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Maths C3 - Trigonometry... Help??

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Original post by Philip-flop
Oh really? Can you explain. I'm not sure I see how the other identities are derived from that. I see them all as separate identities which is why I find it difficult to remember them :frown:


Take the original identity:
cos⁡2x+sin⁡2x≡1\displaystyle \cos^2 x + \sin^2 x \equiv 1

Of course from here, you can rearrange for sine or cosine, or leave it for 1.

If you divide everything by cosine squared you get:

1+sin⁡2xcos⁡2x≡1cos⁡2x⇒1+tan⁡2x≡sec⁡2x\displaystyle 1+\frac{\sin^2 x}{\cos^2 x} \equiv \frac{1}{\cos^2 x} \Rightarrow 1 + \tan^2 x \equiv \sec^2 x

Again, you can rearrange for whatever you need.

Lastly, take the original and divide by sine squared instead. You get:

cos⁡2xsin⁡2x+1≡1sin⁡2x⇒cot⁡2x+1≡csc⁡2x\displaystyle \frac{\cos^2 x}{\sin^2 x} + 1 \equiv \frac{1}{\sin^2 x} \Rightarrow \cot^2 x + 1 \equiv \csc^2 x

Again, you can rearrange for whichever term you need.
Original post by RDKGames
Take the original identity:
cos⁡2x+sin⁡2x≡1\displaystyle \cos^2 x + \sin^2 x \equiv 1

Of course from here, you can rearrange for sine or cosine, or leave it for 1.

If you divide everything by cosine squared you get:

1+sin⁡2xcos⁡2x≡1cos⁡2x⇒1+tan⁡2x≡sec⁡2x\displaystyle 1+\frac{\sin^2 x}{\cos^2 x} \equiv \frac{1}{\cos^2 x} \Rightarrow 1 + \tan^2 x \equiv \sec^2 x

Again, you can rearrange for whatever you need.

Lastly, take the original and divide by sine squared instead. You get:

cos⁡2xsin⁡2x+1≡1sin⁡2x⇒cot⁡2x+1≡csc⁡2x\displaystyle \frac{\cos^2 x}{\sin^2 x} + 1 \equiv \frac{1}{\sin^2 x} \Rightarrow \cot^2 x + 1 \equiv \csc^2 x

Again, you can rearrange for whichever term you need.

Thank you so much! I probably should have known that considering there was an example on it in the book. But things always seem much clearer after hearing another perspective :smile:
Ok so I think I've hit another brick wall :frown: I have no idea how to even start this question :frown:...

Q) Simplify the following expression...

sec⁡4θ−2sec⁡2θtan2θ+tan4θ \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta
Original post by Philip-flop
Ok so I think I've hit another brick wall :frown: I have no idea how to even start this question :frown:...

Q) Simplify the following expression...

sec⁡4θ−2sec⁡2θtan2θ+tan4θ \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta


Use the fact that sec^2(x)=1+tan^2(x)
Original post by Philip-flop
Ok so I think I've hit another brick wall :frown: I have no idea how to even start this question :frown:...

Q) Simplify the following expression...

sec⁡4θ−2sec⁡2θtan2θ+tan4θ \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta


Factorise it. Let x=sec⁡2θx=\sec^2 \theta if you have to and then treat tan as some constant number. Then apply some identities to the bracket to see if it simplifies.

This is like a simple case of a2−2ab+b2=(a−b)2a^2-2ab+b^2=(a-b)^2
(edited 7 years ago)
Reply 85
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Zacken
22
Original post by Philip-flop
Ok so I think I've hit another brick wall :frown: I have no idea how to even start this question :frown:...

Q) Simplify the following expression...

sec⁡4θ−2sec⁡2θtan2θ+tan4θ \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta


You know that (x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2 in your case, that looks exactly like yours with x=sec⁡2θx = \sec^2 \theta and y=tan⁡2θy = \tan^2 \theta.
Original post by Zacken
You know that (x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2 in your case, that looks exactly like yours with x=sec⁡2θx = \sec^2 \theta and y=tan⁡2θy = \tan^2 \theta.

So in this case...

sec⁡4θ−2sec⁡2θtan2θ+tan4θ \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta

becomes...
(sec⁡2θ)2−2sec⁡2θtan2θ+(tan2θ)2(\sec^2 \theta)^2 - 2\sec^2 \theta tan^2 \theta +(tan^2 \theta)^2

which is in the form...
x2−2xy+y2x^2 - 2xy + y^2

Which is equivalent to the form...
(x−y)2(x-y)^2

Therefore the equation can be re-written as...
(sec2θ−tan2θ)2(sec^2 \theta -tan^2 \theta)^2

And then I go on from there?

How did you recognise that the original equation is the same as the form...(x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2...???

