Trig C3
Maths and statistics discussion, revision, exam and homework help.
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Trig C3
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:

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
is
and
but how are you supposed to know which ones to use?
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Re: Trig C3you don't have to use the same option if Cos2x appears in different places...(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. -
Re: Trig C3^^^^^^^(Original post by the bear)
the rhs looks familiar... It based on tan2 + 1 = sec2
this -
Re: Trig C3What are you asking?(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:

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
is
and
but how are you supposed to know which ones to use?
Is it:
or
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Re: Trig C3Of course, although using the second option is my suggestion purely for the denominator - I may not have made that clear in my previous post.(Original post by the bear)
you don't have to use the same option if Cos2x appears in different places...
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Re: Trig C3Edit: you already worked that out(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:

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
is
and
but how are you supposed to know which ones to use?
Last edited by Nayom1; 10-04-2012 at 12:25. -
Re: Trig C3There are various methods, in order I use(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?
- Hope something obvious jumps out
- Change everything on the LHS to sin(x) and cos(x) and look for opportunities to cancel
- Mess about with the RHS a bit to see if it looks nicer
ALWAYS looking at the answer for clues -
Re: Trig C3well i would put that(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:

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
is
and
but how are you supposed to know which ones to use?

so lets use the
identity (as we want to turn the LHS into
's because thats whats on the RHS)
LHS

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
so you want to turn the LHS into
Now you know you can turn
into
's or
's, so as
try turning the
on the numerator into sin, and the
on the denominator into cos 
LHS =
=
as required
Last edited by just george; 10-04-2012 at 12:44. -
Re: Trig C3
My approach here (if I did not see anything obvious and was wanting to just start something) would be

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

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

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 \frac{1}{2}[sec^2x - 1 + (sec^2x -1)] = \frac{1}{2}[2sec^2x - 2] = sec^2x - 1](http://www.thestudentroom.co.uk/latexrender/pictures/a5/a566e66c5113d97d764eebf540a76884.png)
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
HTHLast edited by TenOfThem; 10-04-2012 at 12:45.


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