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MAT multiple choice (easy question issue?)

Can someone give me some advice as to how to tackle this multiple choice question please? Thanks :smile:
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Reply 1
Avatar for RichE
RichE
15
Original post by iMacJack
Can someone give me some advice as to how to tackle this multiple choice question please? Thanks :smile:


Firstly you might consider finding all those u such that cos(u)=1/2.
For what values of x is cos(x)=1/2?
Original post by iMacJack
Can someone give me some advice as to how to tackle this multiple choice question please? Thanks :smile:


Workout inverse cosine of 1/2, then see if sin(x) satisfies that result(s).
Reply 4
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iMacJack
OP
16
Original post by rayquaza17
For what values of x is cos(x)=1/2?


60 degrees edit: didnt realise you wanted all angles, 60 and 300, between 0 and 2pi

Original post by RDKGames
Workout inverse cosine of 1/2, then see if sin(x) satisfies that result(s).


60 degrees, what do you mean?
(edited 7 years ago)
Original post by iMacJack
Can someone give me some advice as to how to tackle this multiple choice question please? Thanks :smile:


I have a feeling that you might be overthinking this one or I am under thinking it XD.

Take arccos of both sides then try taking arcsin of the result. In your attempt to try you should find something rather interesting and foreboding.

Crap @RDKGames too fast 4 me ;-;
(edited 7 years ago)
Original post by Enigmatically
I have a feeling that you might be overthinking this one or I am under thinking it XD.

Take arccos of both sides then try taking arcsin of the result. In your attempt to try you should find something rather interesting and foreboding.

Crap @RDKGames too fast 4 me ;-;


You can't 'take arccos of both sides' (we do not necessarily have arccos⁡(cos⁡(sin⁡x))=sin⁡x\arccos(\cos(\sin x))=\sin x). I think you're slightly 'under thinking' it.
(edited 7 years ago)
Reply 7
Avatar for iMacJack
iMacJack
OP
16
Original post by Enigmatically
I have a feeling that you might be overthinking this one or I am under thinking it XD.

Take arccos of both sides then try taking arcsin of the result. In your attempt to try you should find something rather interesting and foreboding.

Crap @RDKGames too fast 4 me ;-;


Overthinking it on my behalf probably! I am still a bit confused, the cos(sinx) is the bit which confuses me, obviously the inverse of cos(x) = 1/2 is 60 degrees (and 300 degrees from 0-360 degrees) I don't get the sinx part, though
Original post by iMacJack
Overthinking it on my behalf probably! I am still a bit confused, the cos(sinx) is the bit which confuses me, obviously the inverse of cos(x) = 1/2 is 60 degrees (and 300 degrees from 0-360 degrees) I don't get the sinx part, though


I think an easier way to see the solution to this problem (and quickly) is to note that −1≤sin⁡x≤1⇒cos⁡1≤cos⁡(sin⁡x)≤1-1 \leq \sin x \leq 1 \Rightarrow \cos1 \leq \cos(\sin x) \leq 1 and from π>3⇒π3>1⇒cos⁡π3=12<cos⁡1\pi > 3 \Rightarrow \frac{\pi}{3} > 1 \Rightarrow \cos\frac{\pi}{3}=\frac{1}{2} < \cos1 and the result is clear (even easier to see diagrammatically).
Original post by IrrationalRoot
You can't 'take arccos of both sides' (we do not necessarily have arccos⁡(cos⁡(sin⁡x))=sin⁡x\arccos(\cos(\sin x))=\sin x). I think you're slightly 'under thinking' it.


I apologise to all readers of the thread if I am out of line on this one. By taking arccos of both sides I merely implied removing cos from the left hand side and fimding arccos of the other. I may be wrong, but you might be as well? There is an equal sign in this equation, and I believe from whatever meagre maths I may have learnt that whatever you do to the left, do to the right. I think that moving from cos(sin(x)) = 1/2 to sin(x) = arccos(1/2) is somewhat similar to moving from (5x^2)(y)=5y to x^2 =1. Just because we don't often write ((5x^2)(y))/y=5y/y doesn't mean that it doesn't exist as a step in our working towards the solution.

I wholly understand that with trigonometry arccos and cos have different ranges that come into play but aren't your claims an example of 'slightly under thinking it?' :wink:

Not a shot or an attempt to start a flame war, I would just prefer if you provided concrete proof of your claims, as I would also prefer to learn the right way if I am in the dark.

Original post by iMacJack
Overthinking it on my behalf probably! I am still a bit confused, the cos(sinx) is the bit which confuses me, obviously the inverse of cos(x) = 1/2 is 60 degrees (and 300 degrees from 0-360 degrees) I don't get the sinx part, though


sin(x) has to have a value in between 1 and -1 inclusive. arccos(1/2) = pi/3 and pi/3 is greater than 1, so the equation has no solution.
(edited 7 years ago)
Reply 10
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iMacJack
OP
16
All the confusion on the thread has confused the life out of me

So it's just sinx = 60, pi/3 is greater than one so no solutions?


