Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    15
    ReputationRep:
    Can someone give me some advice as to how to tackle this multiple choice question please? Thanks
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by iMacJack)
    Can someone give me some advice as to how to tackle this multiple choice question please? Thanks
    Firstly you might consider finding all those u such that cos(u)=1/2.
    • Community Assistant
    Offline

    17
    ReputationRep:
    Community Assistant
    For what values of x is cos(x)=1/2?
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by iMacJack)
    Can someone give me some advice as to how to tackle this multiple choice question please? Thanks
    Workout inverse cosine of 1/2, then see if sin(x) satisfies that result(s).
    • Thread Starter
    Offline

    15
    ReputationRep:
    (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?
    Offline

    8
    ReputationRep:
    (Original post by iMacJack)
    Can someone give me some advice as to how to tackle this multiple choice question please? Thanks
    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 ;-;
    Offline

    17
    ReputationRep:
    (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). I think you're slightly 'under thinking' it.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (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
    Offline

    17
    ReputationRep:
    (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 \leq \sin x \leq 1 \Rightarrow \cos1 \leq \cos(\sin x) \leq 1 and from \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).
    Offline

    8
    ReputationRep:
    (Original post by IrrationalRoot)
    You can't 'take arccos of both sides' (we do not necessarily have \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?'

    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.
    • Thread Starter
    Offline

    15
    ReputationRep:
    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
    Offline

    17
    ReputationRep:
    (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?'

    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.
    Offline

    8
    ReputationRep:
    (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.
    Offline

    17
    ReputationRep:
    (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 x^2=1 or y=0.
    Offline

    22
    ReputationRep:
    (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 \theta = \sin x then you can focus on \cos \theta = \frac{1}{2} which gives you \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \ldots.

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

    Now you know that \sin x = k is nonsense (in the reals) when k &gt; 1 (or < -1) which is precisely the case here.
    Offline

    8
    ReputationRep:
    (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 x^2=1 or y=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.
    Offline

    22
    ReputationRep:
    (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.
    Offline

    17
    ReputationRep:
    (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 is better notation than \cos^{-1}; the latter can be very misleading.
    Offline

    8
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.