles and the graphs.

regardless of the -x/x using thoses examples you put, notice how tan is negative and cos is positive which is what I am trying to say at the start of this thread.

when you use a negative value of k: sin is negative, cos +ve, tan -ve, - this is what we have now both established with your example of -30 so how can tan(k) stay the same
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mqb2766
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#11
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#11
(Original post by doozy123)
ok I see where you are coming from with the -x. to be clear x is the same value but i am putting -x to indicate that it is as you are saying, x has a negative value since you know k=arcsinx. i guess you could say i am putting the negative infront of the x to treat x the value itself as positive with a negative infront to help with visualisation in terms of triangles and the graphs.

regardless of the -x/x using thoses examples you put, notice how tan is negative and cos is positive which is what I am trying to say at the start of this thread.

when you use a negative value of k: sin is negative, cos +ve, tan -ve, - this is what we have now both established with your example of -30 so how can tan(k) stay the same
Noone has disagreed about tan being negative in quadrant 4. It is (and cos is positive in both quadrants). I noted as such in every reply. The only issue I had was your use of -x which changed the expression. The expression is the same in quadrants 1 and 4, as per the model solution, but the value of x obviously changes sign because sin changes sign in quadrants 1 and 4.

There are only so many times I can say the same thing. If you disagree with the expressions in the model solution also applying in quadrant 4, create an example like in #9, work it through and post your working/confusion.
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doozy123
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#12
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#12
(Original post by mqb2766)
Noone has disagreed about tan being negative in quadrant 4. It is (and cos is positive in both quadrants). I noted as such in every reply. The only issue I had was your use of -x which changed the expression. The expression is the same in quadrants 1 and 4, as per the model solution, but the value of x obviously changes sign because sin changes sign in quadrants 1 and 4.

There are only so many times I can say the same thing. If you disagree with the expressions in the model solution also applying in quadrant 4, create an example like in #9, work it through and post your working/confusion.
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I’ve worked through an example and I see it now
X holds a negative value and when you use an a example tan(k) is still negative which is why it doesn’t change

Thanks for your help - took a bit of understanding of what you were trying to say about the -x

Cheers
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mqb2766
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#13
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#13
(Original post by doozy123)
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I’ve worked through an example and I see it now
X holds a negative value and when you use an a example tan(k) is still negative which is why it doesn’t change

Thanks for your help - took a bit of understanding of what you were trying to say about the -x

Cheers
Glad the sign problem is sorted. Im not sure whether I should mention the cos(-pi/6) expression
sqrt(1 - 1/4) = sqrt(3)/2
not
sqrt(1 - (pi/6)^2)
.... Obviously the values are similar because sin(x) ~ x for small x, but ....
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