Hey there Sign in to join this conversationNew here? Join for free

Trig C3

Announcements Posted on
Study Help needs new mods! 14-04-2014
We're up for a Webby! Vote TSR to help us win. 10-04-2014
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    This has to be my most hated part of a-level maths I've come across. The questions where it asks you to show that one equation is the same as the other equation, it just seems like a load of guesswork to get it to match. I have an example here:

    1-cos2x/1+cos2x = sec^2x-1

    I wouldn't know where to start unless I look at the solution bank for a hint.

    It seems I have to use the fact that cos2x is 1-2sin^2x and 2cos^2x-1 but how are you supposed to know which ones to use?
    • 38 followers
    Offline

    ReputationRep:
    It is easiest to make the denominator as simple as possible as you can then split up the fraction more easily.

    Therefore you should use the second option.
    • 33 followers
    Offline

    ReputationRep:
    the RHS looks familiar... it based on tan2 + 1 = sec2
    • 33 followers
    Offline

    ReputationRep:
    (Original post by marcusmerehay)
    It is easiest to make the denominator as simple as possible as you can then split up the fraction more easily.

    Therefore you should use the second option.
    you don't have to use the same option if Cos2x appears in different places...
    • 40 followers
    Offline

    ReputationRep:
    (Original post by the bear)
    the rhs looks familiar... It based on tan2 + 1 = sec2
    ^^^^^^^

    this
    • 81 followers
    Offline

    ReputationRep:
    (Original post by Crowbar)
    This has to be my most hated part of a-level maths I've come across. The questions where it asks you to show that one equation is the same as the other equation, it just seems like a load of guesswork to get it to match. I have an example here:

    1-cos2x/1+cos2x = sec^2x-1

    I wouldn't know where to start unless I look at the solution bank for a hint.

    It seems I have to use the fact that cos2x is 1-2sin^2x and 2cos^2x-1 but how are you supposed to know which ones to use?
    What are you asking?

    Is it: (1-cos2x)/(1+cos2x) = sec^2x-1

    or

    1-(cos2x)/(1+cos2x) = sec^2x-1
    • 38 followers
    Offline

    ReputationRep:
    (Original post by the bear)
    you don't have to use the same option if Cos2x appears in different places...
    Of course, although using the second option is my suggestion purely for the denominator - I may not have made that clear in my previous post. :yy:
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Crowbar)
    This has to be my most hated part of a-level maths I've come across. The questions where it asks you to show that one equation is the same as the other equation, it just seems like a load of guesswork to get it to match. I have an example here:

    1-cos2x/1+cos2x = sec^2x-1

    I wouldn't know where to start unless I look at the solution bank for a hint.

    It seems I have to use the fact that cos2x is 1-2sin^2x and 2cos^2x-1 but how are you supposed to know which ones to use?
    Edit: you already worked that out
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    I'm just wondering how do you approach questions like this one? Do you have to think of backwards steps from the simplest side or something?

    Nayom1 I had to use the solutions bank because I didn't know what to do first.
    • 40 followers
    Offline

    ReputationRep:
    (Original post by Crowbar)
    I'm just wondering how do you approach questions like this one? Do you have to think of backwards steps from the simplest side or something?
    There are various methods, in order I use

    1. Hope something obvious jumps out
    2. Change everything on the LHS to sin(x) and cos(x) and look for opportunities to cancel
    3. Mess about with the RHS a bit to see if it looks nicer


    ALWAYS looking at the answer for clues
    • 15 followers
    Offline

    ReputationRep:
    Well, all I know is there's got to be an easier way than I did it

    But it worked... Difference of 2 squares method?
    • 1 follower
    Offline

    ReputationRep:
    (Original post by Crowbar)
    This has to be my most hated part of a-level maths I've come across. The questions where it asks you to show that one equation is the same as the other equation, it just seems like a load of guesswork to get it to match. I have an example here:

    1-cos2x/1+cos2x = sec^2x-1

    I wouldn't know where to start unless I look at the solution bank for a hint.

    It seems I have to use the fact that cos2x is 1-2sin^2x and 2cos^2x-1 but how are you supposed to know which ones to use?
    well i would put that  sec^2x = \frac{1}{cos^2x}

    so lets use the  cos2x=2cos^2x-1 identity (as we want to turn the LHS into  cos^2x's because thats whats on the RHS)

    LHS  = \frac{1-(2cos^2x-1)}{1+(2cos^2x-1)}

     = \frac{2-2cos^2x}{2cos^2x}

     = \frac{2}{2cos^2x} - \frac{2cos^2x}{2cos^2x} = \frac{1}{cos^2x} - 1 = sec^2x - 1 as required

    Its not fool proof, but the fact is all the identities will get you there just by different routes generally just try to compare the two sides of the equation, see what you might want to change using identities and then give it a go. Just try chucking something at it and see what happens if your not sure -- personally i like to change secs and cosecs and cots etc into their cos/sin/tan counterparts, because im happier using the identities to convert between them, but thats a personal preference.

    For example instead you might spot that  sec^2x - 1 = tan^2x

    so you want to turn the LHS into  tan^2x

    Now you know you can turn  cos2x into sin^2x's or cos^2x's, so as tan^2x=\frac{sin^2x}{cos^2x} try turning the cos2x on the numerator into sin, and the cos2x on the denominator into cos

    LHS =  \frac{1-(1-sin^2x)}{1+(cos^2x+1)}

    = \frac{sin^2x}{cos^2x} = tan^2x = sec^2x-1 as required
    • 40 followers
    Offline

    ReputationRep:
    My approach here (if I did not see anything obvious and was wanting to just start something) would be

    \frac{1-cos2x}{1+cos2x} = \frac{1-cos^2x+sin^2x}{1+cos^2x-sin^2x}

    hmmmm ... the RHS has sec so I will try to get cos in the denominator


    \frac{1-cos^2x+sin^2x}{1+cos^2x-sin^2x} = \frac{1-cos^2x+sin^2x}{1+cos^2x-(1-cos^2x)} = \frac{1-cos^2x+sin^2x}{2cos^2x}


    hmmmmm ... what happens if I split up the fraction ... will that help


    \frac{1}{2cos^2x} - \frac{cos^2x}{2cos^2x} + \frac{sin^2x}{2cos^2x} = \frac{1}{2}sec^2x - \frac{1}{2} + \frac{1}{2}tan^2x


    hmmmmmm what if I take out the 1/2 and ... hang on I can use tan^2 = sec^2 -1

    \frac{1}{2}[sec^2x - 1 + (sec^2x -1)] = \frac{1}{2}[2sec^2x - 2] = sec^2x - 1


    This is, in no way the quickest method but it uses the general principle of changing to sin and cos and looking to the RHS for clues

    HTH

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By completing the slider below you agree to The Student Room's terms & conditions and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

    You don't slide that way? No problem.

Updated: April 10, 2012
Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.