fiftythree
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Hi

https://photos.app.goo.gl/HFR5JpQRa8XWLqY99

Im stuck on 5 (i) (b)

I tried doing it the normal way but its not working like part a? Very confused. I tried to make simultaneous equations so i could get rid of theta but got stuck here:
1=(rsinalpha)(rcosalpha)
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RDKGames
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(Original post by fiftythree)
Hi

https://photos.app.goo.gl/HFR5JpQRa8XWLqY99

Im stuck on 5 (i) (b)

I tried doing it the normal way but its not working like part a? Very confused. I tried to make simultaneous equations so i could get rid of theta but got stuck here:
1=(rsinalpha)(rcosalpha)
Where's your working out?
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fiftythree
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(Original post by RDKGames)
Where's your working out?
https://photos.app.goo.gl/1v8ddnmADutVo62fA
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RDKGames
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What is your justification behind saying that

\cos \theta = r \sin \theta \cos \alpha

and

\sin \theta = r\cos \theta \sin \alpha

?
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fiftythree
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(Original post by RDKGames)
What is your justification behind saying that

\cos \theta = r \sin \theta \cos \alpha

and

\sin \theta = r\cos \theta \sin \alpha

?
that was just the method i was taught im not sure
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(Original post by fiftythree)
that was just the method i was taught im not sure
A sad mistake students often do is take what their teacher says at face value, which is understandable, but it seems like you fell into the trap of not being taught this properly. It's always good to go over this topic from a different source to see if what you're doing is valid.

So you have that;

\cos \theta - \sin \theta = r\sin \theta \cos \alpha - r\cos \theta \sin \alpha.

Note that \theta is the main (and only) independent variable here. It varies. r and \alpha do not. They are fixed in place, but it's your whole job in a question like this to determine what exactly they are fixed as.

To ensure both sides are equivalent, it is sufficient to ensure that the coefficients of \sin \theta and \cos \theta are the same.

For the coefficient of \cos \theta, it is 1 on the LHS and -r\sin \alpha on the RHS.

Therefore; 1 = -r\sin \alpha.

For the coeff of \sin \theta, it is -1 on the LHS and r\cos \alpha on the RHS.

Therefore; -1 = r\cos \alpha.


It is now sufficient for you to determine what r is (via Pythagorean theorem with the two results we just got) and also what \alpha is (there are infinitely many options for alpha, but this is why you are given the strict range of 0 <\alpha<\dfrac{\pi}{2} in which there is only one value of alpha for you to have).


So have a go at finding these values.
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(Original post by RDKGames)
A sad mistake students often do is take what their teacher says at face value, which is understandable, but it seems like you fell into the trap of not being taught this properly. It's always good to go over this topic from a different source to see if what you're doing is valid.

So you have that;

\cos \theta - \sin \theta = r\sin \theta \cos \alpha - r\cos \theta \sin \alpha.

Note that \theta is the main (and only) independent variable here. It varies. r and \alpha do not. They are fixed in place, but it's your whole job in a question like this to determine what exactly they are fixed as.

To ensure both sides are equivalent, it is sufficient to ensure that the coefficients of \sin \theta and \cos \theta are the same.

For the coefficient of \cos \theta, it is 1 on the LHS and -r\sin \alpha on the RHS.

Therefore; 1 = -r\sin \alpha.

For the coeff of \sin \theta, it is -1 on the LHS and r\cos \alpha on the RHS.

Therefore; -1 = r\cos \alpha.


It is now sufficient for you to determine what r is (via Pythagorean theorem with the two results we just got) and also what \alpha is (there are infinitely many options for alpha, but this is why you are given the strict range of 0 <\alpha<\dfrac{\pi}{2} in which there is only one value of alpha for you to have).


So have a go at finding these values.
Thanks a lot for the reply.

What am i doing wrong?

https://photos.app.goo.gl/opvK5zXvEQt9AUTU9
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(Original post by fiftythree)
Thanks a lot for the reply.

What am i doing wrong?

https://photos.app.goo.gl/opvK5zXvEQt9AUTU9
A couple of things;

(a) you have solved *only* -1 = \sqrt{2}\cos \alpha whereas you also need to consider the fact that this same solution must also satisfy 1 = -\sqrt{2}\sin \alpha.

(b) are you aware that when you solve a trig equation, there are many solutions you can have? Do you understand how to find those different solutions? This is what you would need to do when solving -1 = \sqrt{2}\cos \alpha.