Posted from TSR Mobile
Original post by Enigmatically
I apologise to all readers of the thread if I am out of line on this one. By taking arccos of both sides I merely implied removing cos from the left hand side and fimding arccos of the other. I may be wrong, but you might be as well? There is an equal sign in this equation, and I believe from whatever meagre maths I may have learnt that whatever you do to the left, do to the right. I think that moving from cos(sin(x)) = 1/2 to sin(x) = arccos(1/2) is somewhat similar to moving from (5x^2)(y)=5y to x^2 =1. Just because we don't often write ((5x^2)(y))/y=5y/y doesn't mean that it doesn't exist as a step in our working towards the solution.

I wholly understand that with trigonometry arccos and cos have different ranges that come into play but aren't your claims an example of 'slightly under thinking it?' :wink:

Not a shot or an attempt to start a flame war, I would just prefer if you provided concrete proof of your claims, as I would also prefer to learn the right way if I am in the dark.

Yes whatever you do to the left do to the right. But you can't just remove cos from the left hand side and find arccos of the right. That is not doing the same thing to both sides."cos(sin(x)) = 1/2 to sin(x) = arccos(1/2)" simply isn't valid; it could be wrong depending on the value of sin(x).

Notice how if sin(x) is negative then on the right you have a nonnegative quantity even though you have a negative quantity on the left. That's a 'proof' that you've done something wrong.

This trig equation (and many others) are almost never as simple as just removing cos from one side and putting arccos on the other. Reason being that cos does not technically have an inverse.
Original post by iMacJack
All the confusion on the thread has confused the life out of me

So it's just sinx = 60, pi/3 is greater than one so no solutions?


Posted from TSR Mobile


Yeah. You asked for hints so I didn't want to drop a complete solution and shag the question up for you. BTW, we are working in radians so I think you should change that 60 into radians XD.
Original post by Enigmatically
I think that moving from cos(sin(x)) = 1/2 to sin(x) = arccos(1/2) is somewhat similar to moving from (5x^2)(y)=5y to x^2 =1.


Oh also this happens to be an invalid step too. I get what you were trying to show but I must point out that you get x2=1x^2=1 or y=0y=0.
Reply 14
Avatar for Zacken
Zacken
22
Original post by iMacJack
All the confusion on the thread has confused the life out of me

So it's just sinx = 60, pi/3 is greater than one so no solutions?


Posted from TSR Mobile


I like IrrationalRoot's answer a lot, but to clarify what the others were saying, making the substition θ=sin⁡x\theta = \sin x then you can focus on cos⁡θ=12\cos \theta = \frac{1}{2} which gives you θ=π3,5π3,…\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \ldots.

So you've reduced it to solving sin⁡x=π3,5π3\sin x = \frac{\pi}{3}, \frac{5\pi}{3}.

Now you know that sin⁡x=k\sin x = k is nonsense (in the reals) when k>1k > 1 (or < -1) which is precisely the case here.
Original post by IrrationalRoot
Oh also this happens to be an invalid step too. I get what you were trying to show but I must point out that you get x2=1x^2=1 or y=0y=0.


Sorry about the prior piss up I caused. I overlooked the trivial fact that arccos is not necessarily the inverse of cos. My teacher insisted that I use cos^-1 instead of arccos, thus sometimes I am under the illusion that they are inverses. I am aware that arccos just finds angles in the range -pi/2 to pi/2 inclusive that give a specific value of cos(x), and not all the angles that may satisfy this equation or condition, thus in that regard I think it further solidifies your conclusion that we cannot take it as an inverse as it still leaves the condition partially completed. Haven't done much maths at all this summer and my Backhouse is escaping me. Guess I have to polish up. Thank you very much for refreshing my memory.
(edited 7 years ago)
Reply 16
Avatar for Zacken
Zacken
22
Original post by Enigmatically
Sorry about the prior piss up I caused. I overlooked the trivial fact that arccos is not necessarily the inverse of cos. My teacher insisted that I use cos^-1 instead of arccos, thus sometimes I am under the illusion that they are unversed.I am aware of this fact, just haven't done much maths at all this summer and my Backhouse is escaping me. Guess I have to polish up. Thank you very much for the aid.


Props to you for being so gracious about it, it's nice to see this sort of discussion; true spirit of mathematics and all.
Original post by Enigmatically
Sorry about the prior piss up I caused. I overlooked the trivial fact that arccos is not necessarily the inverse of cos. My teacher insisted that I use cos^-1 instead of arccos, thus sometimes I am under the illusion that they are unversed.I am aware of this fact, just haven't done much maths at all this summer and my Backhouse is escaping me. Guess I have to polish up. Thank you very much for the aid.


Lol no problem. That's why I think that arccos⁡\arccos is better notation than cos⁡−1\cos^{-1}; the latter can be very misleading.
Original post by Zacken
Props to you for being so gracious about it, it's nice to see this sort of discussion; true spirit of mathematics and all.


Thanks. Maths can vary wildly in methodology as you never know if there may be another way to go about a question that has not yet been discovered, but atleast it is objective in the sense that irrespective of the method you get to the same answer. So I try to keep my mind open to see if there is something new I can learn, because none of us knows it all. Nonetheless, I try to discern whatever I am told and take it with spoonfuls of salt occasionally, but you have to know where you screw up so you can fix it.
(edited 7 years ago)

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