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Maths C3 - Trigonometry... Help??

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Original post by Philip-flop
I thought that if sin⁡x=0\sin x=0 then xx can equal either −2π,−π,0,π,2π -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0

No, I know that sinsin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain −π2≤x≤π2 -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x)sin(x) gets "inverted" to create the function arcsinarcsin

That is all correct.
Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down :frown:
C3 - EXE 6E.png

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(edited 7 years ago)
C3 - EXE 6E.png15.5KB
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Original post by Philip-flop
Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down :frown:
C3 - EXE 6E.png



Yes that is correct.

Alternatively just write:

sin⁡(x)=k⇒x=arcsin⁡(k)=α\sin(x)=k \Rightarrow x=\arcsin(k)=\alpha
and
x=π−αx=\pi-\alpha
(edited 7 years ago)
Original post by RDKGames
But 0 is a multiply of π\pi so it's not wrong, it's just a principal solution, and it doesn't have to be an even multiple. It would be a mess to lead OP to general solutions of sine at this point.


Idk why I said even, sorry (but this was obviously a mistake...).
And no, what I'm saying is that I could take the number x=2πx=2\pi and give you the information sin⁡x=0\sin x=0, in which case x=0x=0 is wrong.. This was the point.
It certainly wouldn't be 'a mess to lead OP to general solutions'. I'm not even teaching them general solutions. I was giving an example to illustrate my point, which it did.
Original post by Philip-flop
I thought that if sin⁡x=0\sin x=0 then xx can equal either −2π,−π,0,π,2π -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0



No, I know that sinsin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain −π2≤x≤π2 -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x)sin(x) gets "inverted" to create the function arcsinarcsin


Then it doesn't make sense why you don't understand arcsin⁡(sin⁡x)≢x\arcsin(\sin x) \not\equiv x. This is a direct consequence of what you've just said.
Original post by IrrationalRoot
Idk why I said even, sorry (but this was obviously a mistake...).
And no, what I'm saying is that I could take the number x=2πx=2\pi and give you the information sin⁡x=0\sin x=0, in which case x=0x=0 is wrong.. This was the point.
It certainly wouldn't be 'a mess to lead OP to general solutions'. I'm not even teaching them general solutions. I was giving an example to illustrate my point, which it did.


Ah okay. I do think it would be a mess based on his understanding, and decided to mention it as your point touches upon them in principle with multiples.
1) (cosec2o)^2-(cot2o)^2

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Original post by RDKGames
Ah okay. I do think it would be a mess based on his understanding, and decided to mention it as your point touches upon them in principle with multiples.


I think OP can understand multiples... Anyone who has any knowledge of trig functions will understand sin⁡2kπ=0\sin 2k\pi=0, so this wouldn't confuse OP. It serves as a good example of the fact that sin⁡\sin is many-to-one. If I just say it's 'many-to-one' with no explanation then that would certainly be more confusing and would likely lead to 'a mess' as you put it.
Original post by Philip-flop
I thought that if sin⁡x=0\sin x=0 then xx can equal either −2π,−π,0,π,2π -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0



No, I know that sinsin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain −π2≤x≤π2 -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x)sin(x) gets "inverted" to create the function arcsinarcsin


I must also point out that this: "for it to be a function it must be a one-to-one"
is simply wrong. There seems to still be a misunderstanding here. Unless you're talking about the inverse when you say 'it' in which case ok, but see my previous reply.
Also not sure why so much use of emboldening...
(edited 7 years ago)
Original post by Philip-flop
Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down :frown:
C3 - EXE 6E.png

Attachment not found


I know this is completely off topic, but I love your pi. Could you do a tutorial?:colondollar:
Original post by IrrationalRoot
Then it doesn't make sense why you don't understand arcsin⁡(sin⁡x)≢x\arcsin(\sin x) \not\equiv x. This is a direct consequence of what you've just said.

I'm just confused by the equation. I'm sure it's right but I can't seem to understand the layout of it. Sorry :frown:

Original post by IrrationalRoot
I think OP can understand multiples... Anyone who has any knowledge of trig functions will understand sin⁡2kπ=0\sin 2k\pi=0, so this wouldn't confuse OP. It serves as a good example of the fact that sin⁡\sin is many-to-one. If I just say it's 'many-to-one' with no explanation then that would certainly be more confusing and would likely lead to 'a mess' as you put it.

My knowledge of trig functions is, lets just say limited. I have no idea what you mean by the part where you say sin⁡2kπ=0\sin 2k\pi=0... Where has that K come from??

I know that sin2π=0 sin2 \pi = 0 though

Original post by IrrationalRoot
I must also point out that this: "for it to be a function it must be a one-to-one"
is simply wrong. There seems to still be a misunderstanding here. Unless you're talking about the inverse when you say 'it' in which case ok, but see my previous reply.
Also not sure why so much use of emboldening...

How come you only quoted half of my sentence?