My advice would be to divide one equation by the other; so divide 1 = -\sqrt{2}\sin \alpha by -1 = \sqrt{2}\cos \alpha and solve the remaining equation. This division takes into account that whatever our alpha is, it must satisfy both equations because they both contribute to make up the resulting equation.

Otherwise, you would proceed from where you are right now and determine what the solution(s) are for -1=\sqrt{2}\cos \alpha in the region (0,\frac{\pi}{2}), and then check which of those also satisfy 1=-\sqrt{2}\sin \alpha
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(Original post by RDKGames)
A couple of things;

(a) you have solved *only* -1 = \sqrt{2}\cos \alpha whereas you also need to consider the fact that this same solution must also satisfy 1 = -\sqrt{2}\sin \alpha.

(b) are you aware that when you solve a trig equation, there are many solutions you can have? Do you understand how to find those different solutions? This is what you would need to do when solving -1 = \sqrt{2}\cos \alpha.


My advice would be to divide one equation by the other; so divide 1 = -\sqrt{2}\sin \alpha by -1 = \sqrt{2}\cos \alpha and solve the remaining equation. This division takes into account that whatever our alpha is, it must satisfy both equations because they both contribute to make up the resulting equation.

Otherwise, you would proceed from where you are right now and determine what the solution(s) are for -1=\sqrt{2}\cos \alpha in the region (0,\frac{\pi}{2}), and then check which of those also satisfy 1=-\sqrt{2}\sin \alpha
thanks I got the correct answer! this topic is trickier than i thought
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thanks I got the correct answer! this topic is trickier than i thought
It might be tricky but the fact that this book is incorrect makes learning a whole lot more difficult.

Perhaps you haven't noticed it, but \alpha = \dfrac{\pi}{4} is *not* the right value. In fact, this whole question is impossible because there is no way you can express \cos \theta - \sin \theta as r\sin(\theta - \alpha) if you restrict \alpha \in (0,\frac{\pi}{2}).

Notice that if alpha is in that range, then \cos \alpha > 0. And since r>0 it means that the product r\cos \alpha > 0 ... but when comparing our coefficients we said that -1 = r \cos \alpha. How can something that must be positive be equal to -1 ? It can't therefore there is no such value of \alpha that exists to make this true.

I can explain more if you want but just know this question makes no sense and you can see it for yourself here: https://www.desmos.com/calculator/nfchmdt0hx because you can vary the value of alpha between 0 and pi/2 and you will never make the purple graph line up with the original.
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(Original post by RDKGames)
A sad mistake students often do is take what their teacher says at face value, which is understandable, but it seems like you fell into the trap of not being taught this properly. It's always good to go over this topic from a different source to see if what you're doing is valid.
WHY do you always blame bad teaching? This can be taught well and the student still makes silly errors.
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WHY do you always blame bad teaching? This can be taught well and the student still makes silly errors.
It's either bad teaching, or the student makes mistakes.

They have said it's how they have been taught. A surprising statement but if I assume it to be true then it's just bad teaching hence my comment. I leave it for OP to evaluate it whether it's actually bad teaching or if it's something on their own end.
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(Original post by RDKGames)
It might be tricky but the fact that this book is incorrect makes learning a whole lot more difficult.

Perhaps you haven't noticed it, but \alpha = \dfrac{\pi}{4} is *not* the right value. In fact, this whole question is impossible because there is no way you can express \cos \theta - \sin \theta as r\sin(\theta - \alpha) if you restrict \alpha \in (0,\frac{\pi}{2}).

Notice that if alpha is in that range, then \cos \alpha > 0. And since r>0 it means that the product r\cos \alpha > 0 ... but when comparing our coefficients we said that -1 = r \cos \alpha. How can something that must be positive be equal to -1 ? It can't therefore there is no such value of \alpha that exists to make this true.

I can explain more if you want but just know this question makes no sense and you can see it for yourself here: https://www.desmos.com/calculator/nfchmdt0hx because you can vary the value of alpha between 0 and pi/2 and you will never make the purple graph line up with the original.
hmmm, but when i put alpha = 0.785 indo the desmos graph the 2 graphs overlap and the y-value at this point is 0.

what does this mean?
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(Original post by RDKGames)
It's either bad teaching, or the student makes mistakes.