Are there any other forms that I should be aware of? Or is there a "cheat sheet" of all the kind of forms that I should be familiar with like 'Completing the Square', 'The Difference of Two Squares' etc?

Thanks again for the help Zacken!!


Original post by RDKGames
Factorise it. Let x=sec⁡2θx=\sec^2 \theta if you have to and then treat tan as some constant number. Then apply some identities to the bracket to see if it simplifies.

This is like a simple case of a2−2ab+b2=(a−b)2a^2-2ab+b^2=(a-b)^2

Thank you RDKGames!! I'm slowly getting there with this one :smile:
(edited 7 years ago)
Reply 87
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Zacken
22
Original post by Philip-flop

Therefore the equation can be re-written as...
(sec2θ−tan2θ)2(sec^2 \theta -tan^2 \theta)^2

And then I go on from there?


Well done.

Once you see sec⁡2x\sec^2 x and tan⁡2\tan^2 in the same place, you should automatically write down 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x and then stare at it and re-arrange it and plug it in and use it wherever you can. What do you get from there?

Whenever you see cot⁡2x\cot^2 x anywhere, immediately write down 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x. Especially if csc⁡2x\csc^2 x is involved.

Whenever you see sin⁡2x\sin^2 x or cos⁡2x\cos^2 x, write down the identity connecting those two. sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x. Even if you think they won't be useful. Write them down next to the question and stare at them for two minutes.

So, in this case, can you see what to do?
Reply 88
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Zacken
22
Original post by Philip-flop
How did you recognise that the original equation is the same as the form...(x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2...???

Are there any other forms that I should be aware of? Or is there a "cheat sheet" of all the kind of forms that I should be familiar with like 'Completing the Square', 'The Difference of Two Squares' etc?

Thanks again for the help Zacken!!


Hmmm, this is something that you're going to have to get used to and spot with lots of practice. You should definitely know

(x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2 off the top of your head.

(x−y)2=x2−2xy+y2(x-y)^2 = x^2 -2xy + y^2 should also always be something you should be intimately familiar with.

(x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2 same as above for this one.

Other than that, I don't think there's really anything else you should be extremely familiar with. Just keep practicing. Maybe start your own cheat sheet? Add on all the basic identities, factorising forms, tricks that come up extremely often that you keep forgetting and then whenever you're stuck on a problem, stare at the sheet for a bit. You'll get used to it eventually.
Original post by Philip-flop

Thank you RDKGames!! I'm slowly getting there with this one :smile:


No problem. :smile:

It is best to remember expansions for (a+b)(a−b)(a+b)(a-b), (a+b)2(a+b)^2 and (a−b)2(a-b)^2 as then you can pick up on these patterns fairly quickly by inspection like Zacken and I did.
Original post by Zacken
Well done.

Once you see sec⁡2x\sec^2 x and tan⁡2\tan^2 in the same place, you should automatically write down 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x and then stare at it and re-arrange it and plug it in and use it wherever you can. What do you get from there?

Whenever you see cot⁡2x\cot^2 x anywhere, immediately write down 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x. Especially if csc⁡2x\csc^2 x is involved.

Whenever you see sin⁡2x\sin^2 x or cos⁡2x\cos^2 x, write down the identity connecting those two. sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x. Even if you think they won't be useful. Write them down next to the question and stare at them for two minutes.

So, in this case, can you see what to do?

Yes!! I think I need to get into the habit of writing these Trig Identities down next to my workings so it's easier to think about the next step I should be taking.

Ok. so from...
(sec⁡2θ−tan2θ)2(\sec^2 \theta - tan^2 \theta)^2

I can see that the Trig Identity... 1+tan2θ=sec⁡2θ 1+tan^2 \theta = \sec^2 \theta... can be rearranged to give... sec⁡2θ−tan2θ=1\sec^2 \theta - tan^2 \theta = 1

Therefore...
(sec⁡2θ−tan2θ)2(\sec^2 \theta - tan^2 \theta)^2

Becomes...
(1)2(1)^2

Which simiplified is just...
11
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Zacken
22
Original post by Philip-flop
Yes!! I think I need to get into the habit of writing these Trig Identities down next to my workings so it's easier to think about the next step I should be taking.


I think that would really help.

Which simiplified is just... 11


Yep, well done! By the way, don't feel obliged to LaTeX everything, I'd rather a fast reply than a LaTeX one. :smile:
Original post by Zacken
Hmmm, this is something that you're going to have to get used to and spot with lots of practice. You should definitely know

(x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2 off the top of your head.

(x−y)2=x2−2xy+y2(x-y)^2 = x^2 -2xy + y^2 should also always be something you should be intimately familiar with.

(x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2 same as above for this one.

