# trig c4 question

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Thread starter 3 years ago
#1
are there any values of a for which sec a, cosec a and cot a (for the range 0<(and equal to) a< (and equal to) 360 are all negative? explain your answer

the answer is no for each of the second, third and fourth quadrants a different function is positive
could you explain what this means? is there are another way of reaching this conclusion without using the cast method (if it even involves the cast method)?

the second part of the question are there any values of a for which sec a, cosec a and cot a are all equal? explain your answer
no the graphs of all three of the functions do not intersect at a single point
how does intersection show that the values are equal
could you explain this answer too please?
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3 years ago
#2
(Original post by esmeralda123)
...
First part: You can just say for which values of each of these functions are negative, then show that there is no single value of that is within ALL 3 of these domains. Hence there isn't a value for which they are all negative at once.

Second part: If they intersect at a single point, then these functions take the same value a, and more importantly the same value, hence they would be equal at that point for some 0
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3 years ago
#3
For the first one all three cannot be negative as if cosec(a) and sec(a) are negative then cot(a) will be positive as it’s the same as just cosec(a)/sec(a) so the negatives cancel.

For the second part think of a the definitions of these from the right-triangle. Cosec(a) = hyp/opp, sec(a) = hyp/adj, cot(a) adj/opp. If cosec(a) and sec(a) are equal this means opp = adj. Therefore cot(a) = 1, so now for them all to be equal hyp would have to equal opp and adj. This is impossible in a right triangle so there are no values of a that satisfy this.

For graphs if they intersect at a single point that means values of y and x are equal in all the functions intersecting at that point. You’re plotting y against x where in this case x = a. At a point where x = a on, say, the sec graph y = sec(a) and on cosec graph y = cosec(a). If there is an intersection these are the same point so y = both. So sec(a) = cosec(a)
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3 years ago
#4
You can just simplify it and talk about sin cos tan being all negative (since then 1/... will be negative).

Sin is negative between 180 and 360.
Cos is negative between 90 and 270
Tan is negative between 90 and 180 and 270 and 360.

As you can see there is no overlap of all 3 at once. The CAST diagarm is just a way you can see this graphically.

The graphical proof is that all 3 of them never intersect or touch at a certain point
Your question is how does intersection show they are equal?
I mean, that's just how graphs work. For any graph if two functions intersect or touch each other, then at that value of x, they have the same value, they are equal. That's like basic basic maths.

You could do an algebraic proof. 1 over everything makes it
sin = cos = tan

sin x = tan x
=> sin x - tan x = 0
=> sin x(1 - 1/cos x) = 0
Either sin x =0
so x = 0, 180, 360
tan x undefined at 180
so
x= 0, 360
or
1/cos x =1
cos x= 1
x = 0, 360

ie. sin x = tan x
only at 0 deg and 360 deg within the range specified, when both functions = 0.
However obviously cos x doesn't = 0 at those points.
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Thread starter 3 years ago
#5
(Original post by StayWoke)
You can just simplify it and talk about sin cos tan being all negative (since then 1/... will be negative).

Sin is negative between 180 and 360.
Cos is negative between 90 and 270
Tan is negative between 90 and 180 and 270 and 360.

As you can see there is no overlap of all 3 at once. The CAST diagarm is just a way you can see this graphically.

The graphical proof is that all 3 of them never intersect or touch at a certain point
Your question is how does intersection show they are equal?
I mean, that's just how graphs work. For any graph if two functions intersect or touch each other, then at that value of x, they have the same value, they are equal. That's like basic basic maths.

You could do an algebraic proof. 1 over everything makes it
sin = cos = tan

sin x = tan x
=> sin x - tan x = 0
=> sin x(1 - 1/cos x) = 0
Either sin x =0
so x = 0, 180, 360
tan x undefined at 180
so
x= 0, 360
or
1/cos x =1
cos x= 1
x = 0, 360

ie. sin x = tan x
only at 0 deg and 360 deg within the range specified, when both functions = 0.
However obviously cos x doesn't = 0 at those points.
but how does this show that they are not all negative:

'Sin is negative between 180 and 360.
Cos is negative between 90 and 270
Tan is negative between 90 and 180 and 270 and 360.'

there are curves either side of each other
for them to all be negative what must the graphs be like- should they intersect all at one point ?
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3 years ago
#6
(Original post by esmeralda123)
but how does this show that they are not all negative:

'Sin is negative between 180 and 360.
Cos is negative between 90 and 270
Tan is negative between 90 and 180 and 270 and 360.'

there are curves either side of each other
for them to all be negative what must the graphs be like- should they intersect all at one point ?
For the negative part, they just need to all be below the x axis at the same value of x.
Try and find a value of x where all 3 of them are below the x axis.
That's the graphical way.

Or you can look at what you quoted from me and see that there is no value of x which is
1) between 180 and 360
2) between 90 and 270,
3) between 90 and 180 or 270 and 360.
all of the above^

You can get two of the three required points, but not all 3.
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Thread starter 3 years ago
#7
(Original post by StayWoke)
For the negative part, they just need to all be below the x axis at the same value of x.
Try and find a value of x where all 3 of them are below the x axis.
That's the graphical way.

Or you can look at what you quoted from me and see that there is no value of x which is
1) between 180 and 360
2) between 90 and 270,
3) between 90 and 180 or 270 and 360.
all of the above^

You can get two of the three required points, but not all 3.
so they must intersect at one point below the x axis for them to all be negative
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3 years ago
#8
(Original post by esmeralda123)
so they must intersect at one point below the x axis for them to all be negative
no no, intersection only comes into play in the second part of your OP talking about all of them being the same value.

For them to all be negative, they would all have to be below the axis. No intersection needed.
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Thread starter 3 years ago
#9
(Original post by StayWoke)
no no, intersection only comes into play in the second part of your OP talking about all of them being the same value.

For them to all be negative, they would all have to be below the axis. No intersection needed.
but they are all below the x axis
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3 years ago
#10
(Original post by esmeralda123)
but they are all below the x axis
they all go below the x axis at some point,
but what we're looking for is that they're all below the x axis at the same time.
For a certain value of x, all 3 trig graphs below the axis.
That never happens
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Thread starter 3 years ago
#11
(Original post by StayWoke)
they all go below the x axis at some point,
but what we're looking for is that they're all below the x axis at the same time.
For a certain value of x, all 3 trig graphs below the axis.
That never happens
oh ok thank you
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