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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:

1−cos2x/1+cos2x=sec2x−11-cos2x/1+cos2x = sec^2x-1

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

It seems I have to use the fact that cos2xcos2x is 1−2sin2x1-2sin^2x and 2cos2x−12cos^2x-1 but how are you supposed to know which ones to use?
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.
Reply 2
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the bear
20
the RHS looks familiar... it based on tan2 + 1 = sec2
Reply 3
Avatar for the bear
the bear
20
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...
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by the bear
the rhs looks familiar... It based on tan2 + 1 = sec2


^^^^^^^

this
Reply 5
Avatar for Slowbro93
Slowbro93
20
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=sec2x−11-cos2x/1+cos2x = sec^2x-1

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

It seems I have to use the fact that cos2xcos2x is 1−2sin2x1-2sin^2x and 2cos2x−12cos^2x-1 but how are you supposed to know which ones to use?


What are you asking?

Is it: (1−cos2x)/(1+cos2x)=sec2x−1(1-cos2x)/(1+cos2x) = sec^2x-1

or

1−(cos2x)/(1+cos2x)=sec2x−11-(cos2x)/(1+cos2x) = sec^2x-1
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:
Reply 7
Avatar for Nayom1
Nayom1
2
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=sec2x−11-cos2x/1+cos2x = sec^2x-1

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

It seems I have to use the fact that cos2xcos2x is 1−2sin2x1-2sin^2x and 2cos2x−12cos^2x-1 but how are you supposed to know which ones to use?

Edit: you already worked that out :colondollar:
(edited 12 years ago)
Reply 8
Avatar for Crowbar
Crowbar
OP
5
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.
Reply 9
Avatar for TenOfThem
TenOfThem
18
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
Reply 10
Avatar for wibletg
wibletg
3
Well, all I know is there's got to be an easier way than I did it :redface:

But it worked... Difference of 2 squares method?
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=sec2x−11-cos2x/1+cos2x = sec^2x-1

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

It seems I have to use the fact that cos2xcos2x is 1−2sin2x1-2sin^2x and 2cos2x−12cos^2x-1 but how are you supposed to know which ones to use?


well i would put that sec2x=1cos2x sec^2x = \frac{1}{cos^2x}

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

LHS =1−(2cos2x−1)1+(2cos2x−1) = \frac{1-(2cos^2x-1)}{1+(2cos^2x-1)}

=2−2cos2x2cos2x = \frac{2-2cos^2x}{2cos^2x}

=22cos2x−2cos2x2cos2x=1cos2x−1=sec2x−1 = \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 :smile: 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 :smile: -- 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 sec2x−1=tan2x sec^2x - 1 = tan^2x

so you want to turn the LHS into tan2x tan^2x

Now you know you can turn cos2x cos2x into sin2xsin^2x's or cos2xcos^2x's, so as tan2x=sin2xcos2xtan^2x=\frac{sin^2x}{cos^2x} try turning the cos2xcos2x on the numerator into sin, and the cos2xcos2x on the denominator into cos :smile:

LHS = 1−(1−sin2x)1+(cos2x+1) \frac{1-(1-sin^2x)}{1+(cos^2x+1)}

= sin2xcos2x=tan2x=sec2x−1\frac{sin^2x}{cos^2x} = tan^2x = sec^2x-1 as required
(edited 12 years ago)
My approach here (if I did not see anything obvious and was wanting to just start something) would be

1−cos2x1+cos2x=1−cos2x+sin2x1+cos2x−sin2x\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


1−cos2x+sin2x1+cos2x−sin2x=1−cos2x+sin2x1+cos2x−(1−cos2x)=1−cos2x+sin2x2cos2x\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


12cos2x−cos2x2cos2x+sin2x2cos2x=12sec2x−12+12tan2x\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

12[sec2x−1+(sec2x−1)]=12[2sec2x−2]=sec2x−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
(edited 12 years ago)

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