identities involving Cosec , Sec and Cot

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Alen.m
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#1
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#1
Hi guys,
I've attached you my workings of solving an equation and find all solutions in the given interval but my answers keep coming different than what the text book says, i've doubled checked my workings seem fine to me. any one can help me if I'm wrong at some point?i've used cast diagram to find solutions btw. thanks guys
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problemq
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#2
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You are correct with cos(x) = -1/2 cos(x)=-1/3. But the values in the textbook are correct (how did you obtain such values that you boxed?) Remember to round to at least 2dp. It can be easier to sketch the graph y=cos(x) and look at where the lines y=-1/3 and y=-1/2 intersect with it.
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Dohaeris
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#3
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(Original post by problemq)
You are correct with cos(x) = -1/2 cos(x)=-1/3. But the values in the textbook are correct (how did you obtain such values that you boxed?) Remember to round to at least 2dp. It can be easier to sketch the graph y=cos(x) and look at where the lines y=-1/3 and y=-1/2 intersect with it.
cosx is equal to negative values (-0.5 and -1/3), so aren't you supposed to add/subtract arccos(-0.5) and arccos(-1/3) with pie? You can get the values given by subtracting arccos from 2pie, but I thought you're only supposed to do that when the value cosx equals to is positive.
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problemq
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You need to subtract arcos from 2pi because of the symmetry of the cosine graph. Eg for arcos(-1/2) you obtain a value greater than pi/2, and if you look at the cosine graph you see the other solution is obtained from 2pi - arcos(-1/2) (due to the symmetry) Can you see this?
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Dohaeris
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#5
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(Original post by problemq)
You need to subtract arcos from 2pi because of the symmetry of the cosine graph. Eg for arcos(-1/2) you obtain a value greater than pi/2, and if you look at the cosine graph you see the other solution is obtained from 2pi - arcos(-1/2) (due to the symmetry) Can you see this?
Image


Yeah, I see it now. Do you recommend that for these type of questions, the graph should always be drawn?

Also, doesn't this directly refute the CAST diagram? or is this an exception?
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problemq
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#6
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I do recommend it (for me it is easier) but if you've been taught the CAST diagram then use that. I don't remember the CAST diagram much at all but there must be some misunderstanding of using it (most likely because of the negative signs) here as this problem shouldn't refute the CAST method.
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Dohaeris
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#7
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(Original post by problemq)
I do recommend it (for me it is easier) but if you've been taught the CAST diagram then use that. I don't remember the CAST diagram much at all but there must be some misunderstanding of using it (most likely because of the negative signs) here as this problem shouldn't refute the CAST method.
Yeah, I'm probably misunderstanding something, so I'll go and look it up a bit more. Thanks.

EDIT: I get it now, just a silly mistake in my working.
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Alen.m
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#8
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(Original post by problemq)
You are correct with cos(x) = -1/2 cos(x)=-1/3. But the values in the textbook are correct (how did you obtain such values that you boxed?) Remember to round to at least 2dp. It can be easier to sketch the graph y=cos(x) and look at where the lines y=-1/3 and y=-1/2 intersect with it.
Here's the method i've used which i've doubled checked it over and over to make sure nothings wrong with it. you calculate the angle from cos(x)=-1/2 and cos(x)=-1/3 which are 1.9(radians) and 2.9(radians) then you look for the quadrant where cos is negative which would be second and third quadrant (anticlockwise)then you add and subtract your angle to pi. i can match the answers with text book by taking away the first two angles from 2pi but in that quadrant cos is positive
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Zacken
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(Original post by Alen.m)
Here's the method i've used which i've doubled checked it over and over to make sure nothings wrong with it. you calculate the angle from cos(x)=-1/2 and cos(x)=-1/3 which are 1.9(radians) and 2.9(radians) then you look for the quadrant where cos is negative which would be second and third quadrant (anticlockwise)then you add and subtract your angle to pi. i can match the answers with text book by taking away the first two angles from 2pi but in that quadrant cos is positive
If you want to use that method then you need to ensure that you always do cos^(-1) (POSITIVE angle)

So pi + cos^(-1) (+1/2) and pi - cos^(-1) (+1/2)

Same with the other one.
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Alen.m
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#10
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(Original post by Zacken)
If you want to use that method then you need to ensure that you always do cos^(-1) (POSITIVE angle)

So pi + cos^(-1) (+1/2) and pi - cos^(-1) (+1/2)

Same with the other one.
the text book itself used the same method for finding a trig function of a negative value. Are you saying that i should just ignore the negative sign and calculate the cos of (+1/2) and (+1/3)? Attached you an example of using a cast diagram calculating tan of a negative value
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Zacken
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(Original post by Alen.m)
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the text book itself used the same method for finding a trig function of a negative value. Are you saying that i should just ignore the negative sign and calculate the cos of (+1/2) and (+1/3)? Attached you an example of using a cast diagram calculating tan of a negative value
uh... read the very first bullet point. It clearly dtates "ignore the negative and put +1.33 into the CAST".
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Alen.m
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#12
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(Original post by Zacken)
uh... read the very first bullet point. It clearly dtates "ignore the negative and put +1.33 into the CAST".
That's for when the angle comes out negative but here we have a positive angle ie: cos(-1/2)=1.9 so we dont have to be worry about that . When the trig function comes negative tho we have to find the quadrant where that trig funtion is negative
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Zacken
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#13
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(Original post by Alen.m)
That's for when the angle comes out negative but here we have a positive angle ie: cos(-1/2)=1.9 so we dont have to be worry about that . When the trig function comes negative tho we have to find the quadrant where that trig funtion is negative
Yeah, well, the reality is that you take cos^(-1) (+1/2). The book doesn't need to do that for tangent because it's an odd function, but you need to do it for sine and cosine.
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Alen.m
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#14
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#14
(Original post by Zacken)
Yeah, well, the reality is that you take cos^(-1) (+1/2). The book doesn't need to do that for tangent because it's an odd function, but you need to do it for sine and cosine.
If that's true cos^(-1)(+1/2) which is 1.05 should be included in the answers which is withing the interval as well but it hasnt. Somehow the text book calculated cos^(-1)(-1/2) and took it away from 2pi and the same way for cos^-1(-1/3) and came up with the answers. Any other way you go you come up with at least two different answers compared to text book
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Zacken
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#15
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(Original post by Alen.m)
If that's true cos^(-1)(+1/2) which is 1.05 should be included in the answers which is withing the interval as well but it hasnt. Somehow the text book calculated cos^(-1)(-1/2) and took it away from 2pi and the same way for cos^-1(-1/3) and came up with the answers. Any other way you go you come up with at least two different answers compared to text book
You have two options to solve cos x = -1/2.

Method 1: x = cos^(-1) (-1/2).

Then you use x = cos^(-1/2) = 2pi/3 and x = 2pi - cos^(-1)(-1/2) = 4pi/3.

Method 2: x = cos^(-1) (+1/2).

Then you use x = pi + cos^(-1) (+1/2) and x = pi - cos^(-1) (+1/2) to get x = 4pi/3 and 2pi/3.

Both methods get you the same correct answer. I don't know what you're talking about.
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