Other than that, I don't think there's really anything else you should be extremely familiar with. Just keep practicing. Maybe start your own cheat sheet? Add on all the basic identities, factorising forms, tricks that come up extremely often that you keep forgetting and then whenever you're stuck on a problem, stare at the sheet for a bit. You'll get used to it eventually.

Thank you. I definitely do need to familiarise myself with those equations with their factorising equivalents because when I see one or the other it's like I'm seeing a complete different equation, whereas I should see be able to spot that they are the same thing.

Thank you so much !!! You always manage to clear these things up!

Original post by RDKGames
No problem. :smile:

It is best to remember expansions for (a+b)(a−b)(a+b)(a-b), (a+b)2(a+b)^2 and (a−b)2(a-b)^2 as then you can pick up on these patterns fairly quickly by inspection like Zacken and I did.

Yes it will definitely take me a little bit of time to get used to until I'm able to spot those expansions and immediately think of their factorising equivalent.

Really appreciate your help. I can't thank you enough!
@Zacken and @RDKGames

Do you two know of any Maths books that you would recommend for someone like me? I need something that is "dumbed down" for beginners like me as I feel like there are lots of gaps in my knowledge. I just want to stop being such a nonsense and posting of TSR all the time :colondollar:
Original post by Philip-flop
@Zacken and @RDKGames

Do you two know of any Maths books that you would recommend for someone like me? I need something that is "dumbed down" for beginners like me as I feel like there are lots of gaps in my knowledge. I just want to stop being such a nonsense and posting of TSR all the time :colondollar:


Not really. C3/C4 picks up from from the content of C1/C2 so you might want to go over those 2. As far as mathematical dexterity and quickness goes with tricks, it's just practice really.
Original post by RDKGames
Not really. C3/C4 picks up from from the content of C1/C2 so you might want to go over those 2. As far as mathematical dexterity and quickness goes with tricks, it's just practice really.


Yes, I think once I've finished working through the Edexcel C3 Modular Maths Book I will revisit a lot of the topics from C1 and C2 before I move on to learning the content for C4 :smile:
Seriously why am I always stuck??...

Click to enlarge question below...
C3 - Exe6D Q3.png

For part (a) So far I've got...

cot⁡θ=−3\cot \theta = - \sqrt{3}

cot⁡2θ=−3 \cot^2 \theta = -3

cosec2θ−1=−3\mathrm{cosec} ^2 \theta -1 =-3

cosec2θ=−2\mathrm{cosec} ^2 \theta =-2

sin2θ=1−2 sin^2 \theta = \frac{1}{-2}

But then from there on I know my answer is going to be incorrect :frown:
Reply 97
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Zacken
22
Original post by Philip-flop
Seriously why am I always stuck??...

Click to enlarge question below...
C3 - Exe6D Q3.png

For part (a) So far I've got...

cot⁡θ=−3\cot \theta = - \sqrt{3}

cot⁡2θ=−3 \cot^2 \theta = -3


This bit is incorrect, you've squared both sides, so: (cot⁡θ)2=(−3)2=−3×−3=+3(\cot \theta)^2 = (-\sqrt{3})^2 = -\sqrt{3}\times -\sqrt{3} = +3 but the rest of your method is v. good
Original post by Philip-flop
Seriously why am I always stuck??...

Click to enlarge question below...
C3 - Exe6D Q3.png

For part (a) So far I've got...

cot⁡θ=−3\cot \theta = - \sqrt{3}

cot⁡2θ=−3 \cot^2 \theta = -3

cosec2θ−1=−3\mathrm{cosec} ^2 \theta -1 =-3

cosec2θ=−2\mathrm{cosec} ^2 \theta =-2

sin2θ=1−2 sin^2 \theta = \frac{1}{-2}

But then from there on I know my answer is going to be incorrect :frown:


Second step. Squaring −3-\sqrt{3} would give you +3.
Original post by Philip-flop
Seriously why am I always stuck??...

Click to enlarge question below...
C3 - Exe6D Q3.png

For part (a) So far I've got...

cot⁡θ=−3\cot \theta = - \sqrt{3}

cot⁡2θ=−3 \cot^2 \theta = -3

cosec2θ−1=−3\mathrm{cosec} ^2 \theta -1 =-3

cosec2θ=−2\mathrm{cosec} ^2 \theta =-2

sin2θ=1−2 sin^2 \theta = \frac{1}{-2}

But then from there on I know my answer is going to be incorrect :frown:


Not saying you haven't but you should at least check your working before asking for help. Putting as much effort in as you can before asking for help is much better, since you'll actually learn that way. If you're doing this already then that's good and you can ignore this, but I would think that having checked your working you'd realise that (−3)2≠−3(-\sqrt{3})^2 \not= -3.

Anyway, your method is fine but alternatively you can just find θ\theta by rewriting the original equation as tan⁡θ=−13\tan\theta=-\dfrac{1}{\sqrt{3}}.

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