Let me edit a section of what I said so it is a little clearer...
sin doesn't have an exact inverse as for it (arcsin) to be a function it must be a one-to-one.
(edited 7 years ago)
Original post by asinghj
I know this is completely off topic, but I love your pi. Could you do a tutorial?:colondollar:

Ah schuckkkks :colondollar::colondollar:

Plenty more pi to go around :wink:
LOOL
sin⁡2kπ=0 \sin 2k\pi = 0 just means for every even multiple of π \pi sin will equal 0

i hate latex
(edited 7 years ago)
Original post by Philip-flop
I'm just confused by the equation. I'm sure it's right but I can't seem to understand the layout of it. Sorry :frown:


My knowledge of trig functions is, lets just say limited. I have no idea what you mean by the part where you say sin⁡2kπ=0\sin 2k\pi=0... Where has that K come from??

I know that sin2π=0 sin2 \pi = 0 though


How come you only quoted half of my sentence?

Let me edit a section of what I said so it is a little clearer...
sin doesn't have an exact inverse as for it (arcsin) to be a function it must be a one-to-one.


kk just stands for any integer. So by sin⁡2kπ=0\sin 2k\pi=0 all I'm saying is that sin⁡0=sin⁡2π=sin⁡(−2π)=sin⁡4π=sin⁡(−4π)=...=0\sin0=\sin2\pi=\sin (-2\pi)= \sin 4\pi=\sin (-4\pi)=...=0 since all these angles are full turns of a circle and will thus have the same sines.

I mentioned "Unless you're talking about the inverse when you say 'it' in which case ok." I did acknowledge what you said.
(edited 7 years ago)
Original post by IrrationalRoot
kk just stands for any integer. So by sin⁡2kπ=0\sin 2k\pi=0 all I'm saying is that sin⁡0=sin⁡2π=sin⁡(−2π)=sin⁡4π=sin⁡(−4π)=...=0\sin0=\sin2\pi=\sin (-2\pi)= \sin 4\pi=\sin (-4\pi)=...=0 since all these angles are full turns of a circle and will thus have the same sines.

I mentioned "Unless you're talking about the inverse when you say 'it' in which case ok." I did acknowledge what you said.

Oh right I see. Yeah I get you now. Learn something everyday :smile:

Thanks for being patient with me btw. I really appreciate it!!
Original post by Philip-flop
Oh right I see. Yeah I get you now. Learn something everyday :smile:

Thanks for being patient with me btw. I really appreciate it!!


Good to hear :smile:. Tbh I should be thanking you for being patient with me lol, the number of times today I've had to edit things I've said due to mistakes etc. is ridiculous...
Original post by IrrationalRoot
Good to hear :smile:. Tbh I should be thanking you for being patient with me lol, the number of times today I've had to edit things I've said due to mistakes etc. is ridiculous...


Yes it is ridiculous. Especially for someone like you, after having done so brilliantly in STEP. You should be ashamed of yourself
(edited 7 years ago)
So I've just started Chapter 7 of the Edexcel C3 Modular Maths textbook and have already encountered a problem :frown:

It's about proving the 'addition formula' for cos(A-B) = cosAcosB + sinAsinB
Proving addition formulae (compound angles).png

Ok so I have the following questions to ask about this example...

1.

Why are the coordinates of P and Q... P(cosA, sinA) and Q(cosB, sinB)?

2.

Why is the radius of the circle 1?

3.

Why is the angle POQ equal to (A-B)?? Surely it would be angle POQ = (B-A), is it because using the addition formulae cos(a-b) = cos(b-a)??

4.

How would one know to compare the lengths of PQ^2 from the results of using the cosine rule and the distance between two coordinates for this type of question?

Original post by Philip-flop
So I've just started Chapter 7 of the Edexcel C3 Modular Maths textbook and have already encountered a problem :frown:

It's about proving the 'addition formula' for cos(A-B) = cosAcosB + sinAsinB
Proving addition formulae (compound angles).png

Ok so I have the following questions to ask about this example...

1.

Why are the coordinates of P and Q... P(cosA, sinA) and Q(cosB, sinB)?

2.

Why is the radius of the circle 1?

3.

Why is the angle POQ equal to (A-B)?? Surely it would be angle POQ = (B-A), is it because using the addition formulae cos(a-b) = cos(b-a)??

4.

How would one know to compare the lengths of PQ^2 from the results of using the cosine rule and the distance between two coordinates for this type of question?



1) for P, how far away is it from the center? You need the distance to N for the x coordinate, hence cos(A), Just remember SOHCAHTOA. For the y coordinate, you need the distance to P, which will be sin(A).

2) Same thing for Q.

3) Radius of the circle is 1 either because the question will mention something about the radius, or it simplifies the problem.

4) Idk why it's A-B

5) You have to show that cos(A-B) is cosAcosB + sinAsinB, and from the length of PQ^2 you get an answer and from cosine rule you get another answer, they must be the same, because they both in the form PQ^2 =

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