They have said it's how they have been taught. A surprising statement but if I assume it to be true then it's just bad teaching hence my comment. I leave it for OP to evaluate it whether it's actually bad teaching or if it's something on their own end.
It clearly isn't 'how they've been taught' they are mixing up cos and sin - don't assume his teacher was wrong. I'm sorry to say you do often blame teachers.
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(Original post by fiftythree)
hmmm, but when i put alpha = 0.785 indo the desmos graph the 2 graphs overlap and the y-value at this point is 0.

what does this mean?
Yes you are correct to observe that they intersect at every point where y=0.

In fact, with a little bit more observation you can realise that for \alpha = \dfrac{\pi}{4}, the graph of r\sin(\theta - \alpha) is actually a reflection of the original in the x-axis.

So indeed, for this value of alpha, just change the sign for r as to make it negative. Then the two graph align and it's a valid answer, however of course the question says r>0 so we would be stepping outside its bounds.
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(Original post by Muttley79)
It clearly isn't 'how they've been taught' they are mixing up cos and sin - don't assume his teacher was wrong. I'm sorry to say you do often blame teachers.
I think the problem is that I didnt see an example like this before
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(Original post by RDKGames)
Yes you are correct to observe that they intersect at every point where y=0.

In fact, with a little bit more observation you can realise that for \alpha = \dfrac{\pi}{4}, the graph of r\sin(\theta - \alpha) is actually a reflection of the original in the x-axis.

So indeed, for this value of alpha, just change the sign for r as to make it negative. Then the two graph align and it's a valid answer, however of course the question says r>0 so we would be stepping outside its bounds.
Hi sorry but could u tell me where i went wrong here, its the same question but 5 (ii) (b)

https://photos.app.goo.gl/d4Xuch6SYYE37KKSA
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(Original post by Muttley79)
It clearly isn't 'how they've been taught' they are mixing up cos and sin - don't assume his teacher was wrong. I'm sorry to say you do often blame teachers.
No what they've done is far worse than just 'mixing up cos and sin'! They were effectively using the logic of "if 2 + 1 = 0 + 3 then 2 = 0 and 1 = 3" which really says something is fundamentally wrong in their line of thinking up to a point where I would question whether they have been taught this topic or not by a proper teacher.

I don't always blame teachers but consideringthe fact that this is a site where students come for help, then it is statistically likely that this is the cohort whose teachers are not always good. All you see of myself is my comments on these specific threads, so of course you would suppose that I blame teachers way too often. That's not the case generally, and I am lucky to have had an amazing teacher myself so don't drill it into your head that I'm out here gunning for any and all teachers.

Anyway, as I have said, I leave it up to the OP to really think about whether it's the teaching or themselves which are wrong. That being said, there is no need for you to derail this thread into the typical arguments you like to create. End of discussion.
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(Original post by fiftythree)
Hi sorry but could u tell me where i went wrong here, its the same question but 5 (ii) (b)

https://photos.app.goo.gl/d4Xuch6SYYE37KKSA
Unfortunately it again doesn't make sense.

There is no such alpha you can find to make it hold. Similar argument as above shows this.

0 < \alpha < \frac{\pi}{2} means \cos \alpha > 0. Hence we have that r\cos\alpha > 0, but comparing coefficients we get that -1 = r\cos \alpha which has no solutions.

Here's the situation similar to the above: https://www.desmos.com/calculator/tz9qiovsbo



However, if you would change the requirement from the question to have that \pi < \alpha < \dfrac{3}{2}\pi, then you can answer both questions without trouble. Want to have a go?
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(Original post by RDKGames)
No what they've done is far worse than just 'mixing up cos and sin'! They were effectively using the logic of "if 2 + 1 = 0 + 3 then 2 = 0 and 1 = 3" which really says something is fundamentally wrong in their line of thinking up to a point where I would question whether they have been taught this topic or not by a proper teacher.

I don't always blame teachers but consideringthe fact that this is a site where students come for help, then it is statistically likely that this is the cohort whose teachers are not always good. All you see of myself is my comments on these specific threads, so of course you would suppose that I blame teachers way too often. That's not the case generally, and I am lucky to have had an amazing teacher myself so don't drill it into your head that I'm out here gunning for any and all teachers.

Anyway, as I have said, I leave it up to the OP to really think about whether it's the teaching or themselves which are wrong. That being said, there is no need for you to derail this thread into the typical arguments you like to create. End of discussion.
May I suggest that you make NO comment whatsoever about teachers in future? It really would be better for a forum helper to be neutral.
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