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Trig C3

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

    this
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    (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
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    (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:
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    (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
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    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.
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    (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
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    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?
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    (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
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    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

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Updated: April 10, 2012
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