Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Trig equations - Adding/Subtracting cycles watch

    • Thread Starter
    Offline

    3
    ReputationRep:
    I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

    If for e.g. it said solve for  Sin 2x = 0.5 in between  0 \leq x \leq 360

    As it is 2x it would be between  0 \leq 2x \leq 720 so when I get my values for 2x I add 360 right?

    What about if it was between  0 \leq x \leq  180 and for  -180 \leq x \leq 180 or even  -180 \leq x \leq 0


    and even for if it was  Sin \frac{x}{2} = 0.5
    Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
    Offline

    11
    ReputationRep:
    (Original post by IgorYakov)
    I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

    If for e.g. it said solve for  Sin 2x = 0.5 in between  0 \leq x \leq 360

    As it is 2x it would be between  0 \leq 2x \leq 720 so when I get my values for 2x I add 360 right?

    What about if it was between  0 \leq x \leq  180 and for  -180 \leq x \leq 180 or even  -180 \leq x \leq 0


    and even for if it was  Sin \frac{x}{2} = 0.5
    Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
    When you have a solution in the form a<kx<b
    a/k < x < b/k
    This is another way of solving for x. . .
    Offline

    2
    ReputationRep:
    (Original post by IgorYakov)
    I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

    If for e.g. it said solve for  Sin 2x = 0.5 in between  0 \leq x \leq 360

    As it is 2x it would be between  0 \leq 2x \leq 720 so when I get my values for 2x I add 360 right?

    What about if it was between  0 \leq x \leq  180 and for  -180 \leq x \leq 180 or even  -180 \leq x \leq 0


    and even for if it was  Sin \frac{x}{2} = 0.5
    Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
    I would always suggest that you do a quick sketch graph. That should make things quite clear.

    Personally, if I was given  sin (2x) I would change the variable so that  y = 2x and then you would solve the equation sin y = 0.5 which is more straightforward. Once you have your values for y, it is a simple matter to convert back to x.
    Offline

    11
    ReputationRep:
    Multiply the values in the range by the coefficient near x. Remember that \frac{x}{2} = \frac{1}{2} \times x so in this case coefficient is a half - -180<x<180 will become -90<x/2<90.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by joostan)
    When you have a solution in the form a<kx<b
    a/k < x < b/k
    This is another way of solving for x. . .

    (Original post by Brister)
    I would always suggest that you do a quick sketch graph. That should make things quite clear.

    Personally, if I was given  sin (2x) I would change the variable so that  y = 2x and then you would solve the equation sin y = 0.5 which is more straightforward. Once you have your values for y, it is a simple matter to convert back to x.

    (Original post by JerzyDudek)
    Multiply the values in the range by the coefficient near x. Remember that \frac{x}{2} = \frac{1}{2} \times x so in this case coefficient is a half - -180<x<180 will become -90<x/2<90.

    But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

    Solve Sin (2x) = \frac{1}{2} for 0 \leq x \leq 360

    (2x) = 30 and 150 and
    (2x) = 390 and 510 (added 360 to both values as the new range is  0 \leq 2x \leq 720

    but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

    thanks!
    Offline

    2
    ReputationRep:
    (Original post by IgorYakov)
    But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

    Solve Sin (2x) = \frac{1}{2} for 0 \leq x \leq 360

    (2x) = 30 and 150 and
    (2x) = 390 and 510 (added 360 to both values as the new range is  0 \leq 2x \leq 720

    but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

    thanks!
    Okay, since you are working with sine you will be adding/subtracting 360, because that is the period of sine. However, there are two principal solutions which in your case you correctly identify as 2x = sin^{-1}(0.5) = 30 and 2x = 180 - 30 = 150. The symmetry of the graph of sine will give you this easily.

    If you have to deal with ranges other than 0 &lt; x &lt; 360 then you just keep adding/subtracting 360 until you can no longer find more solutions in that range.

    If you are working with sin (\frac{x}{2}), then you can always write \frac{x}{2} = y. Otherwise you can modify your graph and range accordingly.
    Offline

    11
    ReputationRep:
    (Original post by IgorYakov)
    But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

    Solve Sin (2x) = \frac{1}{2} for 0 \leq x \leq 360

    (2x) = 30 and 150 and
    (2x) = 390 and 510 (added 360 to both values as the new range is  0 \leq 2x \leq 720

    but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

    thanks!
    Not necessarily multiples of 180. Just follow what I said - multiply the range by a number near x. Say you've got a range of 0<x<360 and you have to solve an equation which has 2x in brackets (e.g. sin(2x), cos(2x)), this means you have to double x and thus you double 0 and 360 as well: 0<2x<720. Now to check you can divide the whole inequality by 2 and you'll get you original range.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Brister)
    Okay, since you are working with sine you will be adding/subtracting 360, because that is the period of sine. However, there are two principal solutions which in your case you correctly identify as 2x = sin^{-1}(0.5) = 30 and 2x = 180 - 30 = 150. The symmetry of the graph of sine will give you this easily.

    If you have to deal with ranges other than 0 &lt; x &lt; 360 then you just keep adding/subtracting 360 until you can no longer find more solutions in that range.

    If you are working with sin (\frac{x}{2}), then you can always write \frac{x}{2} = y. Otherwise you can modify your graph and range accordingly.
    Ohh okay, so I only ever add or subtract 360, not 180 or 90 or anything else? Or will I need 180 for tan?
    Offline

    2
    ReputationRep:
    (Original post by IgorYakov)
    Ohh okay, so I only ever add or subtract 360, not 180 or 90 or anything else? Or will I need 180 for tan?
    You will indeed need 180 for tan.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Brister)
    You will indeed need 180 for tan.


    Okay! Thanks
    just to make sure I'm clear on this, if it was 2x for Sin or Cos I would add/subtract 360 so they're inside the range
    but if it was 2x for tan I would +/- 180?
    Offline

    2
    ReputationRep:
    (Original post by IgorYakov)
    Okay! Thanks
    just to make sure I'm clear on this, if it was 2x for Sin or Cos I would add/subtract 360 so they're inside the range
    but if it was 2x for tan I would +/- 180?
    Yes, that is correct.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Brister)
    Yes, that is correct.

    Thank you so much!
    Offline

    12
    ReputationRep:
    (Original post by Brister)
    Yes, that is correct.
    Hi, but if the range given was e.g.  -180 \leq \theta \leq 180 would you +/- 360 even for Tan?
    Offline

    2
    ReputationRep:
    (Original post by Secret.)
    Hi, but if the range given was e.g.  -180 \leq \theta \leq 180 would you +/- 360 even for Tan?
    No dude, changing the range of \theta won't change the behavior of tan(\theta). The period of the tangent function is still 180. The only difference is that you may have less values for \theta which solve the particular equation.
    Offline

    12
    ReputationRep:
    (Original post by Brister)
    No dude, changing the range of \theta won't change the behavior of tan(\theta). The period of the tangent function is still 180. The only difference is that you may have less values for \theta which solve the particular equation.
    Okk thanks!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 29, 2013
Poll
Do you like carrot cake?
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

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

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