C4 trig

One thing i found confusing though is, how just before you get to this step:
http://www.wolframalpha.com/input/?i=%28cosA+%2B+sinA%29+%2F+%28cosA-sinA%29+%3D+tan%28A+%2B%28pi%2F4%29%29

You get this:
http://www.wolframalpha.com/input/?i=square+root+%28%28cos%5E2+x+%2B+2sinxcosx+%2B+sin%5E2+x%29+%2F+%28cos%5E2+x+-2sinxcosx+%2B+sin%5E2+x%29%29

and the denominator can be factorised to either:
(sinx - cosx)^2
or
(cosx - sinx)^2

but only the second one in right. How would you know which one works in the exam without having to do them both fully?

One thing i found confusing though is, how just before you get to this step:
http://www.wolframalpha.com/input/?i=%28cosA+%2B+sinA%29+%2F+%28cosA-sinA%29+%3D+tan%28A+%2B%28pi%2F4%29%29

You get this:
http://www.wolframalpha.com/input/?i=square+root+%28%28cos%5E2+x+%2B+2sinxcosx+%2B+sin%5E2+x%29+%2F+%28cos%5E2+x+-2sinxcosx+%2B+sin%5E2+x%29%29

and the denominator can be factorised to either:
(sinx - cosx)^2
or
(cosx - sinx)^2

but only the second one in right. How would you know which one works in the exam without having to do them both fully?

No no that's not what I'm saying. It is an assumption, anything is an assumption if you don't prove it. What makes the difference is what you're ALLOWED to assume, and you're allowed to quote the formula booklet without proof.

In the same way this is a well known result and can easily be proved if you wanted to.

Anyway we're arguing over something pointless lol.

Well known and allowed to be assumed are two different things (as I said, for identities only stuff in the formula booklet, and in some cases very slight variations, can be quoted without proof). If you prove that identity that's fine but then you might as well do the solution fully (because what you're suggesting is essentially splitting a solution into 2 parts).

Aha yeah I guess so; just want to make sure the OP knows full marks may not be given if you use identities not in the formula booklet.

.

OK fair enough.

I still think you'd get full marks if that was an exam question though .

Continuing this would be a waste of time since we're both right anyhow. So let's call it day .

I disagree, Hassi is right on this. If you just stated that $\tan (A+\dfrac{\pi}{4}) = \tan 2A + \sec 2A$ without proof, you'd almost definitely get no marks. It's not something that the formula book allows you to assume.

Now I'm on my PC I can type it out fully.

Thanks for the solution.

I did it in this way:
$\displaystyle \frac{1+sin2a}{cos2a} = \frac{1+2sinacosa}{cos^2a - sin^2a} = \frac{sin^2a + cos^2a+sinacosa+sinacosa}{cos^2a - sin^2a} = \frac{(sina + cosa)(cosa+sina)}{(cosa - sina)(cosa+sina)} = \frac{sina+cosa}{cosa-sina}[br]$
Dividing top and bottom by cosa,
$[br]\displaystyle \frac{tana+1}{1-tana} = \frac{tana+tan45}{1-tanatan45} = tan\left(a+\frac{\pi}{4}\right) [br][br]$

So are you saying you'd have to prove every identity you use in the exam, for these sorts of questions? E.g. you'd need to show that sin2a= 2cosasina via sin(a+a)?

So are you saying you'd have to prove every identity you use in the exam, for these sorts of questions? E.g. you'd need to show that sin2a= 2cosasina via sin(a+a)?

Ahem he's done it right so far. If he multiplies top and bottom by cosA+sinA and simplifies it reduces to a well known identity.

No because that's not a new identity, that's just values subsituted into an existing identity and collecting like terms (writing 2cosasina instead of sinacosa + sinacosa). These slight variations are allowed, like as far as I know you're allowed to state $1 + tan^2x = sec^2x$ without proof because it's only a slight variations from $cos^2x + sin^2x = 1$

I didn't suggest that at all. sin2a=2cosasina is just a special case of an identity that you're given, you don't need to show anything there, just replace B with an A in the standard sin(A+B) identity and there's literally nothing to be shown as you don't use any further identities to get from sin(A+A) to sin2a=2sinacosa.

Basically, if I asked you to use exactly one of the identities in the formula book or sin^2x+cos^2x=1 in any of it's forms (i.e. any of the identities that you are allowed to assume in an exam) to show that the expression you've stated is true and you couldn't do it, then it's clearly not a direct consequence of anything that you're allowed to assume and hence must be proven before use.

Hang about, you both argued that you can quote without proof a trig identity that is in the formula book, yet now you change on stance on it by allowing variations.

By the same logic I could say tan(A/2+pi/4) is nothing special since you could replace A/2 and pi/4 with x from tan(x+x).

I have always said with slight variations "It is an assumption if you don't prove it. In a level maths you can only assume what's in the formula booklet and slight variations " - that was from 3 hours ago.

I'm not saying you can just 'vary' them however you want. I'm saying there are certain identities you are also allowed to assume true; and the exam board allow this (this is my interpretation, by the way) because they're so similar to ones in the formula booklet that they're basically already given.

In a way you're right. It depends how you define "variation"- that's too vague.

The complete reason is the fact the exam board would want you to memorise them and recall them in the exam.

"Candidates should learn the following formulae, which are not included in the formulae booklet, but which may be required to answer questions."

It is too vague - that was just me saying that certain identities are allowed which aren't on the formula book But I agree with the exam board requesting students memorise them is the proper reasoning here.

Anyway I think the point has been made now, and we've discussed this enough. The OP has figured out how to do the question (without using either of our methods, by the way ) and we're way off